Analytical mechanics, 7ed., Solutions manual

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Instructor’s Solutions Manual To Accompany

Analytical Mechanics 7th Edition Grant R. Fowles - University of Utah George L. Cassiday - University of Utah

ISBN-10: 0-534-49493-5 ISBN-13: 978-0-534-49493-3

Contents

1. Fundamental Concepts: Vectors. 2. Newtonian Mechanics: Rectilinear Motion of a Particle. 3. Oscillations. 4. General Motion of a Particle in Three Dimensions. 5. Noninertial Reference Systems. 6. Gravitation and Central Forces. 7. Dynamics of Systems of Particles. 8. Mechanics of Rigid Bodies: Planar Motion. 9. Motion of Rigid Bodies in Three Dimensions. 10. Lagrangian Mechanics. 11. Dynamics of Oscillating Systems.

CHAPTER 1 FUNDAMENTAL CONCEPTS: VECTORS 1.1

K K (a) A + B = (iˆ + ˆj ) + ( ˆj + kˆ) = iˆ + 2 ˆj + kˆ 1 K K A + B = (1 + 4 + 1) 2 = 6 K K (b) 3 A − 2 B = 3(iˆ + ˆj ) − 2( ˆj + kˆ) = 3iˆ + ˆj − 2kˆ K K (c) A ⋅ B = (1)(0) + (1)(1) + (0)(1) = 1 iˆ ˆj kˆ K K (d) A × B = 1 1 0 = iˆ(1 − 0) + ˆj (0 − 1) + kˆ(1 − 0) = iˆ − ˆj + kˆ

0 1 1 1 K K A × B = (1 + 1 + 1) 2 = 3

K K K 1.2 (a) A ⋅ ( B + C ) = 2iˆ + ˆj ⋅ iˆ + 4 ˆj + kˆ = (2)(1) + (1)(4) + (0)(1) = 6

)(

(

K

K

)

K

( A + B ) ⋅ C = ( 3iˆ + ˆj + kˆ ) ⋅ 4 ˆj = (3)(0) + (1)(4) + (1)(0) = 4 2 1 0 K K K (b) A ⋅ ( B × C ) = 1 0 1 = −8 0 4 0

K

K

K

K

K

K

( A × B ) ⋅ C = A ⋅ ( B × C ) = −8 K K K K K K K K K (c) A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C = 4 iˆ + kˆ − 2 4 ˆj = 4iˆ − 8 ˆj + 4kˆ K

K

K

K

K

K

(

) ( )

K K K

K K K

( A × B ) × C = −C × ( A × B ) = − ( C ⋅ B ) A − ( C ⋅ A) B  = − 0 ( 2iˆ + ˆj ) − 4 ( iˆ + kˆ )  = 4iˆ + 4kˆ  

1

K K 5a 2 A ⋅ B (a )(a ) + (2a )(2a) + (0)(3a) 1.3 cos θ = = = 2 AB a 5 14 5a 2 14a 2

θ = cos −1

5 ≈ 53° 14

1.4

K (a) A = iˆ + ˆj + kˆ K K A = A⋅ A K (b) B = iˆ + ˆj K K B = B⋅B

: body diagonal = iˆ ⋅ iˆ + ˆj ⋅ ˆj + kˆ ⋅ kˆ = 3

: face diagonal = 2

iˆ ˆj kˆ K K K (c) C = A × B = 1 1 1 1 1 0 K K A⋅ B 1−1 (d) cos θ = = =0 AB 3 2

1.5

∴θ = 90D

K K K B = B = A × C = AC sin θ

K K A ⋅ C = AC cos θ = u K K A C = Cx + A u K = 2 A+ A

1.6

B A u ∴ Cx = C cos θ = A ∴ C y = C sin θ =

K K K K B× A u K B× A B  K K Cy = 2 A +   A AB  A  B× A 1 K K B× A A2

K d d dA ˆ d = i (α t ) + ˆj ( β t 2 ) + kˆ (γ t 3 ) = iˆα + ˆj 2 β t + kˆ3γ t 2 dt dt dt dt K d2A ˆ = j 2 β + kˆ6γ t dt 2

2

K K 0 = A ⋅ B = ( q )( q ) + ( 3)( − q ) + (1)( 2 ) = q 2 − 3q + 2

1.7

( q − 2 ) ( q − 1) = 0 ,

q = 1 or 2

K K2 K K K K K K A + B = ( A + B ) ⋅ ( A + B ) = A2 + B 2 + 2 A ⋅ B

1.8

K K 2  A + B  = A2 + B 2 + 2 AB  

K K K K A+ B ≤ A + B

K K Since A ⋅ B = AB cosθ ≤ AB ,

K K K K K K A ⋅ B = AB cosθ = A B cosθ ≤ A B B cos θ ≤ B

K K K K K K K K K Show A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C

1.9

iˆ K A × Bx Cx

or

kˆ K K Bz = ( Ax Cx + Ay C y + Az Cz ) B − ( Ax Bx + Ay By + Az Bz ) C Cz

ˆj By Cy

= ( Ax BxCx + Ay Bx C y + Az Bx Cz − Ax Bx Cx − Ay By Cx − Az Bz Cx ) iˆ

+ ( Ax By Cx + Ay By C y + Az By Cz − Ax Bx C y − Ay By C y − Az Bz C y ) ˆj + ( Ax Bz Cx + Ay Bz C y + Az Bz C z − Ax BxCz − Ay By Cz − Az Bz Cz ) kˆ

( A B C + A B C − A B C − A B C ) iˆ = + ( A B C + A B C − A B C − A B C ) ˆj + ( A B C + A B C − A B C − A B C ) kˆ y

x

y

z

x

z

y

y

x

z

z

x

x

y

x

z

y

z

x

x

y

z

z

y

x

z

x

y

z

y

x

x

z

y

y

z

3

iˆ Ax By Cz − Bz C y

iˆ ( Ay Bx C y − Ay By Cx − Az Bz Cx + Az Bx Cz ) kˆ Az = + ˆj ( Az By Cz − Az Bz C y − Ax Bx C y + Ax By Cx ) Bx C y − By Cx + kˆ A B C − A B C − A B C + A B C ( x z x x x z y y z y z y)

ˆj Ay Bz Cx − Bx Cz

1.10

y = A sin θ 1  Α = 2  xy  + y ( B − x ) = xy + yB − xy = AB sin θ 2 

K K Α = A× B 1.11

iˆ K K K K A ⋅ ( B × C ) = A ⋅ Bx Cx 1.12

x G u

kˆ Ax Bz = Bx Cz Cx

ˆj By Cy

Ay By Cy

Bx Az Bz = − Ax Cz Cx

By Ay Cy

Bz

K K K Az = − B ⋅ ( A × C ) Cz

z G A

G C x

G B

G K K Let A = (Ax,Ay,Az), B = (0,By,0) and C = (0,Cy,Cz) K K G G G Cz is the perpendicular distance between the plane A , B and its opposite. u = B x C is G G G G directed along the x-axis since the vectors B , C are in the y,z plane. u x = B x C = ByCz G G is the area of the parallelogram formed by the vectors B , C . Multiply that area times the K K height of plane A , B = Ax to get the volume of the parallopiped

G G G V = Ax u x = Ax By Cz = A • B x C

(

)

4

1.13

For rotation about the z axis:

iˆ ⋅ iˆ′ = cos φ = ˆj ⋅ ˆj ′, kˆ ⋅ kˆ′ = 1 iˆ ⋅ ˆj ′ = − sin φ

ˆj ⋅ iˆ′ = sin φ For rotation about the y′ axis:

iˆ ⋅ iˆ′ = cos θ = kˆ ⋅ kˆ′, iˆ ⋅ kˆ′ = sin θ kˆ ⋅ iˆ′ = − sin θ  cos θ I  T = 0  sin θ 

ˆj ⋅ ˆj ′ = 1

0 − sin θ   cos φ  1 0   − sin φ 0 cosθ   0

sin φ cos φ 0

0   cos θ cos φ   0  =  − sin φ 1   sin θ cos φ

cosθ sin φ cos φ sin θ sin φ

− sin θ   0  cosθ 

1.14

3 iˆ ⋅ iˆ′ = cos 30D = 2

ˆj ⋅ iˆ′ = sin 30D = 1 2

kˆ ⋅ iˆ′ = 0

1 iˆ ⋅ ˆj ′ = − sin 30D = − 2

ˆj ⋅ ˆj ′ = cos 30D = 3 2

kˆ ⋅ ˆj ′ = 0

ˆj ⋅ kˆ′ = 0 iˆ ⋅ kˆ′ = 0  3 1  3  0  3+   2 2 2  2    Ax′       3 3  A  = − 1   y′   2 2 0   3  =  2 3 − 1   −1  Az ′   0 1     −1   0       K A = 3.232iˆ′ + 1.598 ˆj ′ − kˆ′

kˆ ⋅ kˆ′ = 1

5

1.15

1.

Rotate thru φ about z-axis

φ = 45D

Rφ

2.

Rotate thru θ about x’-axis

Rθ

3.

Rotate thru ψ about z’-axis

θ = 45 ψ = 45D

 1  2   1 Rφ =  − 2   0  

1 2 1 2 0

 1 0  1  Rθ =  0 2  1  0 − 2 

D

 0   0  1  

 0   1  2  1   2

1  1  0  2 2   1  1  Rψ =  − 0 2 2   0 1  0     1 1 1  1 +  2− 2 2 2 2  1 1 1  1 R (ψ ,θ , φ ) = Rψ Rθ Rφ =  − − − + 2 2 2 2 2 2  1 1  −  2 2  1 K   where x′ =  0  and 0   2 2 we have: ψ + β + α 2 = 1

K K Condition is: x′ = Rx K K Since x ⋅ x = 1

After a lot of algebra: α = 1.16

Rψ

1  2   1 α  1      or  0  = R (ψ ,θ , φ )  β   2 0 γ       1   2 α  K   x =β  γ   

1 2 1 2 1 − , β= + , γ = 2 4 2 4 2

K v = vτˆ = ctτˆ v2 c 2t 2 K a = vτˆ + nˆ = cτˆ + nˆ b ρ

6

b K , v = τˆ bc and c K K v ⋅a c bc cos θ = = = va bc 2c 2

K a = cτˆ + cnˆ

at t =

1 2

θ = 45D 1.17

K ˆ ω sin (ω t ) + ˆj 2bω cos (ω t ) v ( t ) = −ib 1

1

K v = ( b 2ω 2 sin 2 ω t + 4b 2ω 2 cos 2 ω t ) 2 = bω (1 + 3cos 2 ω t ) 2 K ˆ ω 2 cos ω t − ˆj 2bω 2 sin ω t a ( t ) = −ib 1

K a = bω 2 (1 + 3sin 2 ω t ) 2

1.18

K v = 2bω ;

t = 0,

at

at

t=

π , 2ω

K v = bω

K ˆ ω cos ω t − ˆjbω sin ω t + kˆ 2ct v ( t ) = ib K ˆ ω 2 sin ω t − ˆjbω 2 cos ω t + kˆ 2c a ( t ) = −ib 1

1

K a = ( b 2ω 4 sin 2 ω t + b 2ω 4 cos 2 ω t + 4c 2 ) 2 = ( b 2ω 4 + 4c 2 ) 2

1.19

K  ˆr + rθeˆθ = bke kt eˆr + bce kt eˆθ v = re K a =  r − rθ 2 eˆr + rθ + 2rθ eˆθ = b ( k 2 − c 2 ) e kt eˆr + 2bcke kt eˆθ K K b 2 k ( k 2 − c 2 ) e 2 kt + 2b 2 c 2 ke 2 kt v ⋅a cos φ = = 1 1 va kt 2 2 2 kt  2 2 2 2 2 2 be ( k + c ) be ( k − c ) + 4c k  

(

cos φ =

1.20

)

(

)

k ( k 2 + c2 )

(k

2

+c

1 2 2

) (k

2

+c

k

=

2

) (k

2

+c

1 2 2

)

, a constant

K  − Rφ ) eˆ + ( 2 Rφ + Rφ ) eˆ +  a = (R zeˆz R φ K a = −bω 2 eˆR + 2ceˆz 1

K a = ( b 2ω 4 + 4c 2 ) 2

7

1.21

1.22

K r ( t ) = iˆ (1 − e − kt ) + ˆje kt K ˆ − kt + ˆjkekt r ( t ) = ike K ˆ 2 e − kt + ˆjk 2 ekt r ( t ) = −ik

K v = eˆr r + eˆφ rφ sin θ + eˆθ rθ

π  1 π K  v = eˆφ bω sin  1 + cos ( 4ω t )   − eˆθ b ω sin ( 4ω t ) 2  2  4 π K π  v = eˆφ bω cos  cos ( 4ω t )  − eˆθ bω sin ( 4ω t ) 2 8  1

2  2 K π  π v = bω cos 2  cos 4ω t  + sin 2 4ω t  8  4  

Path is sinusoidal oscillation about the equator.

1.23

K K v ⋅ v = v2 K K dv K K dv ⋅v + v ⋅ = 2vv dt dt K K 2v ⋅ a = 2vv K K v ⋅ a = vv

8

1.24

1.25

K d K K K dr K K K d K K  r ⋅ ( v × a )  = ⋅ (v × a ) + r ⋅ (v × a ) dt dt dt K K K K K K  dv K   K da   = v ⋅ ( v × a ) + r ⋅  × a  +  v ×   dt      dt K  K K  = 0 + r ⋅ 0 + ( v × a ) d K K K K K K  r ⋅ ( v × a )  = r ⋅ ( v × a ) dt K K v = vτˆ and a = aτ τˆ + an nˆ

K K v ⋅a K K v ⋅ a = vaτ , so aτ = v 1

a 2 = aτ2 + an2 , so an = ( a 2 − aτ2 ) 2

1.26

For 1.14, aτ =

−b 2ω 3 cos ω t ⋅ sin ω t + b 2ω 3 sin ω t ⋅ cos ω t + 4c 2t

(b ω 2

aτ =

2

1

cos 2 ω t + b 2ω 2 sin 2 ω t + 4c 2t 2 ) 2

4c 2t

(b ω 2

2

1 2 2 2

+ 4c t

)

1

 2 2 2 16c 4t 2 2 an =  b ω + 4c − 2 2  b ω + 4c 2 t 2   1 b 2 k ( k 2 − c 2 ) e 2 kt + 2b 2 c 2 ke 2 kt 2 2 2 kt = bke ( k + c ) For 1.15, aτ = 1 kt 2 2 2 be ( k + c ) 1

1

2 2 an = b 2 e 2 kt ( k 2 + c 2 ) − b 2 k 2e 2 kt ( k 2 + c 2 )  = bce kt ( k 2 + c 2 ) 2  

2

1.27

v K K v = vτˆ , a = vτˆ + nˆ

ρ

v 2 v3 K K v × a = v ⋅ an = v =

ρ

ρ

9

1.28

K ˆ sin θ + ˆjb cos θ rD P = ib K ˆ θ cos θ − ˆjbθ sin θ vrel = ib K ˆ θ cosθ − θ 2 sin θ − ˆjb θsin θ + θ 2 cosθ arel = ib

(

)

at the point θ =

π

2 K So, vrel = bθ = v

(

)

K K , vrel = −v

v aD = b b v2 v2 K Now, arel = vrelτˆ + rel nˆ = aDτˆ + nˆ b ρ

θ =

v b

θ =

1

 2 v4  2 K arel =  aD + 2  b   K K K K K K vP = v + vrel and aP = aD + arel

 a  a  v2 v2 K aP = iˆ  aD + b  D cos θ − 2 sin θ   − ˆjb  D sin θ + 2 cosθ  b b b  b   1

 2 v4 2v 2 K aP = aD  2 + 2 cos θ + 2 2 − sin θ  aD b aDb   K aP is a maximum at θ = 0 , i.e., at the top of the wheel.

2v 2 −2sin θ − cos θ = 0 aDb 

v2    aDb 

θ = tan −1  −

2 0 0  x -x 0   x x 0   2 x       = x x 0 − x x 0 = 0 2 x 2 0  Therefore, x = 1 1.29 RR      2 0 0 1 0 0 1  0  0 1     The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2)

10

1.30 (a)

a

θ

b

φ

a = iˆ cos θ + ˆj sin θ b = iˆ cos ϕ + ˆj sin ϕ

(

)(

a ⋅ b = cos (θ − ϕ ) = iˆ cos θ + ˆj sin θ ⋅ iˆ cos ϕ + ˆj sin ϕ

)

cos (θ − ϕ ) = cos θ cos ϕ + sin θ sin ϕ

(b)

(

) (

b × a = kˆ sin (θ − ϕ ) = iˆ cos θ + ˆj sin θ × iˆ cos ϕ + ˆj sin ϕ

)

sin (θ − ϕ ) = sin θ cos ϕ − cos θ sin ϕ

--------------------------------------------------------------------------------------------------------

11

CHAPTER 2 NEWTONIAN MECHANICS: RECTILINEAR MOTION OF A PARTICLE 2.1

1 ( F + ct ) m t 1 F c 2 x = ∫ ( F + ct ) dt = t + t 0 m m 2m t F c 2 F c 3 x=∫  t+ t  dt = t 2 + t 0 m m 2m  6m 

(a) x =

F sin ct m t F F F t x=∫ sin ct dt = − cos ct 0 = (1 − cos ct ) 0 m cm cm t F F 1 x=∫ (1 − cos ct ) dt = 1 − sin ct  0 cm cm  c 

(b) x =

F ct e m F ct t F ct x= e = ( e − 1) cm 0 cm F  1 ct 1  F ct x=  e − − t  = 2 ( e − 1 − ct ) cm  c c  cm

(c) x =

2.2

dx dx dx dx = ⋅ =x dt dx dt dx dx 1 x = ( F + cx ) dx m 1 xdx = ( F + cx ) dx m 1 2 1 cx 2  x = F x+  m 2 2 

(a) x =

1

x 2 x =  ( 2 F + cx )  m 

(b) x = x

dx 1 = F e − cx dx m

12

1 F e − cx dx m 1 2 F − cx F x =− e − 1) = ( (1 − e−cx ) 2 cm cm

xdx =

1

 2F 2 x= 1 − e − cx )  (  cm  dx 1 (c) x = x = ( F cos cx ) dx m F xdx = cos cx dx m 1 2 F x = sin cx cm 2 1

 2F 2 sin cx  x=  cm 

2.3

(a) V ( x ) = − ∫

x

( F + cx ) dx = − F x − x

cx 2 +C 2

F − cx e +C c x F (c) V ( x ) = − ∫ F cos cx dx = − sin cx + C x c x

(b) V ( x ) = − ∫ F e − cx dx = x

2.4

dV ( x ) = −kx dx x 1 V ( x ) = ∫ kx dx = kx 2 0 2 (b) T = T ( x ) + V ( x )

(a) F ( x ) = −

T ( x) = T −V ( x) =

1 k ( A − x2 ) 2

1 2 kA 2 (d) turning points @ T ( x1 ) → 0 ∴ x1 = ± A

(c) E = T =

2.5

x kx 3  1 1 kx 4 V ( x ) = ∫  kx − 2 dx = kx 2 − 0 2 4 A2 A   1 2 1 kx 4 (b) T ( x ) = T − V ( x ) = T − kx + 2 4 A2 (c) E = T

(a) F ( x ) − kx +

kx 3 A2

so

13

(d) V ( x ) has maximum at F ( xm ) → 0 3

kx xm = ± A kxm − m2 = 0 A 1 1 kA4 1 2 V ( xm ) = kA2 − = kA 2 4 A2 4 If E < V ( xm ) turning points exist. Turning points @ T ( x1 ) → 0 let u = x1

2

1 1 ku 2 E − ku + =0 2 4 A2 solving for u , we obtain 1   4E  2   2  u = A 1 ± 1 − 2    kA     or 1

 4E   2  x1 = ± A 1 − 1 − 2    kA    2.6

x = v ( x) =

α

x=−

x

α x2

x=−

α2 x3

mα 2 F ( x ) = mx = − 3 x 2.7

2.8

F ≥ Mg sin θ

F = mx = mx

dx dx

x = bx −3 dx = −3bx −4 dx F = m ( bx −3 )( −3bx −4 ) F = −3mb 2 x −7 2.9

 m m  (a) V = mgx = (.145kg )  9.8 2  (1250 ft )  .3048  = 541J s  ft   

14

(b) T =

T=

1 2 1 1  mg  1 m 2 g 2 mv = mvt = m  = 2 2 2  c2  2 .22 D 2

(.145kg )

2

m   9.8 2  s  

kg ( 2 )(.22 ) ( 2 )(.0366 ) m 2

= 87 J 3

  t  2 3 ∫ Fdx = ∫ −cv dx = −c ∫ v dt = −c ∫  −vt tanh  τ   dt t  t   t  3  1 = cvt τ  − tanh 2   + ∫ tanh   d    τ   τ   τ   2 t  t  3  1 = cvt τ  − tanh 2   + ln cosh    τ   τ   2

t Now tanh 2  τ

  ≅ 1 for t τ    t  t Meanwhile x = ∫ vdt = ∫  −vt tanh    dt = vtτ ln cosh   τ  τ   t x ln cosh   =  τ  vtτ  m x = (1250 ft )  .3048  = 381 m ft   1

 m 2  kg .145 9.8 ( )     mg  m s2      = 34.72 vt =   = kg 2 s  (.22 )(.0732 )   c2    m  1 2

1

 2 1    m 2 (.145kg )   = 3.543s τ = =   (.22 )(.0732 )2 kg  9.8 m    c2 g     m s 2     3.81 2 3 Fdx = .22 .0732 34.72 3.543 − .5 + ( )( ) ( ) ( )   = 454 J ∫ ( 34.72 )( 3.54 )   V − T = 541J − 87 J = 454 J 2.10

F 1F 2 t t, x= 2m m F F 2 t1 , t = t1 For t1 ≤ t ≤ 2t1 : v = t1 , x = 2m m For 0 ≤ t ≤ t1 : v =

15

F 2 F 1 2F 2 t1 + t1 ( t − t1 ) + ( t − t1 ) 2m 2 m m F 2 F 2 F 2 5F 2 t1 + t1 + t1 = t1 At t = 2t1 : x = 2m 2m m m dv dv dx dv c 3 a= = ⋅ = v ⋅ = − v2 dt dx dt dx m 1 − c v 2 dv = − dx m 1 v − xmax c ∫v v 2 dv = ∫0 − mdx 1 c −2v 2 = − xmax m x=

2.11

1

xmax

2.12

2mv 2 = c

Going up: Fx = − mg sin 30 − µ mg cos 30 x = − g ( sin 30 + 0.1cos 30 ) = −5.749 v = v + at

at the highest point v = 0 so tup = −

m s2

v = 0.174v s a

1 2 2 2 xup = v tup + atup2 = 0.174v − .087v = 0.087v m 2 2 Going down: x ′ = 0.087v , v ′ = 0 , a′ = −9.8 ( 0.5 − 0.0866 )

ttotal

2.13

1 2 2 xdown = 0 = 0.087v − 4.0513tdown 2 tdown = 0.207v s = tup + tdown = 0.381v s

At the top v = 0 so e −2 kxmax = Coming down x = xmax

g k

g 2 +v k and at the bottom x = 0

g 2  v 1 g g k 2 1) =   v = −  ( g 2 k k g 2 +v  +v  k k  2

16

v=

vt v

(v

t

2.14

2

+v

1 2 2

)

, vt =

g mg = k c2

Going up: Fx = −mg − c2 v 2 dv c2 2 a = v = − g − kv , k = m dx v x vdv ∫v − g − kv 2 = ∫0 dx −

v 1 ln ( − g − kv 2 ) = x v 2k g + kv 2 = e −2 kx 2 g + kv

g g 2 v 2 =  + v  e −2kx − k k  2 Going down: Fx = −mg + c2 v dv v = − g + kv 2 dx v x vdv = ∫0 − g + kv2 ∫0 dx v 1 ln ( − g + kv 2 ) = x − x 0 2k k 1 − v 2 = e 2 kx e −2 kx g g g  v 2 = −  e −2 kx  e 2 kx k k  2.15

Using

dv = mg − c1v − c2 v 2 dt t dt v dv = ∫0 m ∫0 mg − c1v − c2v 2

m

dx 1 2cx + b − b 2 − 4ac = ln ∫ a + bx + cx 2 b2 − 4ac 2cx + b + b2 − 4ac ,

17

2

−2c2 v − c1 − c1 + 4mgc2

t 1 = ln 2 2 m c1 + 4mgc2 −2c2 v − c1 + c1 + 4mgc2

(

t 2 c1 + 4mgc2 m

)

1 2

2c v + c + ( = ln ( 2c v + c −

2

0

)( c − + 4mgc )( c +

2

1

c1 + 4mgc2

2

1

c1

2

v

2

2

) + 4mgc )

1

c1 + 4mgc2

1

c1

2

2

2

as t → ∞ , 2c2 vt + c1 − c1 + 4mgc2 = 0 c1  c1  mg   +   + 2c2  2c2  c2    Alternatively, when v = vt , dv 2 m = 0 = mg − c1vt − c2 vt dt 2

1 2

vt = −

c1  c1  mg   +   + 2c2  2c2  c2    2

vt = −

2.16

1 2

dv k = − x −2 dx m v x kdx ∫0 vdv = ∫b − mx 2 1 2 k 1 1 v =  −  2 m x b

a=v

1

1

dx  2k  1 1   2  2k  b − x   2 v= =  −  =   dt  m  x b    mb  x   1

 x 2 3  x t 0  mb   mb  0  x  ∫0 dt = ∫b  2k  b − x  dx =  2k  ∫1  b x  d  b   1 −   b x Since x ≤ b , say = sin 2 θ b 1 2

1

1 2

1

 mb3  2 0 sin θ ( 2sin θ cos θ dθ )  2mb3  2 0 2 t = =  ∫− π  ∫− π sin θ dθ k k 2 cos θ   2   2 1

 mb3  2 t =  π  8k 

18

2.17

dv dv = mv = f ( x )i g ( v ) dt dx mvdv = f ( x ) dx g (v)

m

By integration, get v = v ( x ) = If F ( x,t ) = f ( x )i g ( t ) :

dx dt

d 2x d  dx  = m   = f ( x )i g ( t ) 2 dt dt  dt  This cannot, in general, be solved by integration. If F ( v,t ) = f ( v )i g ( t ) : m

dv = f ( v )i g ( t ) dt mdv = g ( t ) dt f (v)

m

Integration gives v = v ( t ) dx = v (t ) dt dx = v ( t ) dt A second integration gives x = x ( t ) 2.18

c1 = (1.55 ×10−4 )(10−2 ) = 1.55 × 10−6 c2 = ( 0.22 ) (10−2 ) = 2.2 × 10−5 2

kg s

kg s 1

 1.55 ×10−6  2 (10−7 ) ( 9.8 )  2 1.55 × 10  vt = − +   + 2 × 2.2 ×10−5  2 × 2.2 ×10−5  2.2 × 10−5    m vt = 0.179 s −6

