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CHAPTER 11 Dynamics of Rigid Bodies 11-1. The calculation will be simplified if we use spherical coordinates: sin cos sin sin cos x r y r z r θ φ θ φ θ = = = (1) z y x Using the definition of the moment of inertia, () 2 ij ij k i j k I r x xx ρ δ dv = (2) we have ( ) ( ) ( ) 2 2 33 2 2 2 2 cos cos I r z dv r r r dr d d ρ ρ θ θ = = φ (3) or, ( ) ( ) 1 2 4 2 33 0 1 0 5 1 cos cos 4 2 5 3 R I r dr d R + = = π d ρ θ θ πρ φ (4) 353

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### Transcript of Dynamics of Rigid Bodies - Cengage€¦ · CHAPTER 11 Dynamics of Rigid Bodies 11-1. The...

CHAPTER 11 Dynamics

of Rigid Bodies

11-1. The calculation will be simplified if we use spherical coordinates:

sin cos

sin sin

cos

x r

y r

z r

θ φ

θ φ

θ

= ==

(1)

z

y

x Using the definition of the moment of inertia,

( ) 2ij ij k i j

k

I r x x xρ δ dv

= − ∑∫ (2)

we have

( )

( ) ( )

2 233

2 2 2 2cos cos

I r z dv

r r r dr d d

ρ

ρ θ θ

= −

= −

∫ φ (3)

or,

( ) ( )1 2

4 233

0 1 0

5

1 cos cos

42

5 3

R

I r dr d

R

+

= −

= ⋅

∫ ∫ ∫π

dρ θ θ

πρ

φ

(4)

353

354 CHAPTER 11

The mass of the sphere is

343

M = Rπρ (5)

Therefore,

233 5

I MR2

= (6)

Since the sphere is symmetrical around the origin, the diagonal elements of I are equal:

211 22 33

25

I I I MR= = = (7)

A typical off-diagonal element is

( )

( )

12

2 2 2sin sin cos cos

I xy dv

r r dr d

ρ

dρ θ φ φ θ

= −

= −

∫ φ (8)

This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is

11

22

33

0 00 00 0

I I

I I

I I

−0− =

− (9)

From (9) and (7), we have

21 2 3

25

I I I MR= = = (10)

11-2.

a) Moments of inertia with respect to the x axes: i

x3 = x3′

R hCM

x1

x1′x2

x2′

It is easily seen that for i ≠ j. Then the diagonal elements become the principal

moments I , which we now calculate.

0ijI = iiI

i

The computation can be simplified by noting that because of the symmetry, . Then, 1 2I I I= ≠ 3

( )2 2 21 21 2 3 1 22

2 2I I

x x x dI I vρ+

= = = + +∫ (1)

DYNAMICS OF RIGID BODIES 355

which, in cylindrical coordinates, can be written as

( )2 2 21 2 0 0 0

22

h Rz hd dz r z rd

πI I r

ρφ= = +∫ ∫ ∫ (2)

where

2

3M MV R

ρπ

= =h

(3)

Performing the integration and substituting for ρ, we find

( )21 2

34

20I I M R h= = + 2 (4)

3I is given by

( )2 2 23 1 2I x x dv r rdr d dzρ ρ= + = ⋅∫ ∫ φ (5)

from which

23

310

I MR= (6)

b) Moments of inertia with respect to the xi′ axes:

Because of the symmetry of the body, the center of mass lies on the 3x′ axis. The coordinates of the center of mass are (0 0,0, )z , where

3

034

x dvz h

dv

′= =∫∫

(7)

Then, using Eq. (11.49),

2ij ij ij i jI I M a a aδ = − −′ (8)

In the present case, and 1 2 0a a= = ( )3 3 4a = h , so that

2 21 1

2 22 2

23 3

9 3 116 20 4

9 3 116 20 4

310

I I Mh M R h

I I Mh M R h

I I MR

= − = +′

= − = +′

= −′

2

2

11-3. The equation of an ellipsoid is

22 231 2

2 2 2 1xx x

a b c+ + = (1)

356 CHAPTER 11

which can be written in normalized form if we make the following substitutions:

1 2 3, ,x a x b x cξ η ζ= = = (2)

Then, Eq. (1) reduces to

2 2 2 1ξ η ζ+ + = (3)

This is the equation of a sphere in the (ξ,η,ζ) system.

If we denote by dv the volume element in the system and by dτ the volume element in the (ξ,η,ζ ) system, we notice that the volume of the ellipsoid is

ix

1 2 3

43

V dv dx dx dx abc d d d

abc d abc

ξ η ζ

τ π

= = =

= =

∫ ∫ ∫

∫ (4)

because dτ∫ is just the volume of a sphere of unit radius.