(10 ) ( 9.8) = 0.211 m −7

Using equation 2.29, vt =

2.2 ×10−5

s

2.19 F ( x ) = − Aeα x = mx

Let u = eα v

or

du = α eα v dv

F ( v ) = − Aeα v = mv

dv =

du du = αv αe αu

dv A = − dt αv e m du αA ∴ 2 =− dt u m

19

Integrating 1 1 A − = αt u u m

and substituting eα v = u

 A  ln 1 + eα v α t  α  m  (b) t = T @ v = 0  A  α v = ln 1 + eα v αT   m  A m 1 − e −α v  eα v = 1 + eα v αT T= αA m dv A vdv A = − dx (c) v = v = − eα v αv dx m e m

(a) v = v −

1

again, let u = eα v

du = α udv

or

dv =

du αu

v=

1

α

ln u

1  du α ln u  α u A = − dx Integrating and solving u m m x = 2 1 − (1 + α v ) e −α v  α A 2.20

d ( mv ) = mv + vm = mg dt 4 but m = ρ π r 3 m = ρ1π r 2 v 3 4 4 4 so (1) π ρ r 3v + π ρ1r 2 v 2 = π 2 = π ρ r 3 g 3 3 3 F=

Now

ρ1 ≈ 10−3 ρ

so, second term is negligible-small

hence v ≈ g and v ≈ gt speed ∝ t but

m = ρ 4π r 2 r = ρ1 π r 2 v or r ≅

1 ρ1 v 4ρ

Hence r ≈

1 ρ1 gt and rate of 4ρ

growth ∝ t The exact differential equation from (1) above is: 2

4 4ρ 4ρ r 4 r + πρ1 πρ r = π ρ rg 3 3 ρ1 ρ1

20

3r 2 ρ = 1 g 4ρ r Using Mathcad, solve the above non-linear d.e. letting

which reduces to: r +

ρ1 ≈ 10−3 and R ≈ 0.01mm (small raindrop). Graphs ρ

show that v ∝ r ∝ t and r ∝ t 2

21

CHAPTER 3 OSCILLATIONS 3.1

x = 0.002sin  2π ( 512 s −1 ) t  [ m ]

m m xmax = ( 0.002 )( 2π )( 512 )   = 6.43   s s 2 2 m m xmax = ( 0.002 )( 2π ) ( 512 )  2  = 2.07 × 104  2  s  s  3.2

x = 0.1sin ω t [m]

When t = 0, x = 0

and

ω = 5 s −1

3.3

m x = 0.1ω cos ω t   s m x = 0.5   = 0.1ω s 2π T= = 1.26 s

ω

x ( t ) = x cos ω t +

x

ω

sin ω t and ω = 2π f

x = 0.25cos ( 20π t ) + 0.00159sin ( 20π t ) [ m ]

3.4

cos (α − β ) = cos α cos β + sin α sin β x = A cos (ω t − φ ) = A cos φ cos ω t + A sin φ sin ω t x = Α cos ω t + Β sin ω t , Α = A cos φ , Β = A sin φ

3.5

1 2 1 2 1 2 1 2 mx1 + kx1 = mx2 + kx2 2 2 2 2 2 2 2 2 k ( x1 − x2 ) = m ( x2 − x1 ) 1

k  x22 − x12  2 = ω =  m  x12 − x22  1 2 1 2 1 2 kA = mx1 + kx1 2 2 2 m x2 x2 − x2 x2 A2 = x12 + x12 = 1 12 22 1 + x12 k x2 − x1 x x −x x  A=   x −x  2 2 1 2 2 2

2 2 2 1 2 1

1 2

22

3.6

3.7

l 1 1 T =π s ≈ 2.5 s =π 9.8 g 2 6

For springs tied in parallel: Fs ( x ) = − k1 x − k2 x = − ( k1 + k2 ) x 1

(k + k ) 2 ω= 1 2   m  For springs tied in series: The upward force m is keq x . Therefore, the downward force on spring k2 is keq x . The upward force on the spring k2 is k1 x′ where x′ is the displacement of P, the point at which the springs are tied. Since the spring k2 is in equilibrium, k1 x′ = keq x . Meanwhile, The upward force at P is k1 x′ . The downward force at P is k2 ( x − x′ ) . Therefore, k1 x′ = k2 ( x − x′ )

x′ =

k2 x k1 + k2

 k x  And keq x = k1  2   k1 + k2  1

ω= 3.8

 k1k2 2 =  m  ( k1 + k2 ) m 

keq

For the system ( M + m ) , − kX = ( M + m ) X The position and acceleration of m are the same as for ( M + m ) :

k xm M +m   k k xm = A cos  t + δ  = d cos t M m M m + +   The total force on m, Fm = mxm = mg − Fr xm = −

23

Fr = mg +

mk mkd k xm = mg + t cos M +m M +m M +m

For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, Fr = 0 at xm = −d : mkd 0 = mg − M +m g ( M + m) d= k 3.9

x = e −γ t A cos (ω d t − φ )

dx = −e −γ t Aω d sin (ω d t − φ ) − γ e −γ t A cos (ω d t − φ ) dt dx = 0 = ω d sin (ω d t − φ ) + γ cos (ω d t − φ ) maxima at dt tan (ω d t − φ ) = −

γ ωd

thus the condition of relative maximum occurs every time that t increases by ti +1 = ti +

2π

2π

ωd

:

ωd

For the i th maximum: xi = e −γ ti A cos (ω d ti − φ ) xi +1 = e

−γ ti +1

A cos (ω d ti +1 − φ ) = e

−γ

2π ωd

xi

2π

−γ xi = e ωd = eγ Td xi +1

3.10

(a)

(b) (c)

c = 3 s −1 2m 2 ω d = ω 2 − γ 2 = 16 s −2

γ=

k = 25 s −2 m 2 ω r = ω d 2 − γ 2 = 7 s −2

ω2 =

∴ω r = 7 s −1 F 48 = Amax = m = 0.2 m Cωd 60.4

tan φ =

2γω r 2γω r ω r 7 = = = 2 2 2 (ω − ω r ) 2γ γ 3

∴φ ≈ 41.4

24

3.11

17 2 β mx = 0 2 3 17 γ = β and ω 2 = β 2 2 2 2 2 2 2 ω r = ω − 2γ = 4 β mx + 3β mx +

(a)

Amax =

(b)

=

ω d2 = ω 2 − γ 2 =

25 2 β 4

∴ω d =

5 β 2

2A 15β 2

e −γ Td =

3.12

F 2mγω d

∴ω r = 2 β

γ=

1 2

1 ln 2 = f d ln 2 Td 1

ω d = (ω 2 − γ 2 ) 2

(a)

1 2 2

So, ω = (ω + γ ) 2 d

1

1

1

1

 2  γ 2  2   ln 2  2  2 f =  fd +    = f d 1 +     2π     2π    f = 100.6 Hz 1

(b)

ω r = (ω d2 − γ 2 ) 2  2  γ 2  2   ln 2 2  2 fr =  fd −    = f d 1 −     2π     2π    f r = 99.4 Hz

3.13

Since the amplitude diminishes by e −γ Td in each complete period, n 1 e −γ Td = = e −1 e γ Td n = 1 ω 1 γ= = d Td n 2π n

(

Now So

)

(

ωd = ω 2 − γ 2 ω = (ω d2 + γ

)

1 2 2

)

1 2

1

1 2  = ω d 1 + 2 2   4π n 

25

2π

1

Td ω d ω  1 2 = = = 1 + 2 2  2π ω d  4π n  T

ω

For large n,

Td 1 ≈ 1+ 2 2 T 8π n 1

3.14

(

(a) ω r = ω − 2γ 2

1 2 2

)

2 2  ω   = ω 2 − 2    = 0.707ω  2   

1

 1 2 2 ω 1 −  ω −γ ωd 4 = (b) Q = =  = 0.866 2γ 2γ ω  2   2  ω  2   ( 2ω ) 2γω 2 2 (c) tan φ = 2 =  2  2 =− 2 3 ω −ω ω − 4ω 1 2 2

(

)

 2  

φ = tan −1  −  = 146.3 3 1

2  2 2 ω  2 2 (d) D (ω ) = (ω − 4ω ) + 4   ( 4ω 2 )  = 3.606 ω 2  2    F F A (ω ) = m = 0.277 D (ω ) mω 2

3.15

A (ω ) ≈

Amaxγ 1

 (ω − ω ) 2 + γ 2  2   1 1 γ for A (ω ) = Amax , = 1 2 2  (ω − ω ) 2 + γ 2  2   2 (ω − ω ) + γ 2 = 4γ 2

ω − ω = ±γ 3 ω =ω ±γ 3

26

3.16

(b) Q =

ω2 −γ 2 ωd = 2γ 2γ

ω2 =

1 , LC

γ=

R 2L

2  1   R   − 2   LC   4 L   L  1 =  2 −  Q=  R   R C  4 2   2L  L ω  R (c) Q =   = C = R R  2γ 

3.17

Fext = F sin ω t = Im  F eiω t 

and x ( t ) is the imaginary part of the solution to: mx + cx + kx = F eiω t

i.e.

i ω t −φ x ( t ) = Im  Ae ( )  = A sin (ω t − φ )

where, as derived in the text, F A= 1 ( k − mω 2 )2 + c 2ω 2  2   and 2γω tan φ = 2 ω −ω2 3.18

Using the hint, Fext = Re ( F e β t ) ,

where β = −α + iω ,

and x(t) is the real part of the solution to: mx + cx + kx = F e β t . x = Ae β t −iφ Assuming a solution of the form: ( mβ 2 + cβ + k ) x =  FA  xeiφ F mα 2 − 2imαω − mω 2 − cα + icω + k = ( cos φ + i sin φ ) A F m (α 2 − ω 2 ) − cα + k = cos φ A F ω ( −2mα + c ) = sin φ A

27

φ = tan −1

ω ( c − 2mα ) m (α 2 − ω 2 ) − cα + k

Using sin 2 φ + cos 2 φ = 1 , 2 F2 2 =  m (α 2 − ω 2 ) − cα + k  + ω 2 ( c − 2mα ) 2 A F A=

{m (α − ω ) − cα + k  + ω ( c − 2mα ) } 2

2

2

2

1 2 2

and x ( t ) = Ae −α t cos (ω t − φ ) + the transient term.

3.19

l  A2  T ≈ 2π 1 −  g 8 

(a)

for A =

(b)

π 4

−

1 2

l ×1.041 g

, T ≈ 2π

4π 2l × 1.084 T2 l 4π 2l gives g = 2 , approximately 8% too small. Using T = 2π g T

(c)

g=

λ A3 ω2 and λ = B=− 32ω 2 6 B A2 = A 192

for A = 3.20

π 4

,

f ( t ) = ∑ cn einω t

B = 0.0032 A n = 0, ±1, ±2, . . .

n

f ( t ) = ∑ cn cos nω t + ∑ cni sin nω t , n = 0, ±1, ±2, . . . n T 2 T − 2 T 2 T − 2

n

and cn =

1 T∫

f ( t ) e − inωt dt ,

cn =

1 T∫

f ( t ) cos ( nω t ) dt −

n = 0, ±1, ±2, . . . i T

T 2 T − 2

∫

f ( t ) sin ( nω t ) dt

The first term on cn is the same for n and − n ; the second term changes sign for

n vs. − n . The same holds true for the trigonometric terms in f ( t ) . Therefore, when

terms that cancel in the summations are discarded:

28

1 T  f ( t ) = c + ∑  ∫ 2T f ( t ) cos ( nω t ) dt  cos nω t − n T 2  T 1  + ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t , − n T 2 

1 T2 n = ±1, ±2, . . ., and c = ∫ T f ( t ) dt T −2 Now, due to the equality of terms in ± n :  2 T2  f ( t ) = c + ∑  ∫ T f ( t ) cos ( nω t ) dt  cos nω t − n T 2  T 2  + ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t , − n T 2  n = 1, 2,3, . . . Equations 3.9.9 and 3.9.10 follow directly.

3.21

f ( t ) = ∑ cn e

inω t

n

T=

2π

ω

so

,

1 T2 cn = ∫ T f ( t ) e− inωt dt , T −2

ω cn = 2π =

π ω π − ω

∫

and n = 0, ±1, ±2, …

f ( t ) e− inω t dt

π  ω  0 − inω t ω − inω t − + e dt e dt  ( ) π  ∫− ∫ 0 2π  ω 



ω  1 −inω t = e 2π  inω 

0

−

π ω

π  1 − inω t ω  − e  inω 0 

1 1 − e + inπ − e− inπ + 1 2π in For n even, einπ = e − inπ = 1 and the term in brackets is zero. For n odd, einπ = e − inπ = −1 4 cn = , n = ±1, ±3, . . . 2π in 4 inω t f (t ) = ∑ e , n = ±1, ±3, . . . n 2π in 4 1 1 inω t − inω t =∑ ( e − e ) , n = 1,3,5, . . . n π n 2i 41 =∑ sin ( nω t ) , n = 1,3,5, . . . n π n 4 1 1  f ( t ) = sin ω t + sin 3ω t + sin 5ω t + … π 3 5 

=

29

3.22

In steady state, x ( t ) = ∑ An e (

i nω t −φn )

n

An =

Fn m

1

(ω 2 − n 2ω 2 )2 + 4γ 2 n 2ω 2  2   4F , n = 1,3,5, . . . and Now Fn = nπ ω 9ω 2 Q = 100 ≈ so γ 2 ≈ 40, 000 2γ 4F 1 A1 = ⋅ 1 mπ 2 2 2 2  9ω ⋅ ω  2 2 ( 9ω − ω ) + 4 40000    F A1 ≈ 2mπω 2 4F 1 A3 = ⋅ 1 3mπ 2 2 2  2  9ω  ( 9ω 2 − 9ω 2 ) + 4      200   400 F A3 ≈ 27 mπω 2 4F 1 A5 = ⋅ 1 5mπ 2 2   2 2  3ω  2 2 − + ω ω ω 9 25 4 5 )  200  ( )  (   F A5 ≈ 20mπω 2 i.e., A1 : A3 : A5 = 1 : 29.6 : 0.1

3.23

(a)

x +ω2x = 0

y=x

ω = 3ω

Thus y = −ω 2 x

x= y

y dy ω x = =− x dx y 2

divide these two equations: (b)

Solving

ydy

ω2

+ xdx = 0 and Integrating

Let 2C = A2 y2 x2 + =1 ω 2 A2 A 2

→

y2 x2 + =C 2ω 2 2

an ellipse

30

3.24

The equation of motion is F ( x ) = x − x 3 = mx . For simplicity, let m=1. Then

(a)

(b)

x = x − x 3 . This is equivalent to the two first order equations … x = y and y = x − x3 The equilibrium points are defined by x − x 3 = x (1 − x )(1 + x ) = 0

Thus, the points are: (-1,0), (0,0) and (+1,0). We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points. We’ll do this in part (c). dy y x − x 3 The energy can be found by integrating = = or dx x y

∫ y dy = ∫ ( x − x ) dx + C 3

or

y 2 x2 x4 = − +C 2 2 4

In other words … E = T + V =

` (c)

y 2  x4 x2  +  −  = C . The total energy C is 2 4 2

constant. The phase space trajectories are given by solutions to the above equation 1

 2 x4 y = ±  x 2 − + 2C  . 2   The upper right quadrant of the trajectories is shown in the figure below. The trajectories are symmetrically disposed about the x and y axes. They form closed paths for energies C0. Thus, (0,0) is a point of unstable equilibrium.

1

0.5

0

0

0.5

1

1.5

2

31

3.25

θ + sin θ = 0

 d θ 2  − cos θ  = 0 dt  2 

⇒

θ2

Integrating:

2

θ θ

= cos θ θ

0

)

0

θ

∴T = 4 ∫

θ 2 = 2 ( cosθ − cosθ

or

dθ 1

 2 ( cos θ − cos θ )  2

Time for pendulum to swing from θ = 0 to θ = θ is Now—substitute sin φ =

θ

sin sin

2

so φ =

θ

and after some algebra …

2

at θ = θ

2

cos θ = 1 − 2sin

and use the identity

π

T 4

2

θ 2

dφ  2θ  1 − sin 2 

1 2

=

θ

dθ

0

2   2θ 2 θ  − 4 sin sin    2 2    

∴T = 4 ∫

dθ

  2θ 2 θ   4  sin 2 − sin 2     

1

1 2

or

π

dφ

2

T = 4∫

(a)

0

1 − α sin 2 φ 

(1 − α sin 2 φ )

(b)

−

1 2

where α = sin 2

1 2

θ 2

1 3 ≈ 1 + α sin 2 φ + α 2 sin 4 φ + … 2 8

π 2

3  1  T = 4 ∫ dφ 1 + α sin 2 φ + α 2 sin 4 φ + … 8  2  0 2

θ θ 3  θ2 ... (c) α = sin ≈ − + … ∼ 2  2 48 4   θ2  T = 2π 1 + + …  16  -----------------------------------------------------------------------------------------------------2

θ

 α 9  T = 2π 1 + + α 2 + …  4 64 

32

CHAPTER 4 GENERAL MOTION OF A PARTICLE IN THREE DIMENSIONS --------------------------------------------------------------------------------------------------------Note to instructors … there is a typo in equation 4.3.14. The range of the projectile is … v 2 sin 2α v 2 sin 2 2α R=x= 0 … NOT ... 0 g g --------------------------------------------------------------------------------------------------------4.1

∂V ˆ ∂V ˆ ∂V −j −k ∂x ∂y ∂z ˆ ˆ + ˆjxz + kxy F = −c iyz

(a) F = −∇V = −iˆ

(

)

(b) F = −∇V = −iˆ2α x − ˆj 2 β y − kˆ 2γ z (c) F = −∇V = ce−(α x + β y +γ z ) iˆα + ˆj β + kˆγ

(

(d) F = −∇V = −eˆr

)

∂V 1 ∂V 1 ∂V − eˆθ − eˆφ ∂r r ∂θ r sin θ ∂φ

F = −eˆr cnr n −1 4.2

(a) ∇× F =

iˆ

ˆj

∂ ∂x x

∂ ∂y y

iˆ

ˆj

∂ ∂x y

∂ ∂y −x

iˆ

ˆj

∂ ∂x y

∂ ∂y x

kˆ ∂ =0 ∂z z

conservative

(b) ∇× F =

kˆ ∂ = kˆ ( −1 − 1) ≠ 0 ∂z z2

conservative

(c) ∇× F =

kˆ ∂ = kˆ (1 − 1) = 0 ∂z z3

conservative

(d)

33

eˆr ∂ 1 ∇× F = 2 r sin θ ∂r − kr − n

4.3

eˆθ r ∂ ∂θ 0

eˆφ r sin θ ∂ = 0 conservative ∂φ 0

(a) ∇× F =

iˆ

ˆj

∂ ∂x xy

∂ ∂y cx 2

kˆ ∂ = k ( 2cx − x ) ∂z z3

2cx − x = 0 1 c= 2

(b)

∇× F =

also

4.4

(a)

iˆ

ˆj

kˆ

∂ ∂x z y

∂ ∂y cxz y2

∂ ∂z x y

 x cx  = iˆ  − 2 − 2  + y   y x cx − 2 − 2 =0 y y cz z + =0 y2 y2

ˆj  1 − 1  + kˆ  cz + z  2 y2   y y y c = −1

implies that

c = −1 as it must

1 E = constant = V ( x, y, z ) + mv 2 2 1 2 at the origin E = 0 + mv 2 1 1 at (1,1,1) E = α + β + γ + mv 2 = mv 2 2 2

34

v2 = v2 −

2 (α + β + γ ) m 1

2  2 v = v 2 − (α + β + γ )  m  

(b)

v2 −

2 (α + β + γ ) = 0 m 1

2 2 v =  (α + β + γ )  m 

(c)

mx = Fx = −

∂V ∂x

mx = −α ∂V my = − = −2 β y ∂y ∂V mz = − = −3γ z 2 ∂z

4.5

(a)

ˆ + ˆjy F = ix ˆ + ˆjdy dr = idx

on the path x = y : (1,1)

∫(

0,0 )

1

1

1

1

0

0

0

0

F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ xdx + ∫ ydy = 1 ˆ dr = idx dr = ˆjdy

on the path along the x-axis: and on the line x = 1 : (1,1)

∫(

0,0 )

1

1

0

0

F ⋅ dr = ∫ Fx dx + ∫ Fy dy = 1

F is conservative. (b)

ˆ − ˆjx F = iy on the path x = y : (1,1)

∫(

0,0 )

1

1

0

0

1

1

0

0

F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ ydx − ∫ xdy (1,1)

and, with x = y

∫(

on the x-axis:

∫(

and, with y = 0 on the x-axis

∫(

on the line x = 1 :

∫(

0,0 )

(1,0 ) 0,0 ) (1,0 ) 0,0 )

(1,1) 1,0 )

1

1

0

0

F ⋅ dr = ∫ xdx − ∫ ydy = 0 1

1

0

0

F ⋅ dr = ∫ Fx dx = ∫ ydx

F ⋅ dr = 0 1

1

0

0

F ⋅ dr = ∫ Fy dy = ∫ xdy

35

(1,1)

∫(

and, with x = 1 on this path

1,0 )

(1,1)

∫(

0,0 )

1

F ⋅ dr = ∫ dy = 1 0

F ⋅ dr = 0 + 1 = 1

F is not conservative. 4.6

From Example 2.3.2, V ( z ) = −mg

re2 ( re + z )

 z V ( z ) = − mgre 1 +   re  From Appendix D,

(1 + x )

−1

−1

= 1 − x + x 2 +…

  z z2 V ( z ) = − mgre 1 − + 2 + …   re re  2 mgz V ( z ) = −mgre + mgz − +… re With − mgre an additive constant,  z V ( z ) ≈ mgz 1 −   re  ∂ F = −∇V = − kˆ V ( z ) ∂z ˆ 1 − z + z  − 1   = − kmg    re    re ˆ 1 − 2 z  F = − kmg   re  

mx = Fx = 0 , mz

my = Fy = 0

 2z  mz = − mg 1 −  re  

 2z  dz = − mg 1 −  dz re  

h 2z  zdz = − g ∫ 1 −  dz 0 z re    1 h2  − v 2z = − g  h −  re  2 

∫

0

v

re v 2z =0 h − re h + 2g 2

36

h=

re 1 2 2re v 2z − re − g 2 2

h=

re re 2v 2 1− z − 2 2 gre

From Appendix D,

( h, z

1

(1 + x ) 2 = 1 +

h=

re re v 2z v4 − + + z +… 2 2 2 g 4 gre

h≈

v 2z  v 2z  + 1  2 g  2 gre 

re )

x x2 − +… 2 8

−1

v2  v2  From Example 2.3.2, h = 1 −  2 g  2 gre  v2  v2  −1 And with (1 − x ) ≈ 1 + x , + 1 h≈   2 g  2 gre 

4.7

For a point on the rim measured from the center of the wheel: ˆ cos θ − ˆjb sin θ r = ib vt θ =ωt = , so r = −iˆv sin θ − ˆjv cos θ b Relative to the ground, v = iˆv (1 − sin θ ) − ˆjv cosθ For a particle of mud leaving the rim: y = −b sin θ and v y = −v cosθ So

v y = v y − gt = −v cos θ − gt

1 y = −b sin θ − v t cosθ − gt 2 2 At maximum height, v y = 0 :

and

t=−

v cos θ g

 v cos θ h = −b sin θ − v  − g  2 2 v cos θ h = −b sin θ + 2g Maximum h occurs for

 1  v cos θ   cos θ − g  −  g  2  

2

dh 2v 2 cos θ sin θ = 0 = −b cos θ − 2g dθ

37

sin θ = −

gb v2

cos 2 θ = 1 − sin 2 θ =

v 4 − g 2b 2 v4

gb 2 v 4 − g 2b 2 gb 2 v 2 + = 2+ v2 2 gv 2 2v 2g Measured from the ground, gb 2 v 2 ′ =b+ 2 + hmax 2v 2g hmax =

 gb  The mud leaves the wheel at θ = sin −1  − 2   v 

4.8

x = R cos φ so t =

and

x = v x t = ( v cos α ) t

R cos φ v cos α 1 1 and y = v y t − gt 2 = ( v sin α ) t − gt 2 2 2

y = R sin φ

R cos φ 1  R cos φ  R sin φ = ( v sin α ) − g  v cos α 2  v cos α  gR cos 2 φ sin φ = tan α cos φ − 2 2v cos 2 α

α

2

φ

2v 2 cos 2 α 2v 2 cos α tan cos sin α φ − φ = ( ) ( sin α cos φ − cos α sin φ ) g cos 2 φ g cos 2 φ From Appendix B, sin (θ + φ ) = sin θ cos φ + cos θ sin φ R=

2v 2 cos α sin (α − φ ) R= g cos 2 φ dR 2v 2 =0=  − sin α sin (α − φ ) + cos α cos (α − φ )  dα g cos 2 φ  Implies that cos α cos (α − φ ) − sin α sin (α − φ ) = 0

R is a maximum for

From appendix B, cos (θ + φ ) = cos θ cos φ − sin θ sin φ so cos ( 2α − φ ) = 0 2α − φ = Rmax =

π 2

α=

π 4

+

φ 2

2v π φ  π φ  cos  +  sin  −  2 g cos φ  4 2  4 2 2

38

π  π φ  π φ  π φ  Now sin  −  = cos  −  −   = cos  +   4 2  4 2  2  4 2  2v 2 π φ  cos 2  +  Rmax = 2 g cos φ  4 2 Again using Appendix B, cos 2θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 v2   π 2v 2  1 π  1   cos  + φ  + 1 Rmax = cos  + φ  +  =   2 2 g cos φ  2   2  2  g cos φ   2 π  Using cos  + θ  = − sin θ , 2  2 v Rmax = (1 − sin φ ) g (1 − sin 2 φ ) 2

Rmax =

v g (1 + sin φ )

4.9 (a) Here we note that the projectile is launched “downhill” towards the target, which is located a distance h below the cannon along a line at an angle φ below the horizon. α is the angle of projection that yields maximum range, Rmax. We can use the α results from problem 4.8 for this problem. We simply have to replace the angle φ in the above φ result with the angle -φ, to account for the downhill h slope. Thus, we get for the downhill range … Rmax 2v0 2 cos α sin (α + ϕ ) R= cos 2 ϕ g The maximum range and the angle is α are obtained from the problem above again by v 2 (1 + sin ϕ ) π and … 2α = − ϕ . replacing φ with the angle -φ … Rmax = 0 2 2 g cos ϕ 2 2 v (1 + sin ϕ ) v0 h We can now calculate α … Rmax = = 0 = 2 sin ϕ g cos ϕ g (1 − sin ϕ )  gh  1 + 2   v0  π  But, from the above … sin ϕ = sin  − 2α  = cos 2α = 1 − 2sin 2 α 2  gh  gh  Thus … 1 − 2 sin 2 α = 2  1 + 2  v0  v0 

Solving for sin ϕ … sin ϕ =

gh v0 2

39

2sin 2 α =

2 gh = 1− 2 2 csc α v0

 1 gh  1 + 2  =  v0  1 + gh v0 2

 gh  Finally … csc 2 α = 2 1 + 2   v0 

(b)

Solving for Rmax … Rmax =

h h h = = 2 sin ϕ 1 − 2sin α 1 − 2 csc 2 α

Substituting for csc 2 α and solving … v2 gh  Rmax = 0 1 + 2  g  v0  4.10 We can again use the results of problem 4.8. The maximum slope range from problem 4.8 is given by … v0 2 h = Rmax = g (1 + sin ϕ ) sin ϕ Solving for sin ϕ … sin ϕ =

gh v0 2

 gh  1 − 2   v0 

Thus …

cos ϕ sin ϕ We can calculate cos ϕ from the above relation for sin ϕ … xmax = Rmax cos ϕ = h

1 2

 gh   gh  cos ϕ = (1 − sin 2 ϕ ) = 1 − 2 2   1 − 2  v0   v0   Inserting the results for sin ϕ and cos ϕ into the above … 1 2

1

xmax

cos ϕ v0 2  gh  2 1 2 =h = −   sin ϕ g  v0 2 

4.11 We can simplify this problems somewhat by noting that the trajectory is symmetric about a vertical line that passes through the highest point of the trajectory. Thus we have the following picture …

40

z

zmax v0 h0

h1

α

x0

δ

x

x1 R

We have “reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range within which Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28 ft) is the height of the ball when Mickey strikes it at home plate. δ is the distance behind home plate where the ball would be hypothetically launched at some angle α to achieve the total range R. x1 (=328 ft) is the distance the ball actually would travel from home plate if not caught by Mickey. (Note, because of the symmetry, v0 is the speed of the ball when it strikes the ground … also at the same angle α at which was launched. We will calculate the value of x0 assuming a time-reversed trajectory!) v 2 sin 2α 2v0 2 sin α cos α = (1) The range of the ball … R = 0 g g 2

R g R tan α −   2 2 2 2v0 cos α  2  g 2 h1 = x1 tan α − (3) The height at x1 … ( x1 ) 2 2 2v0 cos α g tan α = and inserting this into (2) gives … From (1) … 2 2 R 2v0 cos α R R R zmax = tan α − tan α = tan α 2 4 4 4z Thus, R = max and inserting this expression and the first previously derived into (3) tan α

(2) The maximum height … zmax =

(4)

h1 = x1

( x tan α ) tan α − 1

2

4 zmax Let u = x1 tan α and we obtain the following quadratic … u 2 − 4 zmax u + 4 zmax h1 = 0 and solving for u … 1 h   u = 2 zmax 1 ± (1 − h1 zmax ) 2  and letting ε = 1 , we get … zmax  

41

u ≈ zmax ε = h1

or

u ≈ 2 zmax ( 2 − ε ) = 2 zmax ( 2 − .0475 ) = 3.9 zmax . This result is the correct one …

3.9 zmax = 0.821 ∴ α = 39.40 x1 Now solve for x0 using a relation identical to (4) …

Thus, tan α =

h0 = x0

( x tan α ) tan α − 0

2

4 zmax Again we obtain a quadratic expression for u = x0 tan α which we solve as before. This time, though, the first result for u is the correct one to use … u = zmax ε ≈ h0 and we obtain … h x0 = 0 = 11.9 ft tan α 4.12

The x and z positions of the ball vs. time are 1 1 1 x = v t cos θ z = v t cos θ sin θ − gt 2 2 2 2 1 Since vx = v cos θ 2 v2 1 cos 2 θ sin 2θ R= The horizontal range is 2 g dR The maximum range occurs @ =0 dθ 1 1 dR v 2   2 1 =  2 cos θ cos 2θ − cos θ sin θ sin 2θ  = 0 dθ g  2 2 2  1 1 1 Thus, 2 cos 2 θ cos 2θ = cos θ sin θ sin 2θ 2 2 2 2 Using the identities: 2 cos θ = 1 + cos 2θ and sin 2θ = 2sin θ cosθ We get: (1 + cosθ ) ( 2 cos2 θ − 1) = sin θ sin θ cosθ = (1 − cos 2 θ ) cosθ or

(1 + cosθ ) ( 3cos 2 θ

Thus

cos θ = −1 ,

− cosθ − 1) = 0

cosθ =

(

1 1 ± 13 6

)

Only the positive root applies for the θ -range:

(

)

1 cosθ = 1 + 13 = 0.7676 6 Thus (b) for v = 25 m s −1 Rmax = 55.4 m

0 ≤θ ≤

π 2

θ = 39 51′ @ θ = 39 51′

42

(a) The maximum height occurs at

or

1 v cos θ sin θ = gT 2 v2 1 H= cos 2 θ sin 2 θ 2g 2

or at

dz =0 dt 1 v cos θ sin θ 2 T= g

maximum at fixed θ

dH =0 dα dH v 2  1 1  2 1 2 =  2 cos θ sin θ cosθ − cos θ sin θ sin θ  = 0 dα 2 g  2 2 2  Using the above trigonometric identities, we get 1 1 (1 + cosθ ) sin θ cosθ = sin θ sin 2 θ = sin θ (1 − cos 2 θ ) 2 2 or sin θ (1 + cosθ )( 3cos θ − 1) = 0

The maximum possible height occurs @

1 3 The first two roots give minimum heights; the last gives the maximum 1 Thus, H max = 18.9m @θ = cos −1 = 70 32′ 3 There are 3-roots:

4.13

sin θ = 0 ,

cos θ = −1 ,

cos θ =

The trajectory of the shell is given by Eq. 4.3.11 with r replacing x z g z = r − 2 r2 where r = v cos θ z = v sin θ r 2r Thus, z = r tan θ −

g r2 sec 2 θ 2v 2

Since sec2 θ = 1 + tan 2 θ We have: g r2 g r2 2 tan θ − tan θ + + =0 r z 2v 2 2v 2 (r,z) are target coordinates. The above equation yields two possible roots: 1 1  2 4 2 2 2 2 ± − − tan θ = v v 2 gzv g r ( )  gr   The roots are only real if v 4 − 2 gzv 2 − g 2 r 2 ≥ 0 The critical surface is therefore: v 4 − 2 gzv 2 − g 2 r 2 = 0 4.14

If the velocity vector, of magnitude s , makes an angle θ with the z-axis, and its

43

projection on the xy-plane make an angle φ with the x-axis: x = s sin θ cos φ , and Fx = Fr sin θ cos φ = mx y = s sin θ sin φ , and Fy = Fr sin θ sin φ = my z = s cos θ ,

and Fz = − mg + Fr cos θ = mz

Since Fr = −c2 s = −c2 ( x + y 2 + z 2 ) , the differential equations of motion are not 2

2

separable. mx = −c2 s 2 sin θ cos φ = −c2 sx dx dx ds dx m = m ⋅ = ms = −c2 sx dt ds dt ds dx c2 c = − ds = −γ ds , where γ = 2 m x m x ln x − ln x = ln = −γ s x

θ

s

y

x = x e −γ s y = y e −γ s

Similarly 4.15

z

x

φ

z g γ x g  γx  From eqn 4.3.16,  + 2  max + 2 ln 1 − max  = 0 γ x  γ γ  x  2 3 u u From Appendix D: ln (1 − u ) = −u − − − … for u < 1 2 3 3  γx  γx γ 2 xmax 2 γ 3 xmax − + terms in γ 4 ln  1 − max  = − max − 2 3 x  x 2x 3x  2 3 z xmax gxmax gxmax gxmax gγ xmax + − − − + terms in γ 2 = 0 2 3 x 2x 3x γx γx 2 xmax +

3x 3x 2 z xmax − ≈0 gγ 2γ 1

xmax

3x  9 x 2 3x 2 z  2 ≈− ± +  gγ  4γ  16γ 2 1

3 x 3 x  16γ z  2 ± xmax ≈ − 1 +  4γ 4γ  3g  Since xmax > 0 , the + sign is used. From Appendix D: 1

2

 16γ z  2 8γ z 1  16γ z  3 −  1 +  = 1+  + terms in γ 3g  3g 8  3g   3x 3x 2 x z 8 x γ z 2 − + terms in γ 2 xmax = − + + 2 4γ 4γ g 3g

44

2 x z 8x z 2 − γ +… 3g 2 g z = v sin α and 2 x z = v 2 sin 2α :

xmax =

For

xmax =

v 2 sin 2α 4v 3 sin 2α sin α − γ +… 3g 2 g

4.16 y

x

x = A cos (ω t + α ) ,

x = − Aω sin (ω t + α )

from x = 0 , from x = A , y = B cos (ω t + β ) ,

α =0 x = A cos ω t y = −ω B sin (ω t + β )

1 2 1 2 1 2 kB = ky + my 2 2 2 with

B 2 = 16 A2 + Then

1

ω2

( 9ω

2

y = 4 A , y = 3ω A

and ω =

k : m

A2 ) = 25 A2

B = 5A 4 A = 5 A cos β and 3ω A = −5ω A sin β 4  3 β = cos −1   = sin −1  −  = −36.9 5  5

y = 5 A cos (ω t − 36.9

)

Since maximum x and y displacements are ± A and ±5A , respectively, the motion takes place entirely within a rectangle of dimension 2 A and 10A . ∆ = β − α = −36.9 − 0 = −36.9 2 AB cos ∆ From eqn 4.4.15, tan 2ψ = 2 A − B2 4 ( 2 A)( 5 A) cos ( −36.9 ) 10  5  1 = tan 2ψ = =− 2 −24 3 A2 − ( 5 A ) 1

 1

ψ = tan −1  −  = −9.2 2  3 4.17

mx = Fx = −  x = A cos  

∂V = − kx = −π 2 mx ∂x  k t + α  = A cos (π t + α ) m 

45

∂V = −4π 2 mx ∂y y = B cos ( 2π t + β )

my = −

∂V = −9π 2 mz ∂z z = C cos ( 3π t + γ )

mz = −

Since x = y = z = 0 at t = 0 ,

π

α = β =γ = −

π 2

 x = A cos  π t −  = A sin π t 2  x = Aπ cos π t

Since v 2 = x 2 + y 2 + z 2 and x = y = z , v x = = Aπ 3 v A= π 3 v x= sin π t π 3 y = B sin 2π t , y = 2 Bπ cos 2π t v = 2π B y = 3 v B= 2π 3 v sin 2π t y= 2π 3 z = C sin 3π t , z = 3Cπ cos 3π t v = 3Cπ z = 3 v C= 3π 3 v sin 3π t z= 3π 3 Since ω x = π , ω y = 2π , and ω z = 3π the ball does retrace its path. tmin =

2π n1

ωx

=

2π n2

ωy

=

2π n3

ωz

The minimum time occurs at n1 = 1 , n2 = 2 , n3 = 3 .

46

tmin = 4.18

2π

π

=2

Equation 4.4.15 is

tan 2ψ =

2 AB cos ∆ A2 − B 2

Transforming the coordinate axes xyz to the new axes x′y′z ′ by a rotation about the z-axis through an angleψ given, from Section 1.8: x′ = x cosψ + y sin ψ , y′ = − x sinψ + y cosψ or, x = x′ cosψ − y′ sinψ , and y = x′ sinψ + y′ cosψ From eqn. 4.4.10:

x2 2 cos ∆ y 2 x − + 2 = sin 2 ∆ y 2 A AB B

Substituting: 1 x′2 cos 2 ψ − 2 x′y′ cosψ sinψ + y′2 sin 2 ψ ) 2 ( A 2 cos ∆ 2  x′ cosψ sinψ + x′y′ ( cos 2 ψ − sin 2 ψ ) − y′2 cosψ sin ψ  −  AB  1 + 2 ( x′2 sin 2 ψ + 2 x′y′ cosψ sin ψ + y′2 cos 2 ψ ) = sin 2 ∆ B For x′ to be a major or minor axis of the ellipse, the coefficient of x′y′ must vanish. 2 cosψ sinψ 2 cos ∆ 2 cosψ sinψ cos 2 ψ − sin 2 ψ ) + − − =0 ( 2 A AB B2 From Appendix B, 2 cosψ sin ψ = sin 2ψ and cos 2 ψ − sin 2 ψ = cos 2ψ sin 2ψ 2 cos ∆ cos 2ψ sin 2ψ =0 − − + A2 AB B2 1  2 cos ∆  1 tan 2ψ  2 − 2  = A  AB B 2 AB cos ∆ tan 2ψ = 2 A − B2 4.19

Shown below is a face-centered cubic lattice. Each atom in the lattice is centered within a cube on whose 6 faces lies another adjacent atom. Thus each atom is surrounded by 6 nearest neighbors at a distance d. We neglect the influence of atoms that lie at further distances. Thus, the potential energy of the central atom can be approximated as 6

V = ∑ cri −α i =1

2d

47

1

r1 = ( d − x ) + y 2 + z 2  2   2

−

α

α

 2x x2 + y2 + z 2  2 + r = ( d − 2dx + x + y + z ) = d 1 −  d d2   1 n From Appendix D, (1 + x ) = 1 + nx + n ( n − 1) x 2 + … 2 2  α  2 x x + y 2 + z 2  1  α   α  r1−α = d −α 1 −  − +  +  −   − − 1 d2  2  d  2  2  2  −α 1

2

2

2

2 −2

−α

2 2 2 2 2 2 2  2 x  2 x3   2x   x + y + z   x + y + z   + term s in +  −  − 2         d2 d2 d3   d   d     

x3   α  α   4 x2  αx α − 2 ( x 2 + y 2 + z 2 ) +  + 1  2 + terms in 3   r1−α = d −α 1 + d   d 2d 4  2  d   αx α α α   − 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2  r1−α ≈ d −α 1 + d 2d d 2    1

1

r2 = ( − d − x ) + y 2 + z 2  2 =  d 2 + 2dx + x 2 + y 2 + z 2  2   2

−

−α 2

r

r2−α

α

 2 x x2 + y 2 + z 2  2 = d 1 + +  d d2    αx α α α   ≈ d −α 1 − − 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2  d 2d d 2    −α

α α   r1−α + r2−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) x 2  d  d  Similarly:

α α   r3−α + r4−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) y 2  d  d  α α   r5−α + r6−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) z 2  d  d 

 3α   α 2 2α  V ≈ cd −α 6 − 2 ( x 2 + y 2 + z 2 ) +  2 + 2  ( x 2 + y 2 + z 2 )  d d  d   −α −α − 2 2 2 2 2 ≈ 6cd + cd (α − α )( x + y + z )

V ≈ A + B ( x2 + y 2 + z 2 ) 4.20

48

z

ˆ kB

(

F = q E+v×B

ˆjE

ˆ0 iv

(

)

)

ˆ × kB ˆ = iyB ˆ + ˆjy + kz ˆ − ˆjxB v × B = ix ˆ F = iqyB + ˆjq ( E − xB )

y

mx = Fx = qyB qB x−x = y m

x

qB  my = Fy = qE − qxB = qE − qB  x + m 

 y 

2

2

qE qBx  qB  eE eBx  eB  − − + −  y y=  y=− m m  m m m m eE eB +ωx , ω= y +ω2 y = − m m 1  eE  + ω x  + A cos (ω t + θ ) y = 2 − ω  m  y = − Aω sin (ω t + θ ) y = 0 , so θ = 0

1  eE  − +ωx  2  ω  m  1  eE  y = a (1 − cos ω t ) , +ωx  a = 2 − ω  m  qB x=x + y = x − ω y = x − ω a (1 − cos ω t ) m x = ( x − ω a ) + ω a cos ω t y = 0 , so A = −

x = ( x − ω a ) t + a sin ω t x = a sin ω t + bt , mz = Fz = 0 z=zt+z =0

b = x −ωa

4.21

1 2 b mv + mqh = mg 2 2 2 v = g ( b − 2h )

y b h

x

mv 2 = −mg cos θ + R Fr = − b

49

h b h mv 2 mg mg R = mg − =  h − ( b − 2h )  = ( 3h − b ) b b b b the particle leaves the side of the sphere when R = 0 b b above the central plane h = , i.e., 3 3 cos θ =

4.22

1 2 mv + mgh = 0 2 at the bottom of the loop, h = −b 1 2 mv = mgb , 2 v = 2gb

so

h

b

mv 2 Fr = −mg + R = b 2 mv R = mg + = mg + 2mg = 3mg b

v 4.23

From the equation for the energy as a function of s in Example 4.6.2, 1 1  mg  2 E = ms 2 +  s , 2 2  4A  s is undergoing harmonic motion with: "k " g 1 g ω= = = m 4A 2 A Since s = 4 A sin φ , φ increases by 2π radians during the time interval:

 A = 2π  2  ω  g For cycloidal motion, x and z are functions of 2φ so they undergo a complete cycle every time φ changes by π . Therefore, the period for the cycloidal motion is one-half the period for s . 1 A T = T ′ = 2π 2 g ------------------------------------------------------------------------------------------------------

T′ =

2π

50

CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force G acting on him is zero. The scale exerts an upward force, N , whose value is equal to the scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus G G G N + mg − mA0 = 0

G N

N − mg − mA0 = N − mg − m

G A0

g 5 = N − mg = 0 4 4

5 5 mg = W 4 4 W ′ = 150lb. W′ = N =

(a)

(b)

G mg

(b) The acceleration is downward, in the same G direction as g W 3 g N − mg + m   = 0 W ′ = W − = W 4 4 4

W ′ = 90lb.

G G G G 5.2 (a) Fcent = −mω × (ω × r ′ ) G G G For ω ⊥ r ′ , Fcent = mω 2 r ′ eˆr ′

ω = 500 s −1 = 1000π s −1

G 2 Fcent = 10−6 × (1000π ) × 5 eˆr = 5π 2 dynes outward Fcent mω 2 r ′ (1000π ) 5 = = = 5.04 × 104 Fg mg 980 2

(b)

5.3

G G G mg + T − mAD = 0

(See Figure 5.1.2)

g − mg ˆj + T cos θ ˆj + T sin θ iˆ − m   iˆ = 0  10  mg T cos θ = mg , and T sin θ = 10 1 tan θ = , θ = 5.71D 10 mg = 1.005mg T= cos θ

G 5.4 The non-inertial observer thinks that g ′ points downward in the direction of the hanging plumb bob… Thus

51

g G G G g ′ = g − AD = g ˆj − iˆ 10 For small oscillations of a simple pendulum: 1 T = 2π g′ 2

g g ′ = g +   = 1.005 g  10  2

T = 2π 5.5 (a)

1 1 = 1.995π 1.005g g

f = − µ mg is the frictional force acting on the

G A0

G f

box, so G G G f − mA0 = ma′

(b)

(a)

G ( a′ is the acceleration of the box relative to the truck. See G Equation 5.1.4b.) Now, f the only real force acting horizontally, so the acceration relative to the road is f µ mg g a= =− = −µ g = − m m 3 (For + in the direction of the moving truck, the – indicates that friction opposes the forward sliding of the box.) g AD = − (The truck is decelerating.) 2 from above, ma − mA0 = ma′ so g g g a′ = a − AD = − + = 3 2 6

K 5.6 (a) r = iˆ ( xD + R cos Ωt ) + ˆjR sin Ωt K r = −iˆΩR sin Ωt + ˆjΩR cos Ωt K K r ⋅ r = v 2 = Ω 2 R 2 ∴ v = ΩR circular motion of radius R K K K K K ˆ ′ + ˆjy′ (b) r′ = r − ω × r ′ where r ′ = ix ˆ ′ + ˆjy′ = −iˆΩR sin Ωt + ˆjΩR cos Ωt − ω kˆ × ix

(

)

52

= −iˆΩR sin Ωt + ˆjΩR cos Ωt − ˆjω x′ + iˆω y′ x ′ = ω y′ − ΩR sin Ωt y ′ = −ω x′ + ΩR cos Ωt (c) Let u′ = x′ + iy′ here i = −1 ! u ′ = x ′ + iy ′ = ω y′ − ΩR sin Ωt − iω x′ + iΩR cos Ωt iω u ′ = +iω x′ = −ω y′ ∴ u ′ + iω u′ = −ΩR sin Ωt + iΩR cos Ωt = iΩ ReiΩt Try a solution of the form u′ = Ae − iω t + BeiΩt u ′ = −iω Ae− iω t + iΩBeiΩt iω u′ = iω Ae− iω t + iω BeiΩt ∴ u ′ + iω u ′ = i (ω + Ω ) Beiω t

so B =

ΩR ω +Ω

Also at t = 0 the coordinate systems coincide so u ′ = A + B = x′ ( 0 ) + iy′ ( 0 ) = xD + R ΩR ω +Ω ΩR iΩt e + ω +Ω

∴ A = xD + R − B = xD + R −

ω R  − iω t  Thus, u′ =  xD + e ω + Ω  

so, A = xD +

ωR ω +Ω

5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y frame of reference is fixed inertial frame attached to the Sun. (a) In the x, y rotating frame of reference x ( t ) = R cos ( Ω − ω ) t − Rε y ( t ) = − R sin ( Ω − ω ) t

where R is the radius of the asteroid’s orbit and RE is the radius of the Earth’s orbit. Ω is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit. (b)

x ( t ) = − ( Ω − ω ) R sin ( Ω − ω ) t → 0 at t = 0 y ( t ) = − ( Ω − ω ) R cos ( Ω − ω ) t → − ( Ω − ω ) R at t = 0

(c)

K K K K K K K K K K a = A − Aε − Ω × r − 2Ω × r − Ω × Ω × r K Where a is the acceleration of the asteroid in the x, y frame of reference,

53

K K A, Aε are the accelerations of the asteroid and the Earth in the fixed, inertial frame of reference. K K K K K 1st : examine: A − Aε − Ω × Ω × r K K K K K K K K K K = ω × ω × R − Ω × Ω × Rε − Ω × Ω × ( R − Rε ) K K K K K K K K = (ω × ω − Ω × Ω ) × R = − (ω 2 − Ω 2 ) R note: ω = ω kˆ , Ω = Ωkˆ Thus: K K K K a = ( Ω 2 − ω 2 ) R − 2Ω × v Therefore: ˆ + ˆjy = ( Ω 2 − ω 2 ) iR ˆ ˆ ˆ ˆ  ix  cos ( Ω − ω ) t − jR sin ( Ω − ω ) t  −2 jΩx + 2i Ωy

(

)

Thus:  x = ( Ω 2 − ω 2 ) R cos ( Ω − ω ) t + 2Ωy

 y = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx Let  x = ( Ω − ω ) y

and

Then, we have y = ( Ω − ω ) R cos ( Ω − ω ) t +

 y = ( Ω − ω ) x

2Ω y (Ω − ω )

which reduces to

y = − ( Ω − ω ) R cos ( Ω − ω ) t

Integrating … y = − R sin ( Ω − ω ) t → 0 at t = 0 Also, or

− x ( Ω − ω ) = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx x = ( Ω + ω ) R sin ( Ω − ω ) t + 2Ωx x = − ( Ω − ω ) R sin ( Ω − ω ) t

Integrating … x = R cos ( Ω − ω ) t + const

x = R cos ( Ω − ω ) t − Rε → R − Rε at t = 0 5.8

Relative to a reference frame fixed to the turntable the cockroach travels at a constant speed v’ in a circle. Thus y′ v ′2 G G ′ a eˆr ′ . = − ω b Since the center of the turntable is fixed. G ′ x b AD = 0 The angular velocity, ω, of the turntable is constant, so

54

G G ω = ω kˆ′ , with ω = 0 G G G G r ′ = beˆr ′ , so ω × (ω × r ′ ) = −bω 2 eˆr ′

G G G v′ = v′eˆθ ′ , so ω × v′ = −ω v′eˆr ′ G G G G G G G From eqn 5.2.14, a = a′ + 2ω × v′ + ω × (ω × r ′) and putting in terms from above v′ 2 − 2ω v′ − bω 2 ar ′ = − b G G For no slipping F ≤ µ s mg , so a ≤ µ s g

v′ 2 + 2ω v′ + bω 2 ≤ µ s g b 2 vm′ + 2ω bvm′ + b 2ω 2 − bµ s g = 0 vm′ = −ω b ± ω 2b 2 − b 2ω 2 + bµ s g Since v′ was defined positive, the +square root is used. vm′ = −ω b + bµ s g

(b)

5.9

G v′ = −v′eˆθ ′ G G ω × v′ = +ω v′eˆr ′ v′ 2 + 2ω v′ − bω 2 ar ′ = − b 2 ′ v − 2ω v′ + bω 2 ≤ µ s g b vm′ = ω b + bµ s g

G VD 2 VD ˆ iˆ′ As in Example 5.2.2, ω = j ′ and AD = G

ρ

ρ

For the point at the front of the wheel: 2 G G V  r ′ = D ˆj ′ and v′ = −VD kˆ′ b G ω =0 Vb G G V ω × r ′ = D kˆ′ × −bjˆ′ = D iˆ′

ρ

G

G G

ω × (ω × r ′ ) = G G

ω × v′ =

(

)

ρ

2

VD ˆ VDb ˆ VD b ˆ j′ k′× i′ =

ρ

ρ

(

)

ρ

VD ˆ k ′ × −VD kˆ′ = 0

ρ

G VD 2  VD 2 VD 2b  G G G G G ˆ + 2  ˆj ′ a = r ′ + ω × (ω × r ′ ) + AD = i′ +  ρ ρ  b 

55

5.10

(See Example 5.3.3) mω 2 x′ = mx′ x′ ( t ) = Aeω t + Be −ω t x ′ ( t ) = ω Aeω t − ω Be −ω t

Boundary Conditions: l x2 ( 0) = = A + B 2 x ′ ( 0 ) = 0 = ω ( A − B ) ∴A=

l 4

B=

l cosh ω t 2 l l l x′ (T ) = + = cosh ωT 2 2 2

l x ′ ( t ) = ω sinh ω t 2

x′ ( t ) =

(a) (b)

∴ cosh ωT = 2

when the bead reaches the end of the rod

T=

or

1

ω

cosh −1 2 =

1.317

ω

l x ′ (T ) = ω sinh ωT 2 l l = ω sinh cosh −1 2  = ω (1.732 ) = 0.866ω l 2 2 1 l ωl or ω  cosh 2 ωT − 1 2 = 3 = 0.866ω l 2 2

(c)

5.11

l 4

G v′ = 400 ˆj ′ mph = 586.67 ˆj ′ ft ⋅ s −1 G ω = 7.27 ×10−5 cos 41D ˆj ′ + sin 41D kˆ′ s −1 G G ω × v′ = − ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D ) iˆ′ ft ⋅ s −2 G G −2mω × v′ Fcor = Fgrav mg

(

=

)

2 ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D ) 32

Fcor = 0.0017 Fgrav

G G The Coriolis force is in the −ω × v′ direction, i.e., +iˆ′ or east.