The rotational inertia with respect to the passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by

3 -axisx

( )

(

2 23 1 2

2 22 2

MI x x dv

V

Mabc a b d

V

= +

= +

∫ )ξ η τ (5)

In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ :

( )2 2 2 2

2 12 2

0 0 0

2

2

sin

cos sin

2 12

3 5

415

R

a z dv a z r dr r d d

a d d r d

a

a

=

=

=

= × × ×

=

∫ ∫

∫ ∫ ∫π π

θ θ φ

φ θ θ θ

π

π

4 r

(6)

Therefore,

( ) ( )2 22 2 2415

a b d a b2πξ η τ+ = +∫ (7)

and

( )2 23

15

I M a b= + (8)

Since the same analysis can be applied for any axis, the other moments of inertia are

DYNAMICS OF RIGID BODIES 357

( )

( )

2 21

2 22

15

15

I M b c

I M a c

= +

= +

(9)

11-4.

The linear density of the rod is

m

ρ = (1)

For the origin at one end of the rod, the moment of inertia is

3

2

0 3 3m m

I x dxρ= = =∫ 2 (2)

If all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be

(3) 2I ma=

Equating (2) and (3), we find

3

a = (4)

This is the radius of gyration.

11-5.

J M

a

Q

z – a

z

a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short interval of time τ so that

( )t dt= ′ ′∫J F (1)

The equations of motion are

ddt

=p

F (2)

ddt

= ×L

r F (3)

358 CHAPTER 11

which, for this case, yield

( )t dt∆ = =′ ′∫p F J (4)

( )t dt∆ = × = ×′ ′∫L r F r J (5)

Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have

( )CM 0; I z a Jω

= = = = × = −V J L r Jp Mω

ω (6)

so that

CM =J

VM

(7)

and

( )0

Jz a

I ω= −

ωω (8)

where ( ) 20 2 5I M= a .

The velocity of any point a on the ball is given by Eq. (11.1):

CMα α= + ×v V rω (9)

For the point of contact Q, this becomes

( )

CM

51

2

Q aJ

J z aM a

= −

− = −

Jv V ω

(10)

Then, for rolling without slipping, 0Q =v , and we have

( )2 5a z a= − (11)

so that

75

z a= (12)

b) Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinning motion (“English”). For the table not to introduce spurious effects, the rail must be at such a height that the ball will be “reflected” upon collision.

Consider the case in which the ball is incident normally on the rail, as in the diagram. We have the following relationships:

DYNAMICS OF RIGID BODIES 359

y

x

VCM

Before Collision After Collision

Linear Momentum CM

0

x

y

p MV

p

= −

=

CM

0

x

y

p MV

p

= +′

=′

Angular Momentum 0

*

0

x

y

z

L

L

L

=

=

=

0

0

x

y

z

L

L L

L

y

=′

= −′

=′

* The relation between and depends on whether or not slipping occurs. yL CMV

Then, we have

CM2 2xp p J MV∆ = − = = (13)

( )02 2yL L I J z aω∆ = − = = − (14)

so that

( )0 CM2 2I MV z aω = − (15)

from which

2 2

0

CM CM CM

2 25 5

I Ma az a

MV MV Vω ω ω

− = = = (16)

If we assume that the ball rolls without slipping before it contacts the rail, then VCM aω= , and we obtain the same result as before, namely,

25

z a− = a (17)

or,

75

z a= (18)

Thus, the height of the rail must be at a height of ( )2 5 a above the center of the ball.

11-6. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have

22(see Problem 11-1)

5sI MR= (1)

360 CHAPTER 11

For the hollow sphere,

θ

R sin θ

( )2

2 2

0 0

4 3

0

4

sin sin

2 sin

83

hI d R R

R d

R

π π

π

dσ φ θ θ

πσ θ θ

πσ

=

=

=

∫ ∫

θ

or, using , we have 24 R Mπσ =

223hI M= R (2)

Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is

21 12 2

T M y I 2θ= + (3)

where y is the measure of the distance along the plane. The potential energy is

( ) sinU Mg y α= − (4)

where is the length of the plane and α is the angle of inclination of the plane. Now, y = Rθ, so that the Lagrangian can be expressed as

2 22

1 1sin

2 2I

L M y y MgyR

α= + + (5)

where the constant term in U has been suppressed. The equation of motion for y is obtained in the usual way and we find

2

2

singMRy

MR Iα

=+

(6)

Therefore, the sphere with the smaller moment of inertia (the solid sphere) will have the greater acceleration down the plane.

DYNAMICS OF RIGID BODIES 361

11-7.

R

d

x

r

φ

θ

The force between the force center and the disk is, from the figure

k= −F r (1)

Only the component along x does any work, so that the effective force is sinxF kr kxφ= − = − .