5.12

(See Figure 5.4.3) G ω = ω y′ ˆj′ + ω z′ kˆ′ G v′ = vx′iˆ′ + v y′ ˆj ′

56

G G

ω × v′ = −ω z′v y′iˆ′ + ω z′vx′ ˆj′ − ω y′vx′ kˆ′ G G

(ω × v′ )horiz = −ω z′v′y′iˆ′ + ω z′vx′ ˆj′ 1

G G

1

= (ω z2′v y2′ + ω z2′vx2′ ) 2 = ω z′ ( vx2′ + v y2′ ) 2 = ω z ′v′ G G G Fcor = −2mω × v′ G G G G = 2m (ω × v′ )horiz = 2mω z′ v′ , independent of the direction of v′ . Fcor

(ω × v′ )horiz

( )

5.13

horiz

From Example 5.4.1 … 1

1  8h 3  2 ′ xh = ω   cos λ 3  g 

and yh′ = 0 . 1

3 3 2 1 −5 −1  8 × 1250 ft  D ′ xh = ( 7.27 × 10 s )   cos 41 −2 3  32 ft ⋅ s 

xh′ = 0.404 ft to the east. 5.14

From Example 5.4.2: ωH 2 ∆≈ sin λ is the deflection of the baseball towards the south since it vD

was struck due East at Yankee Stadium at latitude λ = 41D N (problem 5.13). v0 is the initial speed of the baseball whose range is H. From eqn 4.3.18b, without air resistance in an inertial reference frame, the horizontal range is … vD2 sin 2α H= g 1

Solving for v0 …

 gH  2 vD =    sin 2α  1

 32 ft ⋅ s −2 × 200 ft  2 −1 vD =   = 113 ft ⋅ s D sin 30   ∴

( 7.27 ×10 ∆≈

−5

s −1 )( 2002 ft 2 )

sin 41D = 0.0169 ft = 0.2 in 113 ft ⋅ s −1 A deflection of 0.2 inches should not cause the outfielder any difficulty.

5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference. Thus … G G G  da  G G  da   =   +ω × a r =   dt  fixed  dt  rot G G G G G G G G G a = r ′ + ω × r ′ + 2ω × r′ + ω × (ω × r ′ )

57

G G G G G G G G G G  da     = r ′ + ω × r ′ + ω × r ′ + 2ω × r ′ + 2ω × r ′  dt rot G G G G G G G G G +ω × (ω × r ′ ) + ω × ω × r ′ + ω × ω × r′

( ) ( ) G G G G G G G G G G G G G G ω × a = ω ×  r ′ + ω × (ω × r ′ ) + 2ω × (ω × r′ ) + ω × ω × (ω × r ′ ) 

G G is ⊥ to ω and r ′ . Let this define a direction nˆ : G G G G ω × r ′ = ω × r ′ nˆ G G G G G Since ω ⊥ nˆ , ω × (ω × r ′ ) is in the plane defined by ω and r ′ and G G G G G G G G ω × (ω × r ′) = ω × nˆ ω × r ′ = ω ω × r ′ . G G G G Since ω ⊥ ω × (ω × r ′ ) … G G G G G G ω × ω × (ω × r ′ )  = ω 2 ω × r ′ G G G G And ω × ω × (ω × r ′ )  is in the direction − nˆ G G G G G G Thus ω × ω × (ω × r ′ )  = −ω 2 (ω × r ′ ) G G G G G G G G G G G ω × a = ω ×  r ′ + ω × ω × r ′ + 2ω × ω × r′ − ω 2 (ω × r ′ )

Now

G G

(ω × r ′ )

(

)

(

)

G G G G G G G G G G G  r =  r ′ + ω × r ′ + 3ω × r′ + 3ω ×  r ′ + ω × (ω × r ′ ) G G G G G G G G +2ω × ω × r ′ + 3ω × ω × r′ − ω 2 (ω × r ′ )

(

5.16

)

(

)

With xD′ = yD′ = zD′ = xD′ = yD′ = 0 , and zD′ = vD′ Equations 5.4.15a – 5.4.15c become: 1 x′ ( t ) = ω gt 3 cos λ − ω t 2 vD′ cos λ 3 y′ ( t ) = 0 1 z′ ( t ) = − gt 2 + vD′t 2 When the bullet hits the ground z ′ ( t ) = 0 , so 2v′ t= D g  8v′ 3   4v ′ 2  1 x′ = ω g  D3  cos λ − ω  D2  vD′ cos λ 3  g   g  4ω vD′ 3 x′ = − cos λ 3g 2 x′ is negative and therefore is the distance the bullet lands to the west.

5.17

With xD′ = yD′ = zD′ = 0 and xD′ = vD cos α yD′ = 0 zD′ = vD sin α we can solve equation 5.4.15c to find the time it takes the projectile to strike the ground …

58

1 z ′ ( t ) = − gt 2 + vD′ t sin α + ω vDt 2 cos α cos λ = 0 2 2vD′ sin α 2v′ sin α or t= ≈ D g − 2ω vD′ cos α cos λ g We have ignored the second term in the denominator—since vD′ would have to be impossibly large for the value of that term to approach the magnitude g g − 2ω vD′ cos α cos λ ≈ g − ω vD′ For example, for λ = 41o and α = 45o or

vD′ ≈

g

ω

≈ 144

km ! s

Substituting t into equation 5.4.15b to find the lateral deflection gives 4ω vD′ 3 y ( t ) = − [ω vD cos α sin λ ] t 2 = − sin λ sin 2 α cos α g2 5.18

Let … G r

y

x

G R0

G aD = acceleration of object relative to Earth G ωD = ωD kˆ = its angular speed G AD = acceleration of satellite G ω = ω kˆ = its angular speed G G G G G G G G aD = a + 2ω × v + ω × (ω × r ) + AD (Equation 5.2.14) G G G G G G G G a = aD − AD − 2ω × v − ω × ω × r

G R

As in problem 5.7 Evaluate the term … G G G G G G G G G G G G G G G G ∆ a = aD − AD − ω × ω × r = ω D × ω D × RD − ω × ω × R − ω × ω × RD − R G G ∆ a = (ω D2 − ω 2 ) RD

(

… given the condition that

but

ωD2 RD3 = ω 2 R 3

G G  R3  ∆ a = −ω 2 RD  1 − 3   RD  G G G G G G RD ⋅ RD = R + r ⋅ R + r = R 2 + r 2 + 2rR cos θ

(

Letting x = cos θ

)(

)

G G  2x  RD ⋅ RD = R 2 + r 2 + 2 Rx ≈ R 2  1 +  for small r R  3

−

3

R3  2 x  2 = 1 +  or and RD3  R 3 G −   G G 2 R x 2   ∆ a = −ω 2 RD 1 −  1 +   ≈ −3ω 2 x D ≈ −3ω 2 x iˆ for small r   R  R    2x 2 RD3 ≈ R 3 1 +  R 

59

)

G G G G ˆ + ˆjy Hence: a = ∆ a − 2ω × v = −3ω 2 x iˆ − 2ω kˆ × ix G  ˆ − 2ω xˆj a = iˆ x + ˆjy = −3ω 2 xiˆ + 2ω yi  So x − 2ω y − 3ω 2 x = 0  y + 2ω x = 0

(

5.19

G G G G mr = qE + q v × B

(

Equation 5.2.14

)

)

G G G G G G G G G  r =  r ′ + ω × r ′ + 2ω × v′ + ω × (ω × r ′ ) G G G G v = v′ + ω × r ′

Equation 5.2.13 q G G G ω =− B so ω = 0 2m G G G G q G G G G G G G mr′ − q B × v′ − B × (ω × r ′ ) = qE + q ( v′ + ω × r ′ ) × B  2 G G G G G G q G G G G G G mr ′ + q v′ × B + (ω × r ′ ) × B = qE + q v′ × B + q (ω × r ′ ) × B 2 G q G G G G mr ′ = qE + (ω × r ′ ) × B 2 G q qB q G G (ω × r ′) × B =   ( r ′)( sin θ )( B ) ∝ B 2 2 2  2m  G G Neglecting terms in B 2 , mr′ = qE 5.20

(

)

(

)

(

)

For x′ = x cos ω ′t + y sin ω ′t y′ = − x sin ω ′t + y cos ω ′t x ′ = x cos ω ′t − xω ′ sin ω ′t + y sin ω ′t + yω ′ cos ω ′t y ′ = − x sin ω ′t − xω ′ cos ω ′t + y cos ω ′t − yω ′ sin ω ′t x ′ = x cos ω ′t + y sin ω ′t + ω ′ y′ y ′ = − x sin ω ′t + y cos ω ′t − ω ′x′  x′ =  x cos ω ′t − xω ′ sin ω ′t +  y sin ω ′t + yω ′ cos ω ′t + ω ′ y ′  y′ = −  x sin ω ′t − xω ′ cos ω ′t +  y cos ω ′t − yω ′ sin ω ′t − ω ′x ′  x′ =  x cos ω ′t +  y sin ω ′t + 2ω ′ y ′ + ω ′2 x′  y′ = −  x sin ω ′t +  y cos ω ′t − 2ω ′x ′ + ω ′2 y′ Substituting into Eqns 5.6.3:  x cos ω ′t +  y sin ω ′t + 2ω ′ y ′ + ω ′2 x′ g g = − x cos ω ′t − y sin ω ′t + 2ω ′ y ′ l l ′ ′ ′   x sin ω t + y cos ω t − 2ω x ′ + ω ′2 y′ g g = + x sin ω ′t − y cos ω ′t − 2ω ′x ′ l l

60

Collecting terms and neglecting terms in ω ′2 : g  g    x + x  cos ω ′t +   y + y  sin ω ′t = 0   l  l    g  g    x + x  sin ω ′t −   y + y  cos ω ′t = 0   l  l    24 hours sin λ 24 T= = 73.7 hours sin19D

5.21

T=

5.22

Choose a coordinate system with the origin at the center of the wheel, the x′ and y′ axes pointing toward fixed points on the rim of the wheel, and the z ′ axis pointing toward the center of curvature of the track. Take the initial position of G the x′ axis to be horizontal in the −VD direction, so the initial position of the y′ axis is vertical. V The bicycle wheel is rotating with angular velocity D about its axis, so … b V G ωl = kˆ′ D b A unit vector in the vertical direction is: Vt Vt nˆ = iˆ′ sin D + ˆj ′ cos D b b At the instant a point on the rim of the wheel reaches its highest point: Vt Vt G  r ′ = bnˆ = b  iˆ′ sin D + ˆj ′ cos D  b b   Since the coordinate system is moving with the wheel, every point on the rim is fixed in that coordinate system. G G r′ = 0 and  r′ = 0 The x′y′z ′ coordinate system also rotates as the bicycle wheel completes a circle around the track: V V  Vt Vt G ω 2 = nˆ D = D  iˆ′ sin D + ˆj ′ cos D  ρ ρ b b  The total rotation of the coordinate axes is represented by: V Vt Vt G G G V  ω = ω1 + ω 2 = D  iˆ′ sin D + ˆj ′ cos D  + kˆ′ D ρ b b b  2 Vt Vt G V  ω = D  iˆ′ cos D − ˆj′ sin D  b ρb  b  2 Vt Vt  V2 G G V  ω × r ′ = D  kˆ′ cos 2 D + kˆ′ sin 2 D  = D kˆ′ b b  ρ ρ 

61

G G

ω × r′ = 0 VDb  ˆ VDt VDt ˆ Vt Vt Vt Vt  − k ′ sin D cos D  + VD  ˆj ′ sin D − iˆ′ cos D   k ′ sin cos ρ  b b b b  b b   2 2 V  Vt Vt V  Vt Vt G G G ω × (ω × r ′ ) = D  kˆ′ sin 2 D + kˆ′ cos 2 D  + D  −iˆ′ sin D − ˆj ′ cos D  b b  b  b ρ  b  2 2 V V G G G ω × (ω × r ′ ) = kˆ′ D − nˆ D b ρ Since the origin of the coordinate system is traveling in a circle of radius ρ : G V2 AD = kˆ′ D G G

ω × r′ =

ρ

G G G G G G G G G G  r =  r ′ + ω × r ′ + 2ω × r′ + +ω × (ω × r ′ ) + AD V2 V2 V2 V2 G  r = kˆ′ D + kˆ′ D − nˆ D + kˆ′ D b ρ ρ ρ V2 V2 G  r = 3 D kˆ′ − D nˆ b ρ

With appropriate change in coordinate notation, this is the same result as obtained in Example 5.2.2. --------------------------------------------------------------------------------------------------

62

CHAPTER 6 GRAVITATIONAL AND CENTRAL FORCES 6.1

4 m = ρ V = ρ π rs3 3 1

 3m  3 rs =    4πρ 

F=

Gmm

( 2rs )

2

2

2

Gm 2  4πρ  3 G  4πρ  3 43 =   =   m 4  3m  4 3  2

F F Gm 2  4πρ  3 3 = =   m W mg 4g  3  1

2

1 F 6.672 × 10−11 N ⋅ m 2 ⋅ kg −2  4π × 11.35 g ⋅ cm −3 1 kg 106 cm3  3 kg = × 1 × × )3   ( W 4 × 9.8 m ⋅ s −2 3 103 g 1 m3  

F = 2.23 × 10−9 W 6.2 (a) The derivation of the force is identical to that in Section 6.2 except here r < R. This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r. GmM R + r  r 2 − R 2  F= Q 1 +  ds s2  4 Rr 2 ∫R − r  R s GmM  R2 − r 2 R2 − r 2  R r R r = + − − + − ( )   θ ψ 4 Rr 2  R+r R−r  P r GmM  2r + R − r − ( R + r )  = 0 F= 4 Rr 2  (b) Again the derivation of the gravitational potential energy is identical to that in Example 6.7.1, except that the limits of integration on s are ( R − r ) → ( R + r ) .

2πρ R 2 R + r ds rR ∫R − r 2πρ R 2 = −G  R + r − ( R − r )  rR  4π R 2 ρ M = −G φ = −G R R For r < R , φ is independent of r. It is constant inside the spherical shell.

φ = −G

6.3

F =−

GMm eˆr r2

63

The gravitational force on the particle is due only to the mass of the earth that is inside the particle’s instantaneous displacement from the center of the earth, r. The net effect of the mass of the earth outside r is zero (See Problem 6.2). 4 M = π r3ρ 3 4 F = − Gπρ mreˆr = − kreˆr 3

r

F

The force is a linear restoring force and induces simple harmonic motion. T=

2π

ω

= 2π

3 m = 2π 4Gπρ k

The period depends on the earth’s density but is independent of its size. At the surface of the earth,

GMm Gm 4 3 = 2 ⋅ π Re ρ Re2 Re 3 4Gπρ g = 3 Re

mg =

T = 2π

6.4

Re 6.38 × 106 m 1 hr = 2π × ≈ 1.4 hr 9.8 m ⋅ s −2 3600 s g

Fg = −

GMm 4 eˆr , where M = π r 3 ρ 2 r 3

The component of the gravitational force perpendicular to the tube is balanced by the normal force arising from the side of the tube. The component of force along the tube is Fx = Fg cos θ The net force on the particle is … 4 F = −iˆ Gπρ mr cos θ 3 r cos θ = x 4 ˆx F = −iˆ Gπρ mx = −ik 3

As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours.

64

6.5

6.6

GMm mv 2 GM v2 = = so 2 r r r for a circular orbit r, v is constant. 2π r T= v 4π 2 r 2 4π 2 3 T2 = r ∝ r3 = 2 v GM 2π r v From Example 6.5.3, the speed of a satellite in circular orbit is …

(a) T =

1

 gR 2  2 v= e   r  3

T=

2π r 2 1

g 2 Re 1

 T 2 gRe2  3 r =  2  4π  1

1

r  T 2 g  3  242 hr 2 × 36002 s 2 ⋅ hr −2 × 9.8 m ⋅ s −2  3 =  =  Re  4π 2 Re   4π 2 6.38 × 106 m  r = 6.62 ≈ 7 Re 3

(b)

T=

2π r 2 1 2

g Re

3

=

2π ( 60 Re ) 2 1 2

g Re

= 2π

603 Re g 1

 2 603 × 6.38 × 106 m = 2π  −2 −2 −2  2 2 2 2  9.8 m ⋅ s × 3600 s ⋅ hr × 24 hr ⋅ day  T = 27.27 day ≈ 27 day

6.7

From Example 6.5.3, the speed of a satellite in a circular orbit just above the earth’s surface is … v = gRe T=

2π Re Re = 2π v g

65

This is the same expression as derived in Problem 6.3 for a particle dropped into a hole drilled through the earth. T ≈ 1.4 hours. 6.8 The Earth’s orbit about the Sun is counter-clockwise as seen from, say, the north star. It’s coordinates on approach at the latus rectum are ( x, y ) = ( ε a, −α ) .

The easiest way to solve this problem is to note that ε = orbit is almost circular! GM S m mv 2 GM S ∴ = and v 2 = 2 r r r with r = α ≈ a ≈ b when ε ≈ 0  GM S  4 m v≈  = 3 ⋅10 s  α  More exactly ml 2 r × v = α v cos β = l , but α = k 2 l α= , Since k = GM S m GM S

1 is small. The 60

(equation 6.5.19) 1

hence l = α v cos β = (α GM S ) 2

1

 GM S  2 1 v= Or   α  cos β The angle β can be calculated as follows:

y 2 x2 + =1 b2 a 2 dy b2 x ∴ =− 2 dx a y

(see appendix C) and at ( x, y ) = ( ε a, −α )

so

dy b 2 ε a b 2 ε b2 2 = 2 = 2 = ε since 1 − ε = dx a α a (1 − ε 2 ) a2

here

dy = tan β = ε dx

and

 GM S  2 1  GM S  2 v= ≈  as before.   α  cos ε  α 

6.9

F ( r ) = Fs + Fd

1

or β ≈ ε (small ε) 1

GMm r2 GM d m Fd = − r2 The net effect of the dust outside the planet’s radius is zero (Problem 6.2). The mass of the dust inside the planet’s radius is: Fs = −

66

4 M d = π r3ρ 3 GMm 4 F ( r ) = − 2 − πρ mGr 3 r 6.10

1 1 − kθ = e r r du k = − e − kθ dθ r

u=

d 2u k 2 − kθ = e = k 2u dθ 2 r From equation 6.5.10 … d 2u 1 + u = k 2u + u = − 2 2 f ( u −1 ) 2 du ml u −1 2 2 f ( u ) = − ml ( k + 1) u 3

f (r ) = −

ml 2 ( k 2 + 1)

r3 The force varies as the inverse cube of r. From equation 6.5.4, r 2θ = l dθ l = 2 e −2 kθ dt r l e 2 kθ dθ = 2 dt r 1 2 kθ lt e = 2 +C r 2k  1  2klt ln  2 + C ′  2k  r  θ varies logarithmically with t.

θ=

k = ku 3 3 r From equation 6.5.10 … d 2u 1 ku + u = − 2 2 ⋅ ku 3 = − 2 2 dθ ml u ml 2 d u  k  + 1 + 2  u = 0 2 dθ  ml 

6.11

f (r ) =

67

k   If 1 + 2  < 0 ,  ml  k   If 1 + 2  = 0 ,  ml  du = C1 dθ u = c1 θ + c2 1 r= c1 θ + c2 k   If 1 + 2  > 0 ,  ml  u = A cos

(

cθ + δ

d 2u − cu = 0 , dθ 2 d 2u =0 dθ 2

)

d 2u + cu = 0 , dθ 2

   k r =  A cos  1 + 2 θ + δ   ml    

6.12

c > 0 , for which u = aebθ is a solution.

c>0

−1

1 1 = r r cos θ du sin θ = dθ r cos 2 θ

u=

d 2u 1  1 2sin 2 θ = +  dθ 2 r  cosθ cos3 θ

 1 =  r cos θ

 2 − 2 cos 2 θ 1 + cos 2 θ 

 1  2  − 1 =  2   r cosθ  cos θ

d 2u = u ( 2r 2u 2 − 1) = 2r 2u 3 − u 2 dθ Substituting into equation 6.5.10 … 1 2r 2u 3 − u + u = − 2 2 f ( u −1 ) ml u −1 2 2 5 f ( u ) = −2r ml u 2r 2 ml 2 f (r ) = − r5 6.13 From Chapter 1, the transverse component of the acceleration is … aθ = rθ + 2rθ If this term is nonzero, then there must be a transverse force given by … f (θ ) = m(rθ + 2rθ ) r = aθ , and θ = bt For f (θ ) = 2mab 2 ≠ 0

Since

f (θ ) ≠ 0 , the force is not a central field.

68

For r = aθ , and the force to be central, try θ = bt n f (θ ) = m  2ab 2 n 2t 2 n − 2 + ab 2 n ( n − 1) t 2 n − 2  For a central field … f (θ ) = 0 2n + ( n − 1) = 0

n=

1 3 1

θ = bt 3 (a)

6.14

Calculating the potential energy  4 a2  dv − = f ( r ) = −k  3 + 5  dr r  r  2 a2  Thus, V = −k  2 + 4  4r  r The total energy is … 1 1  1  9k  9k  2 E = T + V = v2 − k  2 + 2  =  2  − 2 = 0 2  a 4 a  2  2a  4a Its angular momentum is … 9k l 2 = a 2v 2 = = constant = r 4θ 2 2 Its KE is … 2  2 1  dr  2 2  l 2 1 1  dr  2 T = r 2 + r 2θ 2 =  r +  θ =    +r  4 2 2  dθ  2  dθ    r The energy equation of the orbit is … 2  2  2 a2  1  dr  2 l + − T + V = 0 =  r k   2+ 4  4 2  dθ  4r  r  r

(

)

 dr  2 2  9k  2 a2  r k =  + −   2+ 4  4 4r   dθ  r  4r 2

or

 dr  1 2 2   = (a − r )  dθ  9 r = a cos φ

Letting 2

So

 dφ  1   =  dθ  9

then

dφ dr = −a sin φ dθ dθ

1 ∴φ = θ 3

69

1 r = a cos θ ( r = a @θ = 0 ) 3 3π (b) at θ = r → 0 the origin of the force. To find how long it takes … 2 av v l θ= 2= = 1 1 r a 2 cos 2 θ a cos 2 θ 3 3 1 a dt = cos 2 θ dθ 3 v Thus

T=

3π 2

∫ 0

π

3a 2 3π a a 2 1 cos θ dθ = cos 2 φ dφ = ∫ 3 4v v v 0 1

 9k  2 Since v =  2   2a  1

1

3 2  2  2 π a2  2  2 T = πa   =   4 4 k  9k 

(c) Since the particle falls into the center of the force v→∞ (since l = vr⊥ = const )

6.15

From Example 6.5.4 …

v  2r1  =  vc  r1 + r 

1 2

1 2

   2  v  we have V =  Letting V = vc  1+ r   r1   −2 dV 1   r   r   −2 1 +   −   So: = dr1 2V   r1   r12     dV 1 1 r 1 1 r V = Thus = 2  dr1     r  r1 2  r  r1    2  1 +  1 +   r1    r1    r1  r  1+   r1  

70

 dV     V  ≈ 1 r (b)  dr1  = 2r1  dV  = 2 60 1% = 120%! (a) ( )      dr1  2 r1  r1  r  V     r1  The approximation of a differential has broken down – a correct result can be obtained by calculating finite differences, but the implication is clear – a 1% error in boost causes rocket to miss the moon by a huge factor --- ∼ 2! From section 6.5, ε = 0.967 and r = 55 ×106 mi. 1+ ε From equations 6.5.21a&b, r1 = r 1− ε r 1 55 × 10 6 mi 1 AU a = ( r + r1 ) = = × 2 1− ε 1 − 0.967 93 × 10 6 mi a = 17.92 AU 6.16

3

From equation 6.6.5, τ = ca 2

τ = 1yr ⋅ AU τ = 75.9 yr

−

3 2

3

3

×17.92 2 AU 2

From equation 6.5.21a and 6.5.19 … ml 2 α −1 ε = − 1= r0 kr0

ε=

mr v 2 − 1 and k

k = GMm

1

 GM 2 v = ( ε + 1)   r  From Example 6.5.3 we can translate the factor GM into the more convenient GM = ae ve2 … with ae the radius of a circular orbit and ve the orbital speed … 1 2

1

a v   93 ×106 mi 2 v = 1.967 (ε + 1)  =  ( )  ve 6  55 × 10 mi   r  v0 = 1.824 ve Since l is constant … r1v1 = r v r 1− ε 1 − .967 × 1.824 ve = 0.0306 ve v1 = v = v = 1+ ε 1.967 r1 2 e e

ve ≈

2π ae

τ

=

2π × 93 ×106 mi = 66, 705 mph 1 yr × 365 day ⋅ yr −1 × 24 hr ⋅ day −1

v = 1.22 × 105 mph

and

v1 = 2.04 × 103 mph

71

6.17

From Example 6.10.1 … 1

  2 r v 2 2 ε = 1 +  q 2 −  ( qd sin φ )  where q = and d = d ve ae    are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1). φ is the angle between the comet’s orbital velocity and direction vector towards the Sun (see Figure 6.10.1). The orbit is hyperbolic, parabolic, or elliptic as ε is > , = , or < 1 …

2  i.e., as  q 2 −  is > , = , or < 0. R   2 2 2  q −  is > , = , or < 0 as q d is > , = , or < 2. R  Since l is constant, vmax occurs at r and vmin occurs at r1 , i.e. vmax = v and vmin = v1 and form the constancy of l … v1r1 = v r r vmin vmax = v1v = v 2 r1 k r v 2 = ( ε + 1) (See Example 6.5.4) m 2 k  2π a  = GM =  From equation 6.6.5 …  a m  τ  6.18

 2π a  a ( ε + 1) vmin vmax =   ⋅ r1  τ  2

1+ ε . With 2a = r + r1 : 1− ε a ( ε + 1) 1 ( r + r1 )( ε + 1) 1  r  1  1 − ε = =  + 1 ( ε + 1) =  r1 r1 2 2  r1  2  1 + ε

From equation 6.5.21a&b … r1 = r

vmin vmax

 2π a  =   τ 

   + 1 (ε + 1) = 1  

2

6.19 As a result of the impulse, the speed of the planet instantaneously changes; its orbital radius does not, so there is no change in its potential energy V. The instantaneous change in its total orbital energy E is due to the change in its kinetic energy, T, only, so δv δv 1  = 2T δ E = δ T = δ  mv 2  = mvδ v = mv 2 v v 2  δE δv =2 T v

72

But the total orbital energy is k k δE = 2 δa E=− So 2a 2a Since planetary orbits are nearly circular k k V ~− and T~ a 2a δa δE δa and = Thus, δ E ≅ T T a a δa δv =2 We obtain a v

6.20 (a) V =

1

τ

Vdt τ∫ 0

k r From equation 6.5.4, l = r 2θ dθ l r 2 dθ = 2 or dt = dt r l τ 2π kr ∫0 Vdt = −∫0 l dθ From equation 6.5.18a … a (1 − ε 2 ) r= 1 + ε cos θ ka (1 − ε 2 ) 2π τ dθ ∫0 Vdt = − l ∫0 1 + ε cosθ 2π a 2 From equation 6.6.4 … τ = 1− ε 2 l V (r ) = −

k 1 − ε 2 2π dθ ∫ 0 1 + ε cos θ 2π a 2π dθ 2π k 2 ∫0 1 + ε cosθ = 1 − ε 2 , ε < 1 ∴V = − a

V =−

(b)

This problem is an example of the virial theorem which, for a bounded, periodic τ

system, relates the time average of the quantity

∫∑ p⋅r

i

0

to its kinetic energy T. We will

i

derive it for planetary motion as follows: τ τ τ 1 1 1 p ⋅ rdt = mr ⋅ rdt = ∫ ∫ ∫ F ⋅ rdt

τ

0

τ

0

τ

0

73

Integrate LHS by parts τ τ 1 1 1  mr ⋅ r  τ0 − ∫ mr 2 dt = ∫ F ⋅ rdt τ τ τ 0

0

The first term is zero – since the quantity has the same value at 0 and τ . denote time average of the quantity within brackets. Thus 2 T = − F ⋅ r where but

− r ⋅ F = r ⋅∇V = r

dV k = =− V dr r

hence 2 T = − V but hence and Thus:

V V + V = 2 2 k V =2 E but E=− = constant 2a τ k 1 k so 2E = − E = ∫ Edt = E = − a 2a τ0 k 1 k V = − as before and therefore T = − V = 2 2a a E = T + V =−

The energy of the initial orbit is 1 2 k k mv − = E = − 2 2a r k 2 1 v2 =  −  (1) m r a Since ra = a (1 + ε ) at apogee, the speed v1 , at apogee is 6.21

v12 =

k 2 1  k (1 − ε ) −  =  m  a (1 + ε ) a  ma (1 + ε )

To place satellite in circular orbit, we need to boost its speed to vc such that 1 2 k k mvc − = − since the radius of the orbit is ra 2 2ra ra k k vc2 = = mra ma (1 + ε ) Thus, the boost in speed ∆v1 = vc − v1 1

(2)

1  2  k  ∆v1 =   1 − (1 − ε ) 2    ma (1 + ε )  

Now we solve for the semi-major axis a and the eccentricity ε of the first orbit. From (1) above, at launch v = v at r = RE , so

74

k  2 1 −   m  RE a  and solving for a RE noting that … a= 2 RE 2 − mv k k GM E (3) = = gRE mRE RE RE R a= = E = 4.49 ⋅103 km 2  v  1.426 2−  gRE   v2 =

The eccentricity ε can be found from the angular momentum per unit mass, l, equation 6.5.19, and the data on ellipses defined in figure 6.5.1 … 1

 ka (1 − ε 2 )  2  kα  2  l = r θ = v ( RE sin θ ) =   =  m m   where v , θ are the launch velocity, angle 1 2

Solving for ε (using (3) above) v2  v2  2 ε = 1− 2−  sin θ = 0.795 gRE  gRE  ∴ ε = 0.892 2

Inserting these values for a, ε into (2) and using (3) gives 1

(a) (b) 6.22

1   R a  2   ∆v1 =  gRE  E   1 − (1 − ε ) 2  = 4.61 ⋅103 km ⋅ s −1  1 + ε     3 h = a (1 + ε ) − RE = 2.09 ⋅10 km {altitude above the Earth … at perigee}

 −be − br e −br  e − br − = f ′ ( r ) = −k  2 k  2 r3  r2  r f ′(a) 2  = −b +  f (a) a 

 f ′(a)  From equation 6.14.3, ψ = π 3 + a  f (a)  

ψ=

−

2  b +  r 

1 2

= π 3 − ( ab + 2 ) 

−

1 2

π

1 − ab

75

6.23

From Problem 6.9, f ( r ) = −

GMm 4 − πρ mGr 3 r2

2GMm 4 − πρ mG 3 r3 4 4πρ a 3 −3 − − + 2 πρ 2 GMma mG f ′(a) 3 3M = = 4 f ( a ) −GMma −2 − πρ mGa  4πρ a 3  a 1 +  3 3M  

f ′(r ) =

 f ′(a)  From equation 6.14.3, ψ = π 3 + a  f (a)  

 4πρ a  −2 + 3M ψ = π 3 + 4πρ a 3  1+  3M

3

    

−

1 2

1

 1+ c 2 ψ =π   ,  1 + 4c 

6.24

c=

−

1 2

  4πρ a 3   1 4 +     3M   =π   4πρ a 3  + 1   3M  

−

1 2

4πρ a 3 3M

We differentiate equation 6.11.1b to obtain mr = −

For a circular orbit at r = a , r = 0 so dU r =a = 0 dr For small displacements x from r = a , r=x r = x+a and From Appendix D … x2 f ( x + a ) = f ( a ) + xf ′ ( a ) + f ′′ ( a ) + … 2 dU d 2U Taking f ( r ) to be , f ′(r ) = 2 dr dr Near r = a … dU dU d 2U = + x r =a r =a + … dr dr dr 2 d 2U mx = − x 2 r = a dr

dU ( r ) dr

This represents a “restoring force,” i.e., stable motion, so long as

d 2U > 0 at r = a . dr 2

76

6.25

f ′(r ) =

2k 4ε + r3 r5

From equation 6.13.7, the condition for stability is f ( a ) + k ε a  2k 4ε  − +  +  45 ) 2 2 2 .3015 = sin θ − 2sin θ cosθ + cos θ Using the identity 2sin θ cos θ = sin 2θ sin 2θ = 1 − .30152 = 0.9091 Since θ > 45 , 2θ > 90 : 2θ = sin −1 .9091 = 114.62 θ = 57.3

92

P1’

7.18

Conservation of momentum: P1 = P1′ cos φ + P2′ cos (ψ − φ

φ

ψ

P1

)

0 = P1′ sin φ − P2′ sin (ψ − φ )

From Appendix B for sin (α + β ) and cos (α + β ) :

P2’

P1 = P1′ cos φ + P2′ ( cosψ cos φ + sinψ sin φ ) 0 = P1′ sin φ − P2′ ( sin ψ cos φ − cosψ sin φ )

P12 = P1′2 cos 2 φ + P2′2 ( cos 2 ψ cos 2 φ + 2 cosψ cos φ sin φ sinψ + sin 2 ψ sin 2 φ ) +2 P1′ P2′ ( cos 2 φ cosψ + cos φ sinψ sin φ )

0 = P1′2 sin 2 φ + P2′2 ( sin 2 ψ cos 2 φ − 2sinψ cos φ cosψ sin φ + cos 2 ψ sin 2 φ ) −2 P1′ P2′ ( sin φ sinψ cos φ − cosψ sin 2 φ )

Adding: P12 = P1′2 + P2′2 + 2 P1′ P2′ cosψ Conservation of energy: P12 P1′2 P2′2 = + +Q 2m 2m 2m 1 1 2 P1′ P2′ cosψ Q= P12 − P1′2 − P2′2 ) = ( 2m 2m P′ P ′ cosψ Q= 1 2 m

(

)

1 1 m1v12 T1′ = m1v1′ 2 2 2 2 T ′ v′ let r = 1 = 12 … ratio of scattered particle to incident particle energy T1 v1 Looking at Figure 7.6.2 … v1′ ⋅ v1′ = ( v1′ − vcm ) ⋅ ( v1′ − vcm )

7.19

T1 =

2 v1′ 2 = v1′ 2 + vcm − 2v1′ vcm cos φ1 2 + 2v1′vcmγ hence v1′ 2 = v1′ 2 − vcm

2v′v γ v1′ 2 v − + 1 2cm 2 v1 v v1 v1′ = v1 but scattered particle are the same. ∴r =

where γ = cos φ1

2 cm 2 1

v1′ m2 α = = v1 m2 + m1 1 + α

…the center of mass speeds of the incident and

…from equation 7.6.12 where α =

m2 m1

93

vcm 1 m1 = = Equation 7.6.11 v1 m2 + m1 1 + α Thus 1 α2 1 2γ v1′ α2 1 2γ r= − + r2 = − + 2 2 2 2 α + α 1 1 v + ( ) ( ) α α + α + α 1 1 1 1 + + ( ) ( ) ( ) ( ) 1

Simplifying 2γ 12  1 − α  r− r + =0 1+α  1+α  Let x 2 = r and solving the resulting quadratic for x 1 γ 1 2 2   x= + γ − (1 − α )  2 1+α 1+α  Squaring 1  1  2 2 2 2 2 2γ + α − 1 + 2γ (γ + α − 1) 2  r=x = 2  (1 + α )   1  2  2 2 2 2 2 1 2 1 γ α γ γ α + − + + − ( )     And, after a little algebra, we get the desired solution ∆T1 2 2γ  γ + γ 2 + α 2 − 1 = − 2   T1 1 + α (1 + α )

Now

7.20

1 ∆T1 = 1− r = 1− 2 T1 (1 + α )

From Equation 7.6.15 … γ =

m1v1 m1 v1 = v1′ ( m1 + m2 ) m2 (1 + m1 m2 ) v1′

v1 … v1′

Now we solve for 1

 2T  2 v1 =   and now solving for v1′ starting with Equation 7.6.9 …  m1  m2 1 1 v1 we get … µ v1′2 = µ v12 − Q and using v1′ = m1 + m2 2 2 1 1 T −Q = m1v12 −Q = 2 (1 + m1 m2 ) (1 + m1 m2 )   2 T − Q .  m1 (1 + m1 m2 )  (1 + m1 m2 )  Thus, solving for γ … v1′2 =

94

1

γ=

m1 m2

( 2T m1 ) 2 1 2

( 2 m1 ) (1 + m1

  T − Q m2 )   (1 + m1 m2 )  1 2

1 2

Finally…

γ=

7.21

m1 m2

1 1

 Q (1 + m1 m2 )  2 1 −  T  

The time of flight, τ = constant—so τ = v1 τ  γ + γ 2 + α 2 − 1   1+α As an example, let v1 τ = 1 and we have

r but from problem 7.19 above v1′

r = v1′ τ =

r1 = γ α =1 1 α =2 r2 = γ + γ 2 + 3   3 1 α =4 r3 = γ + γ 2 + 15   5 1 α = 12 r4 = γ + γ 2 + 143   13  Below is a polar plot of these four curves.

7.22

pp scattering p–D p – He p–C

From eqn. 7.7.6, Fu − Fg = mv + vm since v = constant, v = 0 m = λ z = λv , λ = mass per unit length Fg = ( λ z ) g  v2  Fu = λ zg + ( λ v ) v = g λ  z +  g 

Fu is equal to the weight of a length z +

v2 of chain. g

95

7.23

4 m = π r3ρ 3 m = 4π r 2 ρ r ∝ π r 2 z where v = z r = kz k a constant of proportionality r = r + kz From eqn. 7.7.6, mg = mv + vm 4 3 4 π r ρ g = π r 3 ρ z + 4π r 2 ρ ( kz ) z 3 3 2 3kz g = z+ r 3z 2 z=g− r z+ k 3z 2 For r = 0 , z = g − z A series solution is used for this differential equation: ∞

z 2 = ∑ an z n n=0

z=

dz dz dz dz 1 d ( z 2 ) = ⋅ =z = dt dz dt dz 2 dz 2 d (z ) = ∑ an nz n −1 dz n

z2 = ∑ an z n −1 z n 1 ∴ z = ∑ an nz n −1 = g − 3∑ an z n −1 2 n n 1 For n = 1 : a1 = g − 3a1 2 2 a1 = g 7 1 For n ≠ 1 : nan = −3an 2 Since n is an integer, an = 0 for n ≠ 1 2 z2 = g z 7 32  g z = g −  g z = z7  7

96

7.24

From eqn. 7.7.6, mg = mv + vm , where m and v refer to the portion of the chain hanging over the edge of the table. m = λ z and v = z where λ is the mass per unit length of chain m = λ z and v = z

dz dz dz dz 1 d ( z 2 ) = ⋅ =z = dt dz dt dz 2 dz 2  1 d (z )  λ z g = λ z  + z (λ z )  2 dz  z=

1 d (z2 ) z2 =g− z 2 dz Because of the initial condition z = b ≠ 0 , a normal power series solution to this differential equation (…as in Prob. 7.22) does not work. Instead, we use the Method of Frobenius …

z=

∞

z 2 = ∑ an z n + s n=0

2

d (z ) = ∑ an ( n + s ) z n + s −1 dz n z2 = ∑ an z n + s −1 z n 1 z = ∑ an ( n + s ) z n + s −1 = g − ∑ an z n + s −1 2 n n

Equality can be attained for an ≠ 0 at n = 0 and n = 3 n ≠ 0,3 …otherwise an = 0 1 a s = −a 2 s = −2 1 z = ∑ an ( n − 2 ) z n −3 = g − ∑ an z n −3 2 n n 1 For n = 3 , a3 = g − a3 2 2 a3 = g 3 1 For all n ≠ 0 , 3: an ( n − 2 ) = − an 2 an = 0 , n ≠ 0,3 . 2 z 2 = a z −2 + gz 3 For n = 0 ,

97

At t = 0 , z = 0 , and z = b a 2 gb 0= 2 + b 3 2 a = − gb3 3 2 b3 2 z 2 = − g 2 + gz 3 z 3 2  b3  2 g At z = a , z 2 = g  a − 2  = 2 ( a 3 − b3 ) a  3a 3  1

 2g 2 z =  2 ( a 3 − b3 )   3a 

7.25

Initially, the upward buoyancy force balances the weight of the balloon and sand. FB − ( M + m ) g = 0

(1)

Let m = m ( t ) − the mass of sand at time t where 0 ≤ t ≤ t  t (2) m = m 1 −   t  The velocity of sand relative to the balloon is zero upon release so V = 0 in equation 7.7.5 … there is no upward “rocket-thrust.”

As sand is released, the net upward force is the difference between the initial buoyancy force, FB, and the weight of the balloon and remaining sand. Let y be the subsequent displacement of the balloon, so equation 7.7.5 reduces to F = ma dv FB − ( M + m ) g = ( M + m ) dt and using (1) and (2) above we get ( M + m ) gt dv m gt = = −g + dt ( M + m ) t − m t (M + m )t − m t whose solution is: ( M + m ) gt ln 1 − m t  dy v= = − gt −   dt m  (M + m )t 

g m   y = C − ∫  gt + ln (1 − kt )  dt , k= k t (M + m   1 gt tdt = C − gt 2 − ln (1 − kt ) − g ∫ k 2 1 − kt Integrating by parts

)

98

1 gt g  1  = C − gt 2 − ln (1 − kt ) − ∫  −1 +  dt k k  2 1 − kt  1 gt gt g = C − gt 2 − ln (1 − kt ) + + 2 ln (1 − kt ) k k k 2 1 g gt = C − gt 2 + 2 (1 − kt ) ln (1 − kt ) + k k 2 but y = 0 at t = 0 so C = 0 gt 1 g y = − gt 2 + 2 (1 − kt ) ln (1 − kt ) k 2 k and at t = t (a) (b) (c)

 M gt 2  2M + m ) m + 2M ( M + m ) ln  2 ( 2m  M +m (M + m ) − m  gt  v= ( M + m ) ln  m  M  H=

letting ε =

  

m 0 : 3cos θ − 2 > 0 2 θ < cos −1 3 The rocket is constrained from sliding forward for Rx < 0 : 2 θ > cos −1 3 Ry =

8.19

mx = −mg sin θ − µ mg cos θ  x = − g ( sin θ + µ cos θ )

x

G f G mg

ω=

5 µ g cos θ t 2 a

θ

1 2 xt : Since acceleration is constant, x = xDt +  2 gt 2 x = vDt − ( sin θ + µ cosθ ) 2 2 Meanwhile ( µ mg cos θ ) a = Iω = ma 2ω 5 5 µ g cos θ ω = 2 a

116

The ball begins pure rolling when v = aω … 5 µ g cos θ v = vD +  xt = vD − g ( sin θ + µ cos θ ) t = a t 2 a 2vD t= g ( 2sin θ + 7 µ cos θ ) At that time: 2 2vD2 g 4vD ( sin θ + µ cos θ ) − x= g ( 2sin θ + 7 µ cosθ ) 2 g 2 ( 2sin θ + 7 µ cos θ )2 x= 8.20

2vD2 ( sin θ + 6 µ cos θ ) g ( 2sin θ + 7 µ cos θ )2

mx = µ mg  x = µg x = µ gt , and x =

1 µ gt 2 2

2 2 ma ω = − µ mga 5 5 µg ω = − 2 a 5 µg ω = ωD − t 2 a Slipping ceases to occur when v = aω … 5 µ gt = aω D − µ gt 2 2 aω D t= 7 g Iω =

 2 aω D  1 x = µg   2  7 µg  x=

2

2 a 2ω D2 49 µ g

 1   1  Let the moments of inertia of A and B be I a  = M a a 2  and I b  = M bb 2  .  2   2  The angular velocity of A is α while that of B is β − α + φ (remember that in two dimensions, angular velocity is the rate of change of an angle between an line or direction fixed to the body and one fixed in space). For rolling contact, lengths traveled along the perimeters of the disks A and B must be equal to the arc length traveled along the track C. aφ = b β = ( a + 2b )(α − φ ) 8.21

117

so that φ =

( a + 2b )α 2 (a + b)

and β =

a ( a + 2b ) α

2b ( a + b ) After some algebra … the angular velocity of B is found to be … aα ω B = β − α + φ = 2b For A , we take moments about O and for B we take moments about its center. Call TA and TB the components of the reaction forces tangent to A and B (the “upward-going” TA acts on disk B . The “downward-going” TA acts on disk A)

Thus

K − TA a = I Aα

(Torque on Disk A)

α −TAb − TB b = − I B β − α + φ = − I B a 2b 1 TA − TB = M B ( a + b ) α − φ = M B aα 2 Eliminate TA and TB α K = 2 ( 4b 2 I A + M B a 2b 2 + a 2 I B ) 4b Integrating this equation gives : α Kt = 2 ( 4b 2 I A + M B a 2b 2 + a 2 I B ) 4b Putting ω A = α at t = tD gives

(

(

)

(Torque on Disk B)

)

(Force on Disk B)

4b 2 KtD ωA = ( 4b2 I A + M B a 2b2 + a 2 I B ) Putting in values for I A and I B gives 4 KtD ωA = 3   a 2  2M A + M B  2  

Since the angular velocity of B is ω B = β − α + φ =

ωB =

a ωA = 2b

aα , we have 2b

2 KtD 3   ab  2 M A + M B  2  

118

From section 8.7 (see Figure 8.7.1), the instantaneous center of rotation is the l point O . If x is the distance from the center of mass to O and is the distance from 2 the center of mass to the center of percussion O′ , then from eqn. 8.7.10 … Ml 2  1  l 2 l I x   = cm =  = 12  M  12 2 M l x= 6 8.22

8.23 In order that no reaction occurs between the table surface and the ball, the ball must approach and recede from its collision with the cushion by rolling without slipping. Using a prime to denote velocity and rotational velocity after the collision: Pˆ ′ = vcm − vcm m Pˆ ′ = ω cm − ( h − a ) ω cm I ′ = aω cm ′ . The conditions for no reaction are vcm = aω cm and vcm

aω cm −

  Pˆ Pˆ = a ω cm − ( h − a )  m I  

1 1 = a (h − a) m I 2 I = ma 2 5 2ma 2 7 h= +a = a 5ma 5 d a= 2 7 h= d 10 8.24

During the collision, angular momentum about the point 0 is conserved: m′v l ′ = Iθ D

m′vDl ′ I After the collision, energy is conserved: 1 2 l l Iθ − mg − m′gl ′ = − mg cos θ D − m′gl ′ cos θ D 2 2 2 2 2 2 1 m′ vD l ′  ml  = g (1 − cos θ D )  + m′l ′  I 2  2 

θ =

119

I=

ml 2 + m′l ′2 3 1

2 2 1   ml   ml 2 ′ ′ ′ ′ θ 2 1 cos − + + vD = g m l m l ( )   D   m′l ′   2  3 

8.25

The effect of rod BC acting on rod AB is impulse + Pˆ ′ . The effect of AB on BC is − Pˆ ′ . mv1 = Pˆ + Pˆ ′ mv2 = − Pˆ ′ l l Iω1 = Pˆ − Pˆ ′ Iω 2 = − Pˆ ′ 2 2 2 ml I= 12 6 ˆ ˆ 6 ω1 = ω 2 = − Pˆ ′ P − P′ ml ml l l vB = v1 − ω1 vB = v2 + ω 2 2 2 Pˆ + Pˆ ′ l  6  ˆ ˆ −   P − P′ vB = m 2  ml  Pˆ ′ l  6 ˆ  vB = − +  − P′  m 2  ml  Pˆ + Pˆ ′ − 3 Pˆ − Pˆ ′ = − Pˆ ′ − 3Pˆ ′

(

)

(

)

(

(

)

)

8 Pˆ ′ = 2 Pˆ Pˆ Pˆ ′ = 4

5Pˆ 4m 9 Pˆ ω1 = 2ml

Pˆ 4m 3Pˆ ω2 = − 2ml

v1 =

v2 = −

Pˆ m ----------------------------------------------------------------------------------------------------------vB = −

120

CHAPTER 9 MOTION OF RIGID BODIES IN THREE DIMENSIONS 9.1 (a) I xx = ∫ ( y 2 + z 2 ) dm

dm = ρ dxdy and m = 2a 2 ρ

z

I xx = ∫

y=a y =0

∫

x=2 a

x=0

(y

2

+ 0 ) ρ dxdy

a

= ∫ 2a ρ y 2 dy 0

I xx =

2 4 ma 2 ρa = 3 3

y

I yy = ∫ ( x 2 + z 2 ) dm = ∫

2a x

a

=∫

a

0

y=a y =0

∫

x=2 a

x=0

x 2 ρ dxdy

8a 3 ρ dy 3

8a ρ 4ma = 3 3 From the perpendicular axis theorem: 5ma 2 I zz = I xx + I yy + 3 4

I yy =

2

I xy = I yx = − ∫ xydm = − ∫ = −∫

a

0

y =a y =0

∫

x=2a

x =0

xy ρ dxdy

4a 2 ρ ydy 2

ma 2 2 I xz = I zx = − ∫ xzdm = 0 = I yz = I zy I xy = I yx = − ρ a 4 = −

2 1 , cos β = , cos γ = 0 5 5 From equation 9.1.10 … ma 2  4  4ma 2  1   ma 2   2   1  I=   +   + 2 − 3 5 3  5   2   5   5  2 = ma 2 15 ω ˆ ˆ G (c) ω = ω iˆ cos α + ˆj cos β = 2i + j 5 From equation 9.1.29 … (b) cos α =

(

)

(

)

121

G  2ω ma 2 ω  ma 2    2ω  ma 2  ω 4ma 2  ˆ L = iˆ  j ⋅ + − + ⋅     − + 3 2 2 3  5 5 5 5        G ω ma 2 L= iˆ + 2 ˆj 6 5 (d) From equation 9.1.32: 1 G G ω ω ma 2 1 T = ω ⋅L = ⋅ ( 2 + 2 ) = ma 2ω 2 2 15 2 5 6 5

(

)

(a)

9.2

G

ω=

ω ˆ ˆ ˆ i + j+k 3

(

)

m ( 2a ) 2 I xx = 2 I rod = 2 = ma 2 = I yy = I zz 12 3 I xy = − ∫ xydm = 0 since, for each rod, either x or 2

a

-a a

-a a

-a

y or both are 0. The same is true for the other products of inertia. From equation 9.1.29: G 2 ω ˆ ˆ ˆ L = ma 2 ⋅ i + j+k 3 3 From equation 9.1.32, 1 ω 2ma 2ω 2 2 2 ma 2ω 2 T= (1 + 1 + 1 ) = 3 2 3 3 3 (b) From equation 9.1.10, with the moments of

(

)

2ma 2 and the products of inertia equal to 0: 3 2ma 2 2 I= cos 2 α + cos 2 β + cos 2 γ ) = ma 2 ( 3 3 (c) For the x-axis being any axis through the center of the lamina and in the plane of the lamina, and the y-axis also in the plane of the lamina … I xx = I yy due to the similar geometry of the mass inertia equal to

distributions with respect to the x- and y-axes. From the perpendicular axis theorem: I zz = I xx + I yy