This corresponds to a potential 2 2x=U k . The kinetic energy of the disk is

2 21 1 32 2 4

T Mx I Mxθ= + = 2 (2)

where we use the result 2 2I MR= for a disk and dx = R dθ. Lagrange’s equations give us

3

02

Mx kx+ = (3)

This is simple harmonic motion about x = 0 with an angular frequency of oscillations

23

km

ω = (4)

11-8.

x3′

x1′x2′

d

h

w

We let be the vertical axis in the fixed system. This would be the axis (i.e., the hinge line) of the door if it were properly hung (no self-rotation), as indicated in the diagram. The mass of the door is M=ρwhd.

3x′

The moment of inertia of the door around the x3′ axis is

23

0 0

13

h wmI dh w d dw

whd= ′ ′ ′∫ ∫ 2Mw= (1)

where the door is considered to be a thin plate, i.e., d w,h.

362 CHAPTER 11

The initial position of the self-closing door can be expressed as a two-step transformation, starting with the position in the diagram above. The first rotation is around the through an angle θ and the second rotation is around the

1 -axisx′

1 -axisx′′ through an angle ψ:

w

h

x3′

x2′x1′

x3′ x3″

x2′

x2″x1′ = x1″

θ

θ

x3′ x3 = x3″w3 = w

x2′x2

x2″x1″ x1

ψψ

The -axes are the fixed-system axes and the are the body system (or rotating) axes which are attached to the door. Here, the Euler angle φ is zero.

1x′ -axesix

The rotation matrix that transforms the fixed axes into the body axes ( )ix x→′ i is just Eq. (11.99)

with φ = 0 and θ → – θ since this rotation is performed clockwise rather than counterclockwise as in the derivation of Eq. (11.99):

cos cos sin sin sinsin cos cos sin cos

0 sin cos

ψ θ ψ θ ψψ θ ψ θ ψ

θ θ

− = − −

λ (2)

The procedure is to find the torque acting on the door expressed in the fixed coordinate system and then to obtain the component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle

3x

0ψ , the rotation is in the direction to decrease ψ. According to Eq. (11.119),

3 3 3 3I N Iω ψ= = (3)

where 1 2 0ω ω= = since . 0φ θ= =

In the body ( system the coordinates of the center of mass of the door are )ix

0

12

R w

h

=

(4)

where we have set the thickness equal to zero. In the fixed ( )ix′ system, these coordinates are

obtained by applying the inverse transformation 1λ− to R; but 1 tλ λ− = , so that

sin

1cos cos sin

2sin cos cos

t

w

w h

w h

ψθ ψ θθ ψ θ

− = = +′ − +

λR R (5)

Now, the gravitational force acting on the door is downward, and in the coordinate system is

ix′

3Mg= −′ ′F e (6)

DYNAMICS OF RIGID BODIES 363

There the torque on the door, expressed in the fixed system, is

1 2 31sin cos cos sin sin cos cos

20 0 1

cos cos sin1

sin2

0

Mg w w h w h

w h

Mg w

= ×′ ′ ′

′ ′ ′ = − − + − +

+ = −

N R F

e e e

ψ θ ψ θ θ ψ θ

θ ψ θψ (7)

so that in the body system we have

2 2cos cos sin cos cos sin1

sin sin2

sin sin

w h w

Mg h

w

θ ψ θ ψ θθ ψθ ψ

+ + = = − −′ ′

λψ

N N (8)

Thus,

31

sin sin2

N Mgw θ ψ− = (9)

and substituting this expression into Eq. (3), we have

23

1 1sin sin

2 3Mgw I Mwθ ψ ψ= = ψ− (10)

where we have used Eq. (1) for . Solving for 3I ψ ,

3

sin sin2

gw

ψ θ= − ψ (11)

This equation can be integrated by first multiplying by ψ :

21 3sin sin

2 2

3sin cos

2

gdt dt

w

gw

= = −

=

∫ ∫ψ ψ ψ θ ψ ψ

θ ψ (12)

where the integration constant is zero since cos 0ψ = when 0ψ = . Thus,

3

sin cosg

wψ θ= ± ψ (13)

We must choose the negative sign for the radical since 0ψ < when cos 0ψ > . Integrating again,

from ψ = 90° to ψ = 0°,

0

02

3sin

cos

Tgddt

ψθ

ψ= −∫ ∫ (14)

364 CHAPTER 11

where T = 2 sec. Rewriting Eq. (14),

2

0

3sin

cosgd

Tw

π

ψ θψ

=∫ (15)

Using Eq. (E. 27a), Appendix E, we find

2

1 2

0

14cos324

d

π

πψ ψ−

Γ = Γ

∫ (16)

From Eqs. (E.20) and (E.23),

1 1 11 0.906

4 4 4

13.624

4

Γ = Γ =

Γ = (17)