I zz = 2 I xx From Table 8.3.1, I zz =

I xx =

ma 2 6

ma 2 12

122

9.3 (a) From equation 9.2.13, tan 2θ =

2 I xy I xx − I yy

ma 2 4ma 2 ma 2 , I yy = , I xy = − 3 3 2 tan 2θ = 1 2θ = 45D θ = 22.5D The 1-axis makes an angle of 22.5D with the x-axis. (b) From symmetry, the principal axes are parallel to the sides of the lamina and perpendicular to the lamina, respectively. From Prob. 9.1, I xx =

9.4 (a) From symmetry, the coordinate axes are principal axes. From Table 8.3.1: z G ω 13 m 2 2 I1 = ( 2a ) + ( 3a )  = ma 2  12 12  10 m 2 I 2 =  a 2 + ( 3a )  = ma 2   12 12 5 m 2 y I 3 =  a 2 + ( 2a )  = ma 2   12 12 ω G ω= ( eˆ1 + 2eˆ2 + 3eˆ3 ) 14 x From equation 9.2.5, 2 2 1 13 2 ω 10 2 4ω 2 5 2 9ω  T =  ma ⋅ + ma ⋅ + ma ⋅  2 12 14 12 14 12 14 

=

7 ma 2ω 2 24

(b) From equation 9.2.4, G ω  2ω  3ω   13  10  5 + eˆ2  ma 2 ⋅ + eˆ3  ma 2 ⋅ L = eˆ1  ma 2 ⋅    14  14  14   12  12  12 G ma 2ω L= (13eˆ1 + 20eˆ2 + 15eˆ3 ) 12 14 G G (1)(13) + ( 2 )( 20 ) + ( 3)(15 )  ω ⋅L cos θ = =  1 1 ω L (12 + 22 + 32 ) 2 ⋅ (132 + 202 + 152 ) 2 =

98

(11,116 )

1 2

= 0.9295

θ = 21.6D 9.5 (a) Select coordinate axes such that the axis of the rod is the 3-axis, its center is at the

123

G origin, and ω lies in the 1, 3 plane. ml 2 From Table 8.3.1, I1 = I 2 = , I3 = 0 12 G ω = ω ( eˆ1 sin α + eˆ3 cos α )

3 G

ω α 2 1 (b)

From equation 9.2.4, G ml 2 ml 2ω ω ( eˆ1 sin α + 0 + 0 ) = eˆ1 sin α L= 12 12 G G ml 2ω sin α L is perpendicular to the rod, and L = 12

G Since ω is constant, from equations. 9.3.5 … N1 = 0 + 0  ml 2  N 2 = 0 + (ω cos α )(ω sin α )    12  N3 = 0 + 0 G ml 2ω 2 sin α cos α N = eˆ2 12 G G N is perpendicular to the rod ( eˆ3 direction) and to L ( eˆ1 direction), and G ml 2ω 2 N = sin α cos α 12

ω G 9.6 From Problem 9.4 … ω = ( eˆ1 + 2eˆ2 + 3eˆ3 ) 14 13 10 5 I1 = ma 2 , I 2 = ma 2 , and I 3 = ma 2 12 12 12 From eqns. 9.3.5: ω2 ma 2 ma 2ω 2 N1 = 0 + ( 2 )( 3) ( 5 − 10 ) = − ( 5) 14 12 28 ω2 ma 2 ma 2ω 2 3 1 13 − 5 = N2 = 0 + ( )( ) ( ) ( 4) 14 12 28 ω2 ma 2 ma 2ω 2 N3 = 0 + (1)( 2 ) (10 − 13) = − 14 12 28 1 2 2 G ma 2ω 2 2 ma ω 5 + 42 + 12 ) 2 = 42 N = ( 28 28 9.7 Multiplying equations 9.3.5 by ω1 , ω 2 , and ω3 ,respectively … 0 = I1ω1ω1 + ω1ω 2ω 3 ( I 3 − I 2 )

0 = I 2ω 2ω 2 + ω1ω 2ω3 ( I1 − I 3 )

124

0 = I 3ω 3ω3 + ω1ω 2ω 3 ( I 2 − I1 )

Adding, 0 = I1ω1ω1 + I 2ω 2ω 2 + I 3ω 3ω3

d 1  I1ω12 + I 2ω 22 + I 3ω 32 )  (  dt  2  d From equation 9.2.5, 0 = [Trot ] dt Trot = constant Multiplying equations 9.3.5 by I1ω1 , I 2ω 2 , and I 3ω 3 , respectively: 0=

0 = I12ω1ω1 + I1ω1ω 2ω 3 ( I 3 − I 2 )

0 = I 22ω 2ω 2 + I 2ω1ω 2ω 3 ( I1 − I 3 ) 0 = I 32ω 3ω 3 + I 3ω1ω 2ω 3 ( I 2 − I1 )

Adding, 0 = I12ω1ω1 + I 22ω 2ω 2 + I 32ω 3ω3 1 d 2 2 0= I1 ω1 + I 22ω 22 + I 32ω32 ) ( 2 dt d From equation 9.2.4, 0 = ( L2 ) dt 2 L = constant 9.8 From equations 9.3.5 for zero torque … 0 = I1ω1 + ω 2ω 3 ( I 3 − I 2 ) 0 = I 2ω 2 + ω1ω3 ( I1 − I 3 )

From the perpendicular axis theorem, I 3 = I1 + I 2 0 = I1ω1 + ω 2ω3 I1 0 = I 2ω 2 − ω1ω 3 I 2 Multiplying by

ω1 I1

and

ω2 I2

, respectively:

0 = ω1ω1 + ω1ω 2ω 3 0 = ω 2ω 2 − ω1ω 2ω 3 Adding, 0 = ω1ω1 + ω 2ω 2 =

1 d ω12 + ω 22 ) ( dt 2

ω12 + ω 22 = constant

If I1 = I 2 , from the third of Euler’s equations … ( I 3ω 3 = 0 )

ω3 = constant 9.9 (a) From symmetry … I s = I 3 and I1 = I 2 = I From the perpendicular axis theorem, I 3 = I1 + I 2 or I s = 2 I

From eqn 9.5.8, Ω = ( 2 − 1) ω cos α

125

For α = 45D , Ω =

ω 2

2π = 2 s = 1.414 s Ω ω G T1 is the period of precession of ω about eˆ3 . From equation 9.6.12,

3

For

G

ω 45o

2π

= 1s , T1 =

 2  I   1   − 1 cos 2 α  = ω 1 +  2 − 1    1 2      I    



φ = ω 1 +  

φ = ω 2

1

1 2

2 s 2

2

2

1 2

5 2

2π T2 =  = 0.4 s = 0.632 s φ G T2 is the period of wobble of eˆ3 about L . I1 = I 2 = I =

(b)

m  2 a 2  17 ma 2 a +  = ⋅ 12  16  16 12

m 2 ma 2 2 2 a + a = ⋅ ( ) 12 12    2  1 15 ω From equation 9.5.8, Ω =  − 1 ω = 2 17 2  17   16  2π 17 = T1 = 2 s = 1.603s Ω 15 I3 = I s =

1

  22 ⋅162   1  2  2 − 1  From equation 9.6.12, φ = ω 1 +    2 17   2     φ = 1.5072 ω 2π 1 T2 =  = s = 0.663s φ 1.5072 9.10

From equation 9.6.10, tan θ =

I tan α . Is

G Since I s > I , θ < α and the angle between the axis of rotation (ω ) and the G invariable line L is α − θ .

( )

From your favorite table of trigonometric identities … tan α − tan θ tan (α − θ ) = 1 + tan α tan θ

126

 I  1 −  tan α Is  tan (α − θ ) =  I 1 + tan 2 α Is

 ( I s − I ) tan α   2  I s + I tan α 

α − θ = tan −1 

Is a maximum. I I tan α For s = 2 , tan (α − θ ) = 2 + tan 2 α I d tan (α − θ ) 1 2 tan 2 α 2 − tan 2 α = − = 2 + tan 2 α ( 2 + tan 2 α )2 ( 2 + tan 2 α )2 d tan α

9.11

α − θ is a maximum for

d tan (α − θ ) = 0 = 2 − tan 2 α d tan α −1 α = tan 2 = 54.7D  1 1 −  2 2 2 = tan (α − θ ) ≤  2 1 4 1+ 2 2  2 D α − θ ≤ tan −1   = 19.5 4  

At the maximum,

( )

G G 9.12 (a) From Problem 9.10, for I s > I , the angle between ω and L is …  ( I s − I ) tan α   2  I s + I tan α 

α − θ = tan −1 

From Problem 9.9(a), I s = 2 I and tan α = tan 45D = 1 1 1 α − θ = tan −1 = tan −1 = 18.4D 3 ( 2 + 1) ma 2 17 ma 2 and I = ⋅ 12 16 12   17     2 − 16     = tan −1 15 = 17.0D −1  α − θ = tan  49  2 + 17   16 

(b) From Prob. 9.9(b), I s = 2 ⋅

127

Is = 1.00327 I G G It follows from equation 9.6.10 that, for I s > I , the angle between ω and L is: From Example 9.6.2, α = 0.2′′ and

9.13

I  tan α   Is 

α − θ = α − tan −1  For α so small,

I  I Iα tan α ≈ α , and tan −1  tan α  ≈ tan α ≈ Is  Is  Is Iα I s − I α −θ ≈ α − = α Is Is .00327 0.2 = 0.00065 arcsec α −θ ≈ 1.00327

From Table 8.3.1, I s =

9.14

3 G

ω0 G

ω

1

ma 2 ma 2 and I = 2 4 G maω G ˆ ωD = ω e3 and Pˆ = eˆ3 4 Selecting the 2-axis through the point of impact, during the collision … G maω ma 2ω G G ∫ Ndt = r × pˆ = aeˆ2 × 4 eˆ3 = 4 eˆ1 G The only non-zero component of N is N1 . 2 During the collision ω 2 = 0 , so integrating G P the first of Euler’s equations 9.3.5 … ∫ N1dt = I1ω1

Immediately after the collision, ma ω  4    =ω 4  ma 2  ( ω3 still is equal to ω , and ω 2 = 0 ) G G After the collision, N = 0 . From Prob. 9.8, with I1 = I 2 , for N = 0 ,

ω1 =

2

ω3 = constant and ω12 + ω 22 = constant .

(ω tan α =

2 1

1

+ ω 22 ) 2

ω3

= constant

Using the values of ωi immediately after the collision …

(ω tan α =

2

1

+ 0) 2

ω

=1

α = 45D

128

G G From equation 9.5.8, with ω = ω D + ω1eˆ1 = ω ( eˆ1 + eˆ3 ) :

Ω = ( 2 − 1)

(

)

2ω ( cos 45D ) = ω

From equation 9.6.12 … 1

φ = 2ω 1 + ( 22 − 1)( cos 2 45D )  2 = ω 5 9.15

G G N = −cω = −c (ω1eˆ1 + ω 2 eˆ2 + ω 3eˆ3 )

Say the 3 axis is the symmetry axis, I 3 = I s , I1 = I 2 = I For the third component of equations 9.3.5 … −cω3 = I sω 3 + 0 3 G ω ω 3 c =− α ω3 Is ω c ln 3 = − t Is (ω 3 )D 2

ct Is

For the first two components of equations 9.3.5 … −cω1 = Iω1 + ω 2ω 3 ( I s − I ) , and

G N

1

ω 3 = (ω 3 )D e

−

−cω 2 = Iω 2 + ω1ω 3 ( I − I s )

Rearranging terms and multiplying by ω1 and ω 2 respectively … c ω1ω1 + ω12 + ω1ω 2ω 3 ( I s − I ) = 0 I c ω 2ω 2 + ω 22 − ω1ω 2ω 3 ( I s − I ) = 0 I c Adding, ω1ω1 + ω 2ω 2 + (ω12 + ω 22 ) = 0 I 1 d 2c ω12 + ω 22 ) = − ( 2 2 I (ω1 + ω 2 ) dt ln

ω12 + ω 22 2c =− t 2 2 (ω1 + ω 2 )D I

(ω

2 1

+ ω 22 ) = (ω12 + ω 22 ) e

(ω12 + ω

−

2 ct I

D

1 2 2 2

)

(ω tan α =

2 1

1

ct

= (ω12 + ω 22 ) 2 e I D

+ω

ω3

1 2 2 2

)

= ( tan α D ) e

1 1  − ct  −   I Is 

129

1 1 For I s > I ,  −  I Is

  > 0 and α decreases with time. 

ma 2 9.16 (a) From table 8.3.1… I s = about the symmetry (spin) axis. 2 a2 G I = m about an axis through cm disk S 4 z the center of mass of the disk and ⊥ a to the spin axis. From the parallel axis theorem, equation 8.3.21, … 2

ma 2 a I disk = + m   about the pivot 4 2 point. 2 ma moment of inertia of I rod =   2 3 the rod about the pivot point. Thus …

o

a/2

45

y x

I = I disc + I rod

2 2  ma 2 2 a  ma = + m    +   = ma 2  2    2  3 3  4

From equation 9.7.10, φ =

I s S + ( I S − 4 MglI cos θ ) 2 s

2

1 2

2 I cos θ

1   2 2   I I cos Mgl 1 4 θ     2 s s φ =    S ±   S − I 2 cos θ  I  I      2 ma Is 1 3 = 2 2 = , cosθ = cos 45D = 2ma I 4 2 3 m  a   m +    9 Ml  2  2  9 = = = cm −1 2 2ma 8a 80 I 3 1  2 2 2   1 3 3 9 cm 1 60 rpm 2         −1   φ =  ( 900 rpm ) ±   ( 900 ) − 4  cm   980 2     −1    s   2   2π s    2 4  80   4     φ = 15.1rpm or 939 rpm

130

(b) From equation 9.7.12, I s2 S 2 ≥ 4MglI  3m   a  I 4 Ml  2   2  3 3 = , = = = cm −1 2 ma Is 3 Is 2a 20 2 4Mgl I  3   4   60 rpm  S ≥ ⋅ = 4  cm −1  ( 980 cm ⋅ s −2 )    −1  Is Is  20   3   2π s  S ≥ 267 rpm

2

2

9.17

(Neglecting the pencil head) From Table 8.3.1, 2 b m  mb 2 2  = Is = 2 8 The moment of inertia of the pencil about an axis thru its center of mass and ⊥ to its symmetry axis from equation  b     2  b2 a 2  a 2 8.3.26 ... I c = m    +  = m  +   4 12   16 12      2

b

a

2

 b2 a 2  a From the parallel axis theorem, I = I c + m   = m  +  2  16 3  4MglI From equation 9.7.12, S 2 ≥ I s2 2 2 a b a  4mg   m  +   2   16 3  S2 ≥ m 2b 4 64 1

1

16  ga  b 2 a 2   2 16  ( 980 )( 20 )  12 202   2 −1 S ≥ 2   +  = 2   +   rad ⋅ s b  2  16 3   3  (1)  2  16 S ≥ 18, 294 rad ⋅ s −1 = 2910 rps

9.18

I s = I rim + I spokes + I hub ma 2 2 3 a ma 2 I spokes = mspokes = , assuming the spokes to be thin rods 3 12 I hub = 0 , assuming its radius is small I rim = mrim a 2 =

131

From the perpendicular axis theorem, equation 8.3.14, I = From equation 9.10.14, S 2 >

Is 2

Imga I s ( I s + ma 2 )

1

 2 1   1 6 mga g 2  = S>   2  2  7 ma    19 a  2    + ma    12  

For rolling without slipping, v = aS 1

  30   2 1 6 × 32 ×      6 ga  2  2   ft ⋅ s −1 = 3.55 ft ⋅ s −1 v>  =  19   19 ×12    ma 2 If the spokes and hub are neglected, I s = 2 1

 2 1   2 1 mga g    = S>   2  2  ma    3a  2    + ma    2   1

  30   2 1 32 ×    ga  2   2   ft ⋅ s −1 = 3.65 ft ⋅ s −1  v>  =  3   3 × 12    G 9.19 From equations 9.3.5 with N = 0 … I1ω1 + ( I 3 − I 2 ) ω 2ω 3 = 0 I 2ω 2 + ( I1 − I 3 ) ω1ω 3 = 0 I 3ω 3 + ( I 2 − I1 ) ω1ω 2 = 0

Differentiating the first equation with respect to t: I1ω1 + ( I 3 − I 2 )(ω 2ω 3 + ω 2ω3 ) = 0 From the second and third equations: (I − I ) (I − I ) ω 2 = 3 1 ω1ω 3 , and ω 3 = 1 2 ω1ω 2 I2 I3

132

(I − I )  (I − I ) I1ω1 + ( I 3 − I 2 )  1 2 ω1ω 22 + 3 1 ω1ω 32  = 0 I2  I3   ( I − I )( I − I )   ( I − I )( I − I )  ω1 + K1ω1 = 0 , K1 = −ω 22  3 2 2 1  + ω 32  3 2 3 1  I1 I 3 I1 I 2    

(a) For ω3 large and ω 2 = 0 , K1 > 0 so ω1 + K1ω1 = 0 is the harmonic oscillator equation. ω1 oscillates, but remains small. Motion is stable for initial rotation about the 3 axis if the 3 axis is the principal axis having the largest or smallest moment of inertia. (b) For ω3 = 0 and ω 2 large, K1 < 0 so ω1 + K1ω1 = 0 is the differential equation for kt − kt exponential growth of ω1 with time: ω1 = Ae 1 + Be 1 . Motion is unstable for the initial rotation mostly about the principal axis having the median moment of inertia.

9.20 I xy = ∑ mi xi yi = 0 since either xi or yi is zero for all six particles. Similarly, all the i

other products of inertia are zero. Therefore the coordinate axes are principle axes. 2 2 I xx = ∑ mi ( yi2 + zi2 ) = m 0 + 0 + b 2 + ( −b ) + c 2 + ( −c )    i

(0,0,c)

I xx = 2m ( b 2 + c 2 )

(0,b,0)

I yy = 2m ( a 2 + c 2 ) I zz = 2m ( a 2 + b 2 )

(a,0,0)

b 2 + c 2 I  I = 2m  0  0 

ω

G

ω=

(a + b + c 2

2

1 2 2

)

( aeˆ1 + beˆ2 + ceˆ3 ) =

G IG From equation 9.1.28, L = I ω b 2 + c 2 G 2mω  0 L= 1  2 2 2 2 ( a + b + c )  0

0 a +c 0 2

0 a2 + c2 0

ω 1

( a 2 + b2 + c2 ) 2

  0  a 2 + b 2  0

a  b     c 

 a    0  b  2 a + b 2   c  0

2

133

 a ( b2 + c 2 )   G 2mω 2 2   a c L= b + ( ) 1 2 2 2 2  ( a + b + c ) c ( a 2 + b2 )   G 1 G From equation 9.1.32, T = ω ⋅ L 2 a ( b2 + c 2 )   1 2mω 2 b ( a 2 + c 2 )  T= a b c [ ] 2 ( a 2 + b2 + c2 )   c ( a 2 + b 2 )    2 mω  a 2 ( b2 + c 2 ) + b2 ( a 2 + c 2 ) + c 2 ( a 2 + b2 ) T= 2 2 2   a +b +c 2mω 2 T= 2 a 2b 2 + a 2 c 2 + b 2 c 2 ) 2 2 ( a +b +c 9.21 For Problem 9.1:  1  3  I 1 2 I = ma −  2   0  

1 2 4 3

−

0

 0  0,   5  3

 2

ω   1 ω= 5  G

0  

1  1   3 − 2 0   2 G I G ma 2ω  1 4   L = Iω = 0 1 −   2 3 5    0  5  0  0   3  2 1 1  3−2  6 1     2 2 ma ω  4  ma ω  1  ma 2ω   2 = −1 + = = 3 5  5  3  6 5       0  0   0      1 1 G G 1 ω ma 2ω T = ω ⋅L = ( 2 1 0 )  2  2 2 5 6 5 0  

=

ma 2ω 2 ma 2ω 2 2 + 2 + 0 = ( ) 60 15

134

For Problem 9.4:  13 0 0  1 I ma 2  ω   G  I = 0 10 0  , 2 ω= 12  14     0 0 5  3 13 0 0   1   13  2 G I G ma 2ω     ma ω   L = Iω = 0 10 0   2  = 20 12 14  12 14       0 0 5 3  15   13  1 G G 1 ω ma 2ω T = ω ⋅L = (1 2 3)  20  2 2 14 12 14  15    =

ma 2ω 2 7 (13 + 40 + 45) = ma 2ω 2 24 (14 ) 24

9.22 Since the coordinate axes are axes of symmetry, they are principal axes and all products of inertia are zero. G From Table 8.3.1, z ω m m 2 2 I xx = ( 2b ) + ( 2c )  = ( b 2 + c 2 )  3 12  m m I yy = ( a 2 + c 2 ) , I zz = ( a 2 + b 2 ) 3 3 2 2 y b + c 0 0  2c I m  0  I =  0 a2 + c2 3 2a  0 0 a 2 + b 2   x 2b a  ω ω G b  ω= aeˆ1 + beˆ2 + ceˆ3 ) = 2 1 ( 2 2   (a + b + c ) c  ( a 2 + b2 + c 2 ) 2   G IG From equation 9.1.28, L = I ω b 2 + c 2 0 0  a  G mω   2 2 0 0   b  L= a +c 1  3 ( a 2 + b 2 + c 2 ) 2  0 0 a 2 + b 2   c 

a ( b2 + c 2 )   G mω 2 2  L= b ( a + c ) 1   3 ( a 2 + b2 + c2 ) 2  2 2  c ( a + b )  1 G G From equation 9.1.32, T = ω T ⋅ L 2

135

a ( b2 + c 2 )   1 mω 2 b ( a 2 + c 2 )  T= a b c [ ] 2 3 ( a 2 + b2 + c2 )   c ( a 2 + b 2 )    T=

mω 2  a 2 ( b 2 + c 2 ) + b 2 ( a 2 + c 2 ) + c 2 ( a 2 + b 2 ) 2 2 2   6(a + b + c )

T=

mω 2 a 2b 2 + a 2 c 2 + b 2 c 2 ) 2 2 2 ( 3( a + b + c )

With the origin at one corner, from the parallel axis theorem: m ( b2 + c 2 ) 4m 2 2 I xx = + m (b2 + c2 ) = (b + c ) 3 3 4m 2 2 4m 2 I yy = a +c ), I zz = ( ( a + b2 ) 3 3 I xy = − ∫ xydm = − ∫ xy ρ dV I xy = − ρ ∫

x=2a

x =0

∫

y =2b y =0

∫

z =2c

z =0

xydxdydz = −8 ρ a 2b 2 c

m = ρ ( 2a )( 2b )( 2c ) = 8 ρ abc , so I xy = − mab

I xy = − mac ,

I yz = − mbc

4 2 2  3 (b + c )  I I = m  −ab    − ac 

−ab 4 2 2 (a + c ) 3 −bc

9.23 (See Figure 9.7.1) Lz = ( Lx′ ) z + ( Ly′ ) z + ( Lz ′ ) z

(

   −bc    4 2 2  + a b ( ) 3 − ac

)

Lz = Ly′ sin θ + Lz′ cos θ = Iφ sin θ sin θ + ( I s S ) cos θ

9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ = θ D where V (θ D ) is a minimum of V (θ ) …

dV (θ ) =0 dθ θ =θ D

( L − Lz′ cosθ ) V (θ ) = z 2 2 I sin θ

2

+ mgl cos θ

(See equation 9.8.7)

136

dV dθ

− cosθ D ( Lz − Lz′ cos θ D ) + Lz′ sin 2 θ D ( Lz − Lz ′ cosθ D ) = − mgl sin θ D = 0 I sin 3 θ D 2

θ =θ D

let γ = Lz − Lz′ cos θ D Then γ 2 cos θ D − γ Lz ′ sin 2 θ D + mglI sin 4 θ D = 0 and solving for γ 1     Lz ′ sin θ D  4mglI cos θ D 2  γ= 1 ± 1 −   2 cos θ D   L2z′     now Lz ′ is large since ψ is large and the precession rate is small, so we can expand the term in square root above and use the (-) solution since γ must be positive … 2

 2mglI cos θ D  1 − 1 +  L2z′   2 mglI sin θ D γ = Lz − Lz′ cosθ D ≈ Lz′ From equations 9.7.2, 9.7.5 and 9.7.7 … Lz = Iφ 2 sin 2 θ 0 + I s φ cos θ 0 + ψ cosθ 0 L cos θ = I (φ cosθ + ψ ) cosθ

γ≈

Lz ′ sin 2 θ D 2 cos θ D

(

z′

0

s

)

0

and … so …

0

γ = ( I sin 2 θ D + I s cos 2 θ D ) φ + I s ψ cosθ D − I s φ cos 2 θ D − I s ψ cosθ D = Iφ sin 2 θ D ≈

mglI sin 2 θ D mglI sin 2 θ D = Lz′ I s ψ + φ cos θ D

(

)

and since ψ >> 0 , we can ignore

the φ term in the denominator and we have … mgl φ ≈ I sψ mgl = 0 and φ = the top will precess without nutation Hence, if ψ large, θ θ =θ1 θ =θ1 I sψ at θ1 = θ D the place where V (θ ) = min .

9.25

a b1 + b2 = 3   2 1 b1 = sin 30D = b2 2 a a b1 = b2 = 2 3 3 a 2 = H 2 + b22 a 2 2a 2 H = a −b = a − = 3 3 2

2

2 2

2

137

2 a 3 Thus, the coordinates of the 4 atoms are:

H=

( 0, 0, H )

Oxygen:

a   a    −b1 , , 0  ;  −b1 , − , 0  2   2   ( b2 , 0, 0 )

Hydrogen: Carbon:

(a) The axes 1, 2, 3 are principal axes if the products of inertia are zero. a    −a   − I xy = ∑ mi xi yi = 0 +  −b1  +  −b1   + 0 ≡ 0 2    2  − I yz = 0 + 0 + 0 + 0 ≡ 0 − I xz = 0 + 0 + 0 + 0 ≡ 0

The 1,2,3 axes are principal axes.

(b) Find principal moments 2

2

a  a  67 2 I1 = I xx = ∑ mi ( y + z ) = 16 H + 1  + 1  = a 6 2 2 89 I 2 = I yy = ∑ mi ( xi2 + zi2 ) = 16 H 2 + 1( b12 ) + 1( b12 ) + 12b22 = a 2 6 2 2  14 a   a  I 3 = I zz = ∑ mi ( xi2 + yi2 ) = 0 +  b12 +    +  b12 +    + b22 = a 2     3 2   2   2 i

 67  6  I = 0    0  

2 i

2

 0  89 0  a2  6  14  0  3 0

(c) I 3 < I1 < I 2 therefore rotation about the1-axis is unstable (see discussion Sec. 9.4) ----------------------------------------------------------------------------------------------------------

138

CHAPTER 10 LAGRANGIAN MECHANICS 10.1

x ( t ) = x ( 0, t ) + αη ( t )

Solution …

x ( t ) = x ( 0, t ) + αη ( t )

where x ( 0, t ) = sin ω t and x ( 0, t ) = ω cos ω t

1 2 1 1 mx V = kx 2 = mω 2 x 2 2 2 2 t2 m J (α ) = ∫ ( x 2 − ω 2 x 2 ) dt 2 t1

T= so:

t

=

m 2 2 2 (ω cos ω t + αη ) − ω 2 ( sin ω t + αη )  dt ∫  2 t1 t

t

t

2 m2 α 2m 2 2 J (α ) = ∫ ω 2 ( cos 2 ω t − sin 2 ω t ) dt + mαω ∫ (η cos ω t − ωη sin ω t ) dt + (η − ω 2η 2 ) dt ∫ 2 t1 2 t1 t1

Examine the term linear in α : t2

∫ (η cos ω t − ωη sin ω t ) dt = η ( t ) cos ω t t1

t2 t1

t2

t2

t1

t1

+ ∫ ωη sin ω tdt − ∫ ωη sin ω tdt ≡ 0

(1 term vanishes at both endpoints: η ( t2 ) = η ( t1 ) = 0 ) st

t

so

t

2 2 1 1 J (α ) = mω 2 ∫ cos 2ω tdt + mα 2 ∫ (η 2 − ω 2η 2 ) dt 2 2 t1 t1

t

2 1 1 = mω [sin 2ω t2 − sin 2ω t1 ] + mα 2 ∫ (η 2 − ω 2η 2 ) dt 4 2 t1

which is a minimum at α = 0 10.2

V = mgz 1 T = m ( x 2 + y 2 + z 2 ) 2 1 L = T − V = m ( x 2 + y 2 + z 2 ) − mgz 2 ∂L ∂L ∂L = mx , = my , = mz ∂x ∂y ∂z d  ∂L    = mx , dt  ∂x 

d  ∂L    = my , dt  ∂y 

d  ∂L    = mz dt  ∂z 

139

∂L ∂L = = 0, ∂x ∂y

∂L = −mg ∂z ∂L d  ∂L  −  =0 ∂qi dt  ∂qi  mx = const my = const

From equations 10.4.5 … mx = 0 , my = 0 , mz = − mg

10.3 Choosing generalized coordinate x as linear displacement down the inclined plane (See Figure 8.6.1), for rolling without slipping … x ω= a 2 1 1 1 12 7   x  T = mx 2 + Iω 2 = mx 2 +  ma 2    = mx 2 2 2 2 25   a  10 For V = 0 at the initial position of the sphere, V = −mgx sin θ 7 L = T − V = mx 2 + mgx sin θ 10 d  ∂L  7 ∂L 7 = mx ,   = mx dt  ∂x  5 ∂x 5 ∂L = mg sin θ ∂x 7 mx = mg sin θ 5 5  x = g sin θ 7 5 xcm = g sin θ From equations 8.6.11 - 8.6.13 …  7 10.4 (a) For x the distance of the hanging block below the edge of the table: 1 1 T = mx 2 + mx 2 = mx 2 and V = −mgx 2 2 L = T − V = mx 2 + mgx d  ∂L  ∂L = 2mx ,    = 2mx  ∂ x dt  ∂x  x ∂L = mg ∂x 2mx = mg g  x= 2

140

1 m′  m′g 2  x  x (b) T = mx 2 + m′x 2 =  m +  x 2 and V = −mgx −  m′  g = −mgx − x 2 2  2l  l  2 m′  m′g 2  L = T − V =  m +  x 2 + mgx + x 2  2l  d  ∂L  ∂L x = ( 2m + m′ ) x ,   = ( 2m + m′ )  dt  ∂x  ∂x m′g ∂L x = mg + l ∂x m′g x ( 2m + m′) x = mg + l g  ml + m′x   x=   l  2m + m′  The four masses have positions: m1 : x1 + x2 m2 : l1 − x1 + x2 m3 : l2 − x2 + x3 m4 : l2 − x2 + l3 − x3

10.5

V = − g  m1 ( x1 + x2 ) + m2 ( l1 − x1 + x2 ) + m3 ( l2 − x2 + x3 ) + m4 ( l2 − x2 + l3 − x3 )  1 2 2 m1 ( x1 + x2 ) + m2 ( − x1 + x2 ) 2 2 2 + m3 ( − x2 + x3 ) + m4 ( − x2 − x3 )  

T=

L = T −V =

1 1 1 2 2 2 m1 ( x1 + x2 ) + m2 ( − x1 + x2 ) + m3 ( − x2 + x3 ) 2 2 2

1 2 + m4 ( − x2 − x3 ) + gx1 ( m1 − m2 ) + gx2 ( m1 + m2 − m3 − m4 ) + gx3 ( m3 − m4 ) + const. 2 ∂L = m1 ( x1 + x2 ) − m2 ( − x1 + x2 ) ∂x1 = ( m1 + m2 ) x1 + ( m1 − m2 ) x2

d ∂L = ( m1 + m2 )  x1 + ( m1 − m2 )  x2 dt ∂x1 ∂L = g ( m1 − m2 ) ∂x1

( m1 + m2 ) x1 + ( m1 − m2 ) x2 = g ( m1 − m2 )

141

∂L = m1 ( x1 + x2 ) + m2 ( − x1 + x2 ) − m3 ( − x2 + x3 ) − m4 ( − x2 − x3 ) ∂x2 ∂L = g ( m1 + m2 − m3 − m4 ) ∂x2 ( m1 − m2 ) x1 + ( m1 + m2 + m3 + m4 ) x2 + ( m4 − m3 ) x3 = g ( m1 + m2 − m3 − m4 ) ∂L = m3 ( − x2 + x3 ) − m4 ( − x2 − x3 ) ∂x3 ∂L = g ( m3 − m4 ) ∂x3

( m4 − m3 ) x2 + ( m3 + m4 ) x3 = g ( m3 − m4 ) m1 = m , m2 = 4m , m3 = 2m , and m4 = m : 3  5mx1 − 3mx2 = −3mg , x1 = (  x2 − g ) 5 −3mx1 + 8mx2 − mx3 = 2mg 1  − mx2 + 3mx3 = mg , x3 = (  x2 + g ) 3 Substituting into the second equation: 9 9 1 1 −  x2 + g + 8 x2 −  x2 − g = 2 g 5 5 3 3 88 8 g   x2 = x2 = g , 15 15 11 3  10  6  x1 =  − g  = − g 5  11  11 1  12  4  x3 =  g  = g 3  11  11

For

Accelerations: m1 : m2 :

m3 : m4 :

5 g 11 7 −  x1 +  x2 = g 11 3 −  x2 +  x3 = g 11 5 −  x2 −  x3 = − g 11  x1 +  x2 = −

10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the ball is calculated in the same way as for the block in Example 10.5.6.   ′ cos θ v 2 = x 2 + x ′2 + 2 xx The ball also rotates with angular velocity ω so …

142

T=

1 2 1 2 1 mv + Iω + Mx 2 2 2 2

x ′ 2 . I = ma 2 5 a 1 1 1   ′ cosθ ) + mx ′2 + Mx 2 T = m ( x 2 + x ′2 + 2 xx 2 5 2 ′ V = −mgx sin θ , for V = 0 at the initial position of the ball. 1  7  1   ′ cos θ  + Mx 2 + mgx′ sin θ L = T − V = m  x 2 + x ′2 + 2 xx 2  5  2 d  ∂L  7 ∂L 7 = mx ′ + mx cos θ ,   = mx′ + mx cos θ ∂x′ 5 dt  ∂x ′  5 ∂L = mg sin θ ∂x′ 7 mx′ + mx cos θ = mg sin θ 5 5  x′ = ( g sin θ −  x cos θ ) 7 d  ∂L  ∂L = mx + mx ′ cos θ + Mx , x + mx′ cos θ   = ( m + M )  ∂x dt  ∂x  ∂L =0 ∂x ( m + M ) x + mx′ cosθ = 0

For rolling without slipping, ω =

5 5 mg sin θ cos θ − mx cos 2 θ = 0 7 7 5mg sin θ cos θ  x= 5m cos 2 θ − 7 ( m + M )

( m + M ) x +

10.7 Let x be the slant height of the particle … v 2 = x 2 + x 2ω 2 1 1 T = mv 2 = m ( x 2 + x 2ω 2 ) 2 2 x V = mgx sin θ = mgx sin ω t 1 L = T − V = m ( x + x 2ω 2 ) − mgx sin ω t θ = ωt 2 d  ∂L  ∂L = mx ,   = mx dt  ∂x  ∂x ∂L = mxω 2 − mg sin ω t ∂x mx = mxω 2 − mg sin ω t  x − ω 2 x = − g sin ω t

143

The solution to the homogeneous equation  x − ω 2 x = 0 , is x = Aeω t + Be −ω t Assuming a particular solution to have the form x p = C sin ω t ,

−ω 2C sin ω t − ω 2C sin ω t = − g sin ω t g C= 2ω 2 g x = Aeω t + Be −ω t + sin ω t 2ω 2 At time t = 0 , x = xD and x = 0 xD = A + B g 0 = ω A −ωB + 2ω 1 g  A =  xD − 2  2 2ω  1 g  B =  xD +  2 2ω 2  g 1 g 1   x = xD  ( eω t + e −ω t )  − 2  ( eω t − e −ω t )  + 2 2  2ω  2  2ω sin ω t From Appendix B, we use the identities for hyperbolic sine and cosine to obtain g g x = xD cosh ω t − 2 sinh ω t + sin ω t 2ω 2ω 2 10.8 In order that a particle continues to move in a plane in a rotating coordinate system, it is necessary that the axis of rotation be perpendicular to the plane of motion. G For motion in the xy plane, ω = kˆω . G G G G ˆ + ˆjy + kˆω × ix ˆ + ˆjy v = v′ + ω × r ′ = ix G v = iˆ ( x − ω y ) + ˆj ( y + ω x )

(

)

(

)

1 G G m 2 mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 ) 2 2 L = T −V d  ∂L  ∂L = m ( x − ω y ) , x − ω y ) ,   = m (  ∂x dt  ∂x  ∂L ∂V = m ( yω + ω 2 x ) − ∂x ∂x ∂V m (  x − ω y ) = m ( yω + ω 2 x ) − ∂x 2 Fx = m (  x − 2ω y − ω x )

T=

∂L = m ( y + ω x ) , ∂y

d  ∂L  y + ω x ) ,   = m (  dt  ∂y 

144

∂L ∂V = m ( − xω + ω 2 y ) − ∂y ∂y m (  y + ω x ) = m ( − xω + ω 2 y ) − Fy = m (  y + 2ω x − ω 2 y )

∂V ∂y

G G For comparison, from equation 5.3.2 ( A0 = 0 and ω = 0 ) G G G G G G G F = ma′ + 2mω × v′ + mω × (ω × r ′ ) G ˆ + ˆjy + 2m ω kˆ × ix ˆ + ˆjy + m ω kˆ × ω kˆ × ix ˆ + ˆjy  F = m ix   2 Fx = m (  x − 2ω y − ω x )

(

)

(

)

(

)

Fy = m (  y + 2ω x − ω 2 y )

10.9 Choosing the axis of rotation as the z axis … G G G G ˆ + ω kˆ × ix ˆ ˆ + ˆjy + kz ˆ + ˆjy + kz v = v′ + ω × r ′ = ix G ˆ v = iˆ ( x − ω y ) + ˆj ( y + ω x ) + kz

(

)

(

)

1 G G m 2 mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 + z 2 ) 2 2 L = T −V d  ∂L  ∂L = m ( x − ω y ) x − ω y ) ,   = m (  ∂x dt  ∂x  ∂L ∂V = m ( yω + ω 2 x ) − ∂x ∂x ∂V m (  x − ω y ) = m ( yω + ω 2 x ) − ∂x 2 Fx = m (  x − 2ω y − ω x )

T=

∂L d  ∂L  = m ( y + ω x ) , y + ω x )   = m (  ∂y dt  ∂y  ∂L ∂V = m ( − xω + ω 2 y ) − ∂y ∂y ∂V m (  y + ω x ) = m ( − xω + ω 2 y ) − ∂y Fy = m (  y + 2ω x − ω 2 y )

d  ∂L  ∂L ∂L ∂V = mz , = mz , =−   dt  ∂z  ∂z ∂z ∂z ∂V mz = − = Fz ∂z G G From equation 5.3.2 ( A0 = 0 and ω = 0 ) …

145

G G G G G G G F = ma′ + 2mω × v′ + mω × (ω × r ′ ) G ˆ + 2mω kˆ × ix ˆ + mω kˆ × ω kˆ × ix ˆ  ˆ + ˆjy + kz ˆ + ˆjy + kz ˆ + ˆjy + kz F = m ix   2 Fx = m (  x − 2ω y − ω x )

(

)

(

)

(

)

Fy = m (  y − 2ω x − ω 2 y ) Fz = m z

10.10

1 m r 2 + r 2θ 2 2 1 2 V = k ( r − lD ) − mgr cos θ 2 m 2 2 2 k 2 L = T −V = r + r θ − ( r − lD ) + mgr cos θ 2 2 d  ∂L  ∂L = mr ,   = mr dt  ∂r  ∂r ∂L = mrθ 2 − k ( r − lD ) + mg cos θ ∂r

(

T=

r

)

(

θ

mr = mr 2 − k ( r − lD ) + mg cos θ

)

∂L ∂L = mr 2θ = −mgr sin θ  ∂θ ∂θ d mr 2θ = − mgr sin θ dt

(

)

a a ( 2θ + sin 2θ ) y = (1 − cos 2θ ) 4 4 at θ = 0 x& y = 0 1 T = m ( x 2 + y 2 ) V = mgy 2 a aθ x = θ + θ cos 2θ = (1 + cos 2θ ) 2 2 a y = θ sin 2θ 2 2 2 ma θ  mga 2 L = T −V = (1 + cos 2θ ) + sin 2 2θ  − (1 − cos 2θ )  8 4 ma 2θ 2 mga = (1 − cos 2θ ) [1 + 2 cos 2θ + 1] − 8 4 ma 2θ 2 mga cos 2 θ  − sin 2 θ  = 2 2  where we used the trigonometric identies … 2 cos 2 θ = 1 + cos 2θ and 2sin 2 θ = 1 − cos 2θ

10.11 (See Example 4.6.2)

x=

(

)

146

s = a sin θ so s = aθ cos θ ms2 mg 2 1 2 1 2 mg L= − s = ms − ks where k = a 2 2a 2 2 The equation of motion is thus k g  s + s = 0 or  s+ s=0 -a simple harmonic oscillator m a

Let

Coordinates:

10.12

x = a cos ω t + b sin θ y = a sin ω t − b cos θ

x = − aω sin ω t + bθ cos θ y = aω cos ω t + bθ sin θ

L = T −V =

1 m ( x 2 + y 2 ) − mgy 2

m 2 2  a ω + b 2θ 2 + 2bθaω sin (θ − ω t )  − mg ( a sin ω t − b cos θ ) 2 d ∂L = mb 2θ + mbaω θ − ω cos (θ − ω t )  dt ∂θ ∂L = mbθaω cos (θ − ω t ) − mgb sin θ ∂θ ∂L d ∂L − = 0 is The equation of motion ∂θ dt ∂θ a g θ − ω cos (θ − ω t ) + sin θ = 0 b b Note – the equation reduces to equation of simple pendulum if ω → 0 . =

(

)

Coordinates: x = l cos ω t + l cos (θ + ω t )

10.13

y = l sin ω t + l sin (θ + ω t ) x = −ω l sin ω t − θ + ω l sin (θ + ω t )

(

(

)

)

y = ω l cos ω t + θ + ω l cos (θ + ω t ) 2 1 1 m ( x 2 + y 2 ) = ml 2 ω 2 + θ + ω  + ...   2 2 ... ml 2ω θ + ω sin ω t ( sin θ cos ω t + sin ω t cos θ ) + cos ω t ( cos θ cos ω t − sin ω t sin θ ) 

(

L =T =

(

)

)

147

2 1 2 2 ml ω + θ + ω  + ml 2ω θ + ω cos θ   2 ∂L d ∂L − =0 ∂θ dt ∂θ ml 2 θ − ωθ sin θ + ml 2ω θ + ω sin θ = 0 θ + ω 2 sin θ = 0

(

=

(

(a) (b)

)

)

(

(

)

)

The bead executes simple harmonic motion (θ > 0 ) about a point diametrically

opposite the point of attachment. (c)

The effective length is " l " =

g

ω2

G 10.14 v = ˆjat + lθ iˆ cos θ + ˆj sin θ

(

) = ilˆ θ cos θ + ˆj ( at + lθ sin θ ) 1 G G m T = mv ⋅ v = ( l θ cos θ + a t 2 2 2

2

2

2 2

+ 2atlθ sin θ + l 2θ 2 sin 2 θ

)

1  V = mg  at 2 − l cos θ  2   at 2  m 2 2 − l cosθ  L = T −V = l θ + a 2t 2 + 2atlθ sin θ − mg  2  2 

(

)

d  ∂L  ∂L 2   = ml 2θ + matl sin θ ,    = ml θ + mal sin θ + matlθ cos θ  ∂θ dt  ∂θ  ∂L = matlθ cos θ − mgl sin θ ∂θ ml 2θ + mal sin θ + matlθ cos θ = matlθ cosθ − mgl sin θ a+g θ + sin θ = 0 l For small oscillations, sin θ ≈ θ a+g θ + θ =0 l 2π l TD = = 2π ω a+g

1 2 1 2 mx + Iθ 2 2 1 12  T = mx 2 +  ma 2 θ 2 2 25  V = −mgx

10.15 (a)

T=

and

I=

2 2 ma 5

148

1 2 1 2 2 mx + ma θ + mgx 2 5 T x The equation of constraint is … f ( x,θ ) = x − aθ = 0 The 2 Lagrange equations with multipliers are … ∂L d ∂L ∂f − +λ =0  x dt x x ∂ ∂ ∂ θ = θt ∂L d ∂L ∂f − +λ =0  ∂θ dt ∂θ ∂θ ∂L d ∂L ∂L ∂f = mx = mx = mg λ =λ dt ∂x ∂x ∂x ∂x mg − mx + λ = 0 and from the θ-equation … mg 2 − ma 2θ − λ a = 0 5  x and substituting into the above … Differentiating the equation of constraint … θ = a 5 2  x = g and λ = − mg 7 7 (b) The generalized force that is equivalent to the tension T is … ∂f 2 Qx = λ = λ = − mg ∂x 7

L = T −V =

G 10.16 For v′ the velocity of a differential mass element, dm, of the spring at a distance x′ below the support … m′ G x′ G dm = dx′ v′ = v , x x 2 m′ 1 x 1  x′  m′ 1 2 1 2 2 T = mv + ∫ ( v′) dm = mx + ∫0  x  dx′ 0 2 2 2 2 x  x 1 1 m′ 2 T = mx 2 + x 2 2 3 m′ 1 2 V = k ( x − l ) − mgx − ∫ gx′dm′ 0 2 x 1 m′ 2 = k ( x − l ) − mgx − ∫ gx′ dx′ 0 x 2 ′ 1 m 2 V = k ( x − l ) − mgx − gx 2 2 ′ 1 m  1 m′  2  L = T − V =  m +  x 2 − k ( x − l ) +  m +  gx 2 3  2 2   ∂L  m′  d  ∂L   m′  =  m +  x , =  m +   x,   ∂x  dt  ∂x   3  3

149

For

∂L m′   = −k ( x − l ) +  m +  g ∂x 2   m′  m′    x = −k ( x − l ) +  m +  g  m +   3  2   g m′  y = x −l − m+ , k 2  m′   y + ky = 0  m +   3  

The block oscillates about the point x = l +

g m′  m+  … k 2 

with a period … m′   m+   2π 3  TD = = 2π  ω k 10.17 Note: 4 objects move – their coordinates are labeled xi : The coordinate of the movable, massless pulley is labeled x p .

Two equations of constraint: f1 ( x1 , x p ) = x p − ( l − x1 ) = 0

f 2 ( x2 , x3 , x p ) = ( x2 + x3 ) − ( 2 x p + l ′ ) = 0

1 1 1 m1 x12 − m1 gx1 + mx22 − m2 gx2 + m3 x32 − m3 gx3 2 2 2 ∂f ∂L d ∂L − + ∑λj j = 0 ∂qi dt ∂qi ∂qi j

L = T −V =

Thus: (1) − m1 g − m1  x1 + λ1 = 0 − m2 g − m2  x2 + λ2 = 0 (2) − m3 g − m3  x3 + λ2 = 0 (3) Now – apply Lagrange’s equations to the movable pulley – note m p → 0 So: (4)

λ1

∂f1 ∂f + λ2 2 = 0 or λ1 − 2λ2 = 0 ∂x p ∂x p

Now - x p can be eliminated between f1 and f 2 f 2 = 2 f1 = x2 + x3 + 2 x1 − ( 2l + l ′ ) = 0

(5) Thus  x2 +  x3 + 2  x1 = 0 With a little algebra – we can solve (1) –(5) for the 5 unknowns x1 , x2 , x3 , λ1 , and λ2

150

2g

λ1 =

λ2 =

 1 1 1 1  +  +    m1 4  m2 m3   As the check, let m2 = m3 = m and m1 = 2m .

g  1 1 1 1  +  +    m1 4  m2 m3  

Thus, there is no acceleration and λ1 = 2mg 10.18 (See Example 5.3.3) (a) r = reˆr  ˆr + rθeˆθ r = re

Constraint: f (θ ) = θ − ω t = 0 so … θ = ω and θ = 0 1 T = m r 2 + r 2θ 2 = L 2 ∂L d ∂L ∂L d ∂L ∂f − = 0 and − +λ =0  ∂r dt ∂r ∂θ dt ∂θ ∂θ  r = ω 2r −2mrrθ − mr 2θ + λ = 0 r = Aeωt + Be −ω t r ( 0) = 0

(

)

ωt

r = ω Ae − ω Be

−ω t

A=

r ( 0 ) = ω l

λ = 2mrrω

so … A + B = 0 ω A − ω B = ωl 2A = l

apply constraint

l 2

l ω t −ω t (e − e ) 2 r = l sinh ω t   r = ω l cosh ω t 

B=−

l 2

thus .. r =

now r = l at t = T so (b)

λ = 2mω 2l 2 sinh ω t cosh ω t T=

1

ω

sinh −1 1 =

0.88

ω

There are 2 ways to calculate F … (i) See Example 5.3.3 … F = 2mω x ′ l ωl and x ′ = ( e ω t + e −ω t ) x ′ = ( e ω t − e −ω t ) 2 2 2 F = 2mω l cosh ω t ∂f (ii) λ is the generalized force, in this case – a torque T, acting on the bead ∂θ ∂f T = rF = λ ∂θ

151

λ ∂f 2mω 2l 2 sinh ω t cosh ω t = r ∂θ l sinh ω t 2 = 2mω l cosh ω t

F=

10.19 (See Example 4.6.1) The equation of constraint is

m 2 2 2 V = mgr cos θ r + r θ 2 m 2 2 2 L= r + r θ − mgr cos θ 2 ∂L d ∂L ∂f − +λ =0 ∂r dt ∂r ∂r ∂f ∂f =1 =0 ∂r ∂θ

T= θ

r

a

Thus

∂L d ∂L ∂f =0 − +λ  ∂θ dt ∂θ ∂θ mrθ 2 − mg cosθ − mr + λ = 0 mgr sin θ − mr 2θ − 2mrrθ = 0

f ( r ,θ ) = r − a = 0

(

)

(

)

Now r = a , r =  r = 0 so 2 maθ − mg cos θ + λ = 0 mga sin θ − ma 2θ = 0 g sin θ and a g so ∫ θdθ = a ∫ sin θ dθ or hence, λ = mg ( 3cosθ − 2 )

θ =

and when λ → 0

dθ dθ 2  θ g g = − cosθ + 2 a a

θ = θ

particle falls off hemisphere at

2 θ D = cos −1   3 10.20 Let x2 mark the location of the center of curvature of the movable surface relative to a fixed origin. This point defines the position of m2. r1 marks the position of the particle of mass m1. r is the position of the particle relative to the movable center of curvature of m2. 1 1 Kinetic energy T = m2 x22 + m1r1 2 2 2 r1 ⋅ r1 = ( x2 + r ) ⋅ ( x2 + r ) = x22 + r 2 + 2 x2 ⋅ r

but r = reˆr + rθeˆθ 1 1  ˆr + rθeˆθ  m2 x22 + m1  x22 + r 2 + r 2θ 2 + 2 x2 ⋅ re  2 2 1 1 = m2 x22 + m1  x22 + r 2 + r 2θ 2 + 2 x2 rθ sin θ − 2 x2 r cos θ  2 2

∴T =

(

)

152

V = − mgy = −mgr sin θ

Potential energy L = T −V

Equation of constraint f ( r ,θ ) = r − a = 0

∴

∂f ∂f ∂f =1 =0 =1 ∂θ ∂x2 ∂r

∂L d  ∂L  ∂f −  =0 +λ ∂xi dt  ∂xi  ∂xi d x2 − m1  x1 − m1  rθ sin θ − r cosθ  = 0 − m2  x2 ) dt d m2  x2 − m1  x1 − m1  rθ sin θ + rθsin θ + rθ 2 cosθ −  r cosθ + rθ sin θ  = 0 dt using constraint r = a = constant ; r =  r =0 m1  x2 = − a θsin θ + θ 2 cos θ (1) m1 + m2  θ −    sin θ + x rθ cosθ + gr cosθ = 0 θ) − r 2θ − 2rr x r sin θ − x r sin θ − x rθ cosθ + xr

Lagrange’s equations

(

2

)

2

2

2

using constraint r = a = constant ; r =  r =0)  x g −θ − 2 sin θ + cos θ = 0 a a r)

− r +  x2 cos θ − x2θ sin θ + rθ 2 + x2θ sin θ + g sin θ +

Apply constraint…

 x2 cos θ + aθ 2 + g sin θ +

λ

(2)

λ m1

=0 m1 Now, plug solution for x2 (1) into equation (2): m1 g θ − θsin θ + θ 2 cosθ sin θ − cosθ = 0 m1 + m2 a g θ(1 − f m1 sin 2 θ ) − θ 2 f m1 sin θ cosθ − cosθ = 0 a m1 where f m1 = m1 + m2 dθ dθ dθ  dθ d  1  2  = =θ = Now θ =  θ  dt dθ dt dθ dθ  2  So – let x = θ 2 dx 2g 1 − f m1 sin −2 θ  − 2 xf m1 sin θ cosθ − cosθ = 0 dθ a Let y = 1 − f m1 sin 2 θ

(

y

=0

(3)

)

dx dy 2 g d ( sin θ ) +x = dθ dθ a dθ

153

2g Hence: d ( yx ) = d ( sin θ ) and a

yx (θ =θ )

(

∫

D

yx θ = 0

)

d ( yx ) =

2g a

sin θ

∫ d ( sin θ ) 0

x = θ 2 = 0 at θ = 0D so 2g yx (θ ) = sin θ a 2g sin θ x (θ ) = θ 2 = a 1 − f m1 sin 2 θ  Now we can solve for θ and plug θ 2 , θ into (3) to obtain λ (θ ) but

 d  1 2  g d  sin θ    θ = dθ  2  a dθ 1 − f m1 sin 2 θ  After some algebra – yields (1 + fm1 sin 2 θ ) g θ = cosθ 2 a (1 − fm1 sin 2 θ )

θ =

The solution for λ (θ ) is thus determined from equation (3) g λ θ 2 − f m θsin θ cos θ + θ 2 cosθ  + sin θ = − a m1a 1

collecting terms:

θ 2 (1 − f m cos 2 θ ) − θ f m sin θ cosθ + 1

1

g λ sin θ = − a m1a

Plug θ , θ into the above --- plus --- a lot of algebra yields … 2 2 λ (θ ) 2 f m2 sin θ − f m1 sin θ cos θ 1 − f m1 sin θ  − = + sin θ 2 m1 g 1 − f m1 sin 2 θ  m2 where f m2 = m1 + m2 As a check, let m2 → ∞ so it is immoveable … then f m2 → 1 and f m1 → 0 and we have −

λ (θ ) m1 g

→ 3sin θ

… which checks

1 2 1 mv = m R 2 + R 2φ 2 + z 2 2 2 m 2 L = T −V = R + R 2φ 2 + z 2 − V 2 d  ∂L  ∂L ∂V ∂L  , = mRφ 2 − = mR = mR ,   ∂R ∂R dt  ∂R  ∂R

10.21 (a)

(

T=

(

)

)

154

 = mRφ 2 − ∂V mR ∂R 2   mR − mRφ = QR d  ∂L  ∂L ∂V 2   =−    = mR φ + 2mRRφ , ∂φ ∂φ dt  ∂φ  ∂V mR 2φ + 2mRRφ = − = Qφ ∂φ d  ∂L  ∂L ∂V ∂L =− = mz ,    = mz , ∂z ∂z dt  ∂z  ∂z ∂V mz = − = Qz ∂z G G G For F = ma , using the components of a from equation 1.12.3:  − Rφ 2 , FR = m R Fφ = m 2 Rφ + Rφ , Fz = m z ∂L = mR 2φ , ∂φ

(

)

(

)

From Section 10.2, since R and z are distances, QR and Qz are forces. However, since φ is an angle, Qφ is a torque. Since Fφ is coplanar with and perpendicular to R ,

Qφ = RFφ (b)

… and all equations agree.

v 2 = r 2 + r 2θ 2 + r 2φ 2 sin 2 θ m 2 2 2 2 2 2 L = T −V = r + r θ + r φ sin θ − V 2 d  ∂L  ∂L ∂L ∂V = mr , = mrθ 2 + mrφ 2 sin 2 θ − = mr ,   dt  ∂r  ∂r ∂r ∂r ∂V = Qr mr − mrθ 2 − mrφ 2 sin 2 θ = − ∂r d  ∂L  ∂L ∂L ∂V 2   = mr 2θ , = mr 2φ 2 sin θ cosθ −    = mr θ + 2mrrθ ,  ∂θ ∂θ dt  ∂θ  ∂θ ∂V = Qθ mr 2θ + 2mrrθ − mr 2φ 2 sin θ cosθ = − ∂θ ∂L d  ∂L  2  2 2  2 2   2 φ sin θ , = mr    = mr φ sin θ + 2mrrφ sin θ + 2mr θφ sin θ cosθ ,  ∂φ dt  ∂φ  ∂L ∂V =− ∂φ ∂φ   sin θ cosθ = − ∂V = Q mr 2φsin 2 θ + 2mrrφ sin 2 θ + 2mr 2θφ φ ∂φ Qr is a force Fr . Qθ and Qφ are torques.

(

)

Since φ is in the xy plane, the moment arm for φ is r sin θ ; i.e. Qφ = r sin θ Fφ .

Qθ = rFθ . From equation 1.12.14 …

155

( ) m ( rθ + 2rθ − rφ sin θ cosθ ) = F   cosθ ) = F m ( rφsin θ + 2rφ sin θ + 2rθφ m  r − rφ 2 sin 2 θ − rθ 2 = Fr 2

θ

φ

The equations agree. ∂V ∂V ∂V = = 0 and = −F . ∂θ ∂φ ∂r For spherical coordinates, using equation 1.12.12: v 2 = r 2 + r 2φ 2 sin 2 θ + r 2θ 2 m 2 2 2 2 L = T −V = r + r φ sin θ + r 2θ 2 − V ( r ) 2 d  ∂L  ∂L ∂L ∂V = mr , = mrφ 2 sin 2 θ + mrθ 2 − = mr ,   dt  ∂r  ∂r ∂r ∂r 2 2 2 mr − mrφ sin θ − mrθ = Fr d  ∂L  ∂L 2   ∂L = mr 2φ 2 sin θ cosθ = mr 2θ ,    = mr θ + 2mrrθ ,  dt  ∂θ  ∂θ ∂θ 2  2 2 mr θ + 2mrrθ − mr φ sin θ cos θ = 0 ∂L = mr 2φ sin 2 θ ,  ∂φ d  ∂L  2  2 2   2    = mr φ sin θ + 2mrrφ sin θ + 2mr φθ sin θ cosθ dt  ∂φ  ∂L =0 ∂φ   sin θ cos θ = 0 mr 2φsin 2 θ + 2mrrφ sin 2 θ + 2mr 2φθ

10.22 For a central field, V = V ( r ) , so

(

)

10.23 Since θ = α = constant, there are two degrees of freedom, r and θ . vr = r , vφ = rφ sin α

1 2 1 mv = m r 2 + r 2φ 2 sin α 2 2 V = mgr cos α m 2 2 2 2 L = T −V = r + r φ sin α − mgr cos α 2 d  ∂L  ∂L ∂L = mrφ 2 sin 2 α − mg cos α = mr , = mr ,   dt  ∂r  ∂r ∂r 2 2  mr = mrφ sin α − mg cos α ∂L ∂L 2  2 φ sin α , =0 = mr ∂φ ∂φ

(

T=

(

)

)

156

d mr 2φ sin α = 0 dt mr 2φ sin α = A = constant

(

Say

)

A2 − mg cos α mr 3 d dr 1 d 2  r r = r = r = 2 dr dt dr m A 2 dr − mg cos α dr d ( r 2 ) = 2 m r3 mr 2 A2 =− − mgr cos α + C 2 2mr 2 The constant of integration C is the total energy of the particle: kinetic energy due to the component of motion in the radial direction, kinetic energy due to the component of motion in the angular direction, and the potential energy. A2 U (r ) = + mgr cos α 2mr 2 For φ ≠ 0 , A ≠ 0 , and turning points occur at r = 0

mr =

A2 Then 0 = − − mgr cos α + C 2mr 2 A2 ( mg cos α ) r 3 − Cr 2 + = 0 2m The above equation is quadratic in r ( A 2 ∝ r 4 ) and has two roots.

l2 =0, 2m defines the turning points. For the particle to remain on a single horizontal circle, there 2 must be two roots with r = rD . Thus ( r − rD ) divided into the above expression leaves a

10.24 Note that the relation obtained in Problem 10.23, ( mg cos α ) r 3 − Cr 2 +

term that is linear in r.

r + ( 2rD − c )

r 2 − 2rrD + rD2 r 3 − Cr 2

+

A 2 2m

r 3 − 2r 2 rD + rrD2

( 2rD − C ) r 2 − rD2 r ( 2rD − C ) r 2 − 2rD ( 2rD − C ) r + rD2 ( 2rD − C )  2rD ( 2rD − C ) − rD2  + A 2 2m − rD2 ( 2rD − C ) For the remainder to vanish, both terms must equal zero. 4rD2 − 2rDC − rD2 = 0 3 C = rD 2

157

A2 − 2rD3 + CrD2 = 0 2m r A2 = D 2 2mrD 2 And with A = mr 2φ sin α D

D

1

 2 1 φD =   2  mrD sin α 

A2 − mg cos α mr 3 For small oscillations about rD , r = rD + ε .  r = ε From Prob. 10.23, mr =

−3

1 1 ε = 1 +  r 3 rD3  rD  A 2  3ε mε = 3 1 − mrD  rD

≈

1  3ε  1 −  rD3  rD 

  − mg cos α  2 3A A2 − mg cos α mε + 4 ε = mrD mrD3 2π

m 2 rD4 2π T= = 2π = 2  ω φD sin α 3A

1 3

1 2 G G m  x + yA  y + zA  z) mv + qv ⋅ A = ( x 2 + y 2 + z 2 ) + q ( xA 2 2 d  ∂L  dA ∂L = mx + qAx , = mx + q x   dt  ∂x  dt ∂x dAx ∂A ∂A ∂A Using the hint, = x x + y x + z x dt ∂x ∂y ∂z ∂Ay  ∂A ∂L ∂A  = q  x x + y + z z  ∂x ∂x ∂x   ∂x ∂A  ∂A  ∂A ∂A ∂A  ∂A  mx + q  x x + y x + z x  = q  x x + y y + z z  ∂y ∂z  ∂x ∂x   ∂x  ∂x

10.25 L =

  ∂Ay ∂Ax   ∂Ax ∂Az mx = q  y  − −  − z  ∂y   ∂z ∂x   ∂x G G G G = q  y ∇ × A − z ∇ × A   z y  G G G = q v × ∇ × A   x

(

)

(

(

  

)

)

158

G G mx = q v × B

(

)

x

Due to the cyclic nature of the Cartesian coordinates, i.e., iˆ × ˆj = kˆ … G G my = q v × B y G G mz = q v × B z G G G Altogether, mr = q v × B

( (

(

) )

10.26 V = mgz

)

1 m ( x 2 + y 2 + z 2 ) 2 p p p x = x similarly, y = y and z = z m m m

T=

∂T = mx , ∂x 1 H = T +V = ( px2 + p y2 + pz2 ) + mgz 2m ∂H px = = x ∂px m ∂H px = constant = 0 = − p x , ∂x d p x = ( mx ) = mx = 0 similarly, p y = constant , or my = 0 dt ∂H p y = = y ∂p y m px =

∂H pz = = z ∂pz m d ∂H p z = ( mz ) = mz = − mg = mg = − p z , dt ∂z These agree with the differential equations for projectile motion in Section 4.3. 10.27 (a) Simple pendulum …

V = −mgl cos θ

1 2 2 ml θ 2 p θ = θ2 ml T=

∂T = ml 2θ ,  ∂θ pθ2 − mgl cos θ H = T +V = 2ml 2 p ∂H = θ2 = θ ∂pθ ml ∂H = mgl sin θ = − pθ ∂θ pθ =

159

(b) Atwood’s machine … 1 1 1 x 2 m1 x 2 + m2 x 2 + I 2 [Includes the pulley] 2 2 2 a p x = I    m1 + m2 + 2  a  

V = −m1 gx − m2 g ( l − x )

p=

T=

I  ∂T  =  m1 + m2 + 2  x , a  ∂x  p2

− ( m1 − m2 ) gx − m2 gl I   2  m1 + m2 + 2  a   p ∂H = = x I  ∂p   m1 + m2 + 2  a   ∂H p = ( m1 − m2 ) g = − ( m1 − m2 ) g = − p , ∂x

H = T +V =

(c) Particle sliding down a smooth inclined plane … 1 V = −mgx sin θ T = mx 2 2 p ∂T p= x = = mx , m ∂x 2 p − mgx sin θ H = T +V = 2m ∂H p = = x ∂p m ∂H p = mg sin θ = − mg sin θ = − p , ∂x 10.28 (a)

F r

L = T (qi , qi ) − T (qi , t ) …note, potential energy is time dependent. k ∂V so F =− V = ∫ Fdr = − e − β t r ∂r 1 G G 1 T = mv ⋅ v = m r 2 + r 2θ 2 2 2

(

)

160

∂L = mr ∂r ∂L pθ = = mr 2θ ∂θ pr =

pr m p θ = θ 2 mr r =

p2 p2 k H = ∑ p i qi − L = pr r + pθθ − r − θ 2 − e − β t 2m 2mr r   Substituting for r and θ … pθ 2 pr 2 k + − e −βt H= 2 2m 2mr r

(b) H = T + V … which is time-dependent (c) E is not conserved 10.29 Locate center of coordinate system at C.M. The potential is independent of the center of mass coordinates. Therefore, they are ignorable.

(r1,θ1)

(r2,θ2) H = ∑ p i qi − L

L = T −V 1 1 1 2 = m1 r12 + r12θ12 + m2 r2 2 + r2 2θ2 2 − k ( r1 + r2 − l ) 2 2 2 where l is the length of the relaxed spring and k is the spring constant. p ∂L p1r = = mr1 r1 = 1r ∂r1 m1 p ∂L p1θ =  = mr12θ1 θ1 = 1θ 2 m1r1 ∂θ1 … and similarly for m2

(

)

(

)

p 2 p 2 p1r 2 p 2 1 + 1θ 2 + 2 r + 2θ 2 − k (r1 + r2 − l ) 2 2m1 2m1r1 2m2 2m2 r2 2 Equations of motion … First, θ1 , θ2 are ignorable coordinates, so p1θ , p2θ are each conserved. ∂H r1 = − p1r = k ( r1 + r2 − l ) = − m1 ∂r1 ∂H r2 = − p 2 r = k ( r1 + r2 − l ) = − m2  ∂r2 p1r = p 2 r ∆p r = 0 ∆pr = constant. The radial momenta are equal and opposite. H=

161

10.30 L =

1 2 1 2 mx − kx 2 2

t2 t2 t2 1 1  0 = δ ∫ Ldt = ∫ δ Ldt = ∫ δ  mx 2 − kx 2  dt t1 t1 t1 2 2 

0=∫

t2

t1

d δx dt

δ x =

∫

t2

t1

( mxδ x − kxδ x ) dt

t2

 = ∫ mx mxδ xdt t1

t2 d  (δ x ) (δ x ) dt = ∫t1 mxd dt

Integrating by parts:

∫

t2

t1

 = mxδ x mxδ xdt

t2 t1

t2

− ∫ δ x d ( mx ) t1

δ x = 0 at t1 and t2 d ( mx ) =

∫

t2

t1

d ( mx ) dt = mx dt dt t2

mxδ x dx = − ∫ δ x mx dt t1

0=∫

t2

t1

( −mxδ x − kxδ x ) dt

mx + kx = 0

10.31 (a) L = − m0 c 2 1 −

v2 −V c2

1

Let γ =

1−

v2 c2

∂L = γ m0 x ≡ px This is the generalized momentum for part (b). ∂x Thus, Lagrange’s equations for the x-component are … d ∂L ∂L d ∂V − = px − =0 dt ∂x ∂x dt ∂x … and so on for the y and z components. (b) H = ∑ vi pi − L i

p but … vi = i γ m0 So … H = ∑ i

H=

pi 2 c 2 m0 c 2 + +V γ m0c 2 γ

p 2 c 2 m0 c 2 + +V γ m0 c 2 γ

162

H= (c)

1 p 2 c 2 + m0 2 c 4 ) + V 2 ( γ m0c

Now, if T = γ m0 c 2 then we have …  p 2c 2  p c + m0 c = m0 c 1 + 2 4   m0 c   γ 2 m0 2 v 2 c 2  2 4 2 2 2 = m0 2 c 4  1 +  = m0 c (1 + γ v c ) 2 4 m0 c   2 2

2 4

2 4

   v2 c2  1 = m0 c  1 + = m0 2c 4  = γ 2 m0 2 c 4 2 2  2 2   1− v c   1− v c  Thus … H = γ m0 c 2 + V = T + V 2 4

 m0 c 2 1 v2 2 ≈ m c 1 + + ...   0 2 2 2 1− v c  2c  1 T ≈ m0 v 2 + m0 c 2 2 ------------------------------------------------------------------------------------------------------------

(d)

T=

163

Chapter 11 11.1 (a)

(b)

(c)

k 2 k2 x + 2 x 2 k At equilibrium, V ′ = kx − 2 x x = k1 3 2k 2 V ′′ = k + 3 x ′′ V x = k1 3 = k + 2k = 3k > 0

Stable

V ( x) = kxe −bx V ′ = ke− bx − bkxe− bx ke − bx − bkxe− bx = 0 At equilibrium 1 x= b V ′′ = −bke −bx − bkxe − bx + b 2 kxe −bx V ′′ x =1 b = −2bke−1 + bke−1 = −bke−1 < 0

Unstable

V ( x) =

V′ = 0

V ( x) = k ( x 4 − b 2 x 2 )

V ′′ = k ( 4 x 3 − 2b 2 x ) At equilibrium

x = 0, ± b

k ( 4 x 3 − 2b 2 x ) = 0

2

V ′′ = k (12 x − 2b 2 ) 2

(d)

V ′′ x =0 = −2kb 2 < 0

Unstable

V ′′ x =±b

Stable

2

= k ( 6b 2 − 2b 2 ) = 4kb 2 > 0

for case (a) ω 2 =

3k m

for case (c) at x = ± b 11.2

T=

2

2π

ω

ω2 =

= 2π

4kb 2 m

m 2π s = 3k 3 T=

2π

ω

= 2π

m =π s 4kb 2

V ( x, y ) = k ( x 2 + y 2 − 2bx − 4by )

164

∂V = 2k ( x − b) ∂x at equilibrium x=b and

∂V = 2k ( y − 2b) ∂y y = 2b

∂V ∂V ∂ 2V = 2k = 2 k = 0 ∂x 2 ∂x∂y ∂y 2 k11 = 2k > 0 k12 = k21 = 0 k22 = 2k 2

k11 k21

2

k12 = 4k 2 − 0 > 0 k22

The equilibrium is stabe. 11.3

1 V ( x) = − kx 2 2 dV ( x) dx F ( x) = − = kx = mx = mx dx dx   kx dx = mx dx x

v

x0

0

k 2 v2 x − x02 ) = m ( 2 2

∫ kxdx = ∫ mx dx

12 dx = k m ( x 2 − x02 ) dt x t dx = ∫x x 2 − x 2 1 2 ∫0 α dt ; where α = k m 0 ( 0)

v=

)

(

ln x + x 2 − x02 − ln x0 = α t x + x 2 − x02 = x0 eα t

x 2 − x02 = x02 e 2α t − 2 xx0 eα t + x 2 x = x0

eα t + e −α t = x0 cosh α t 2

11.4

Let the length of the unstretched, elastic cord be d. Then

2l

y

d = 2 l 2 + y2 1 2 V = k ( d − 2l ) − mgy 2

m

165

V=

(

)

k 4l 2 + 4 y 2 − 8l l 2 + y 2 + 4l 2 − mgy 2

)

(

V = 2k 2l 2 + y 2 − 2l l 2 + y 2 − mgy The first term, 4kl 2 , is an additive constant to the potential energy, so with appropriate adjustment of the zero reference point …

)

(

V ( y ) = 2k y 2 − 2l l 2 + y 2 − mgy −1 2 dV ( y ) = 2k  2 y − 2 yl ( l 2 + y 2 )  − mg   dy At equilibrium, the above expression is zero, so … 4kly 4ky − − mg = 0 l 2 + y2 4kly 4ky − mg = l 2 + y2

16k 2 y 2 − 8kmgy + m 2 g 2 =

16k 2l 2 y 2 l 2 + y2

16k 2 y 4 − 8kmgy 3 + (16k 2l 2 + m2 g 2 − 16k 2l 2 ) y 2 − 8kl 2 mgy + l 2 m2 g 2 = 0 y 4 mg 3 m 2 g 2 2 mg m2 g 2 =0 − y + y − y + 16k 2l 4 2kl 2 16k 2l 2 l 4 2kl 4 y mg and letting u = a= 4kl l 4 3 2 2 u − 2au + a u − 2au + a 2 = 0 11.5

V = mg ( h1 + h2 )

a

h1 = b cos θ

h2 = d sin (θ + ϕ ) h2 = d ( sin θ cos ϕ + cos θ sin ϕ )

φ

h2

d

θ

cos ϕ = θ

h1

b

bθ d

sin ϕ =

a d

h2 = bθ sin θ + a cos ϕ

V = mg ( a + b ) cos θ + bθ sin θ  V ′ = mg ( a + b )( − sin θ ) + b sin θ + bθ cos θ 

166

V ′ = mg [ − a sin θ + bθ cos θ ] V ′′ = mg [b cos θ − bθ sin θ − a cos θ ] V0′′ = mg ( b − a )

Equilibrium … Stable Unstable a b 11.6

cos θ = 1 − sin θ = θ −

θ2 2!

+

θ4 4!

− ...

θ3 θ5

− ... 5!   θ2 θ4    θ3 θ5 V = mg ( a + b )  1 − + − ...  + bθ  θ − + − ...   2! 4! 3! 5!      a − b 2 a − 3b 4   θ + θ − ... V = mg  a + b − 2 24  

For a = b,

3!

+

a   V = mg  2a − θ 4 +... 12  

mga 3 θ + ... terms in higher order of θ 3 V ′′ = − mga θ 2 + ... V ′′′ = − 2mga θ + ... V ′′′′ = − 2mga < 0 ∴ Equilibrium is unstable

V′=−

11.7

a h2

h1

φθ θ b

The center of mass (CM) of the hemisphere is 3 a from the flat side (see Equation 8.1.8). 8 The height of CM) above the point of contact between the two hemispheres is designated by h2 in the figure. h1 is the height of the point of contact above the ground.

3   V = mg ( h1 + h2 ) = mg b cos θ + a cos θ − a cos (θ + ϕ )  and aϕ = bθ 8  

167

 3 bθ    V = mg ( h1 + h2 ) = mg ( a + b ) cos θ − a cos  θ +   8 a     3a  a + b    a + b    o V ′ = mg  − ( a + b ) sin θ +   sin    θ   Equilibrium occurs at θ = 0 a a 8       2  3a  a + b    a + b   V ′′ = mg  − ( a + b ) cos θ +   cos   θ  8  a    a     2  3a  a + b   mg V0′′= mg  − ( a + b ) +  ( a + b )( 3b − 5a )  = 8  a   8a  V0′′ > 0 for 3b > 5a

Therefore, the equilibrium is stable for a
> x, r0 2 + x 2 ≈ r0 2 ,

ε≈

1 2x and r ≈ r0 (1 + ε cos φ ) 2 r0

Thus, for small ε , the expression for the potential energy, V, can be approximated … 2  GM e m a  1  2 x  3  2x  2   dx − + 1 cos cos V ≈− φ φ      2ar0 ∫− a  2  r0  8  r0    2 3 GM e  3cos φ 2a  V ≈−  2a + 0 +  3  2ar0  2r0 2

GM e  a 2 cos 2 φ  V ≈− 1 +  r0  2r0 2  GM e  a 2  GM e a 2 = V′≈ 2 cos φ sin φ sin 2φ   r0  2r0 2  2r03

170

Equilibrium @ φ = 0 GM e a 2 cos 2φ r03 1 M = I = ma 2 3

V ′′ ≈

V0′ ≈

GM e a 2 r03

V0′′ 3GM e = M r03

ω= T0 =

and

2π

ω

= 2π

r03 3GM e

11.13

The amplitude of the symmetric component is A1 and the amplitude of the anti-symmetric component is A2 (See Equations 11.3.19a through 11.3.20b). 1 2 A12 = [ x1 (0) + x2 (0)] 4 A0 1 A1 = [ x1 (0) + x2 (0) ] = 2 2 1 2 A2 2 = [ x2 (0) − x1 (0) ] 4 A0 1 ∴ A1 = A2 A2 = [ x2 (0) − x1 (0) ] = 2 2 From Equation 11.3.18 the solution for x1 is … A0 x1 (t ) = ( cos ω1t + cos ω2t ) (The phase δ 2 is 180o, which insures that x1 (0) = A0 ) 2 A0  (ω1 + ω2 ) t cos (ω1 − ω2 ) t  x1 (t ) =  2 cos  2  2 2  ω + ω2 ω − ω2 and ∆ = 1 ω= 1 Letting … 2 2 x1 (t ) = A0 ( cos ω t cos ∆t ) From Equation 11.3.18, the solution for x2 is … A0 x2 (t ) = ( cos ω1t − cos ω2t ) 2 A0  (ω1 + ω2 ) t sin (ω1 − ω2 ) t  x2 (t ) =  2sin  2  2 2  x2 (t ) = A0 ( sin ω t sin ∆t )

171

11.14

At time t = 0 and short times thereafter … cos ∆t ≈ 1 and sin ∆t ≈ 0 . Thus, … x1 ≈ A0 cos ω t and x2 ≈ 0 . This situation occurs again when ∆t = 2π . 1 1   1  k + 2 k ′  2  k  2  ∆= =   −  2 2  m   m     1 1   1  k  2  2 k ′  2  ∆ =   1 + − 1   2  m   k   

ω2 − ω1

1

k′ 1  2k ′   2k ′  2 for k ′ m . We also assume that the spring is “slack”, m+M

k g