And from Eqs. (E.20) and (E.24),

3 3 31 0.919

4 4 4

31.225

4

Γ = Γ =

Γ = (18)

Therefore,

2

0

3.6242.62

2 1.225cosd

π

ψ πψ

= =∫ (19)

Returning to Eq. (15) and solving for sin θ,

( 22sin 2.62

3wgT

θ = × ) (20)

Inserting the values for g, w(= 1m), and T(= 2 sec), we find

( )1sin 0.058θ −=

or,

3.33θ ≅ ° (21)

DYNAMICS OF RIGID BODIES 365

11-9.

a

O

R

Q P

C C′

θ

y

x

The diagram shows the slab rotated through an angle θ from its equilibrium position. At equilibrium the contact point is Q and after rotation the contact point is P. At equilibrium the position of the center of mass of the slab is C and after rotation the position is C′.

Because we are considering only small departures from θ = 0, we can write

QP Rθ≅ (1)

Therefore, the coordinates of C′ are (see enlarged diagram below)

= + ′r OA AC (2)

so that

sin cos2

cos sin2

ax R R

ay R R

θ θ θ

θ θ θ

= + −

= + +

(3)

C

Q

C′

P

O

A

θ

θ

Consequently,

366 CHAPTER 11

cos cos sin2

cos sin2

sin cos sin2

sin cos2

ax R R R

aR

ay R R R

aR

θ θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

= + − +

= +

= − + + +

= − +

θ

from which

2

2 2 2 2

4a

x y R 2θ θ

+ = + (4)

The kinetic energy is

( )2 21 12 2

T M x y I 2θ= + + (5)

where I is the moment of inertia of the slab with respect to an axis passing through the center of mass and parallel to the z-axis:

( )2 2112

I M a= + (6)

Therefore,

( ) 21

12

T f θ θ= (7)

where

( )2

2 21 4

af M Rθ

Iθ= +

+ (8)

The potential energy is

( )2U Mgy f θ= = − (9)

where

( )2 cos sin2a

f Mg R Rθ θ θ θ = − + + (10)

and where Eq. (3) has been used for y.

The Lagrangian is

( ) ( )21

12

L f f2θ θ= + θ (11)

The Lagrange equation for θ is

DYNAMICS OF RIGID BODIES 367

0d L Ldt θθ

∂ ∂− =∂∂

(12)

Now,

( )1L

f θ θθ∂

=∂

( ) ( )1 1

22 2 2 22

4

d Lf f

dt

aM R I MR

∂= +

= + + +

θ θ θ θθ

θ θ θ θ (13)

( ) ( )21 2

2 2

12

sin cos sin2

Lf f

aMR Mg R R R

∂= +′ ′

+ − − = +

θ θ θθ

θ θ θ θ θ θ (14)

Combining, we find

22 2 2 2 sin cos sin 0

4 2a a

M R I MR Mg R R Rθ θ θ θ θ θ θ θ + + + − + − −

= (15)

For the case of small oscillations, 2θ θ and 2θ θ , so that Eq. (15) reduces to

22 0

4

aMg R

MaI

θ

− θ+ =

+ (16)

The system is stable for oscillations around θ = 0 only if

22

2 0

4

aMg R

MaI

ω

− = >

+ (17)

This condition is satisfied if 2 0R a− > , i.e.,

2a

R > (18)

Then, the frequency is

( )

22 2

21

4 12

aMg R

MaM a

ω

− =

+ + (19)

Simplifying, we have

368 CHAPTER 11

( )2 2

122

4

ag R

− =

+ (20)

According to Eqs. (9) and (10), the potential energy is

( ) cos sin2a

U Mg R Rθ θ θ θ = + + (21)

This function has the following forms for 2R a> and 2R a< :

U(θ) U(θ)

θ

−π/2 π/2

Ra

>2 R

a<

2

Mg Ra

+

2

θ

To verify that a stable condition exists only for 2R a> , we need to evaluate 2U 2θ∂ ∂ at θ = 0:

sin cos2

U aMg Rθ θ θ

θ∂ = − + ∂

(22)

2

2 cos cos sin2

U aMg R Rθ θ θ θ

θ∂ = − + − ∂

(23)

and

2

20

2U

Mg Rθθ =

∂ = −∂a

(24)

so that

2

2 0 if 2

UR

θ∂

> >∂

a (25)

11-10.

z

m

When the mass m is at one pole, the z component of the angular momentum of the system is

225zL I MRω ω= = (1)

After the mass has moved a distance vt = Rθ along a great circle on the surface of the sphere, the z component of the angular momentum of the system is

DYNAMICS OF RIGID BODIES 369

2 2 22sin

5zL MR mR θ φ = + (2)

where φ is the new angular velocity. Since there is no external force acting on the system, angular momentum must be conserved. Therefore, equating (1) and (2), we have

2

2 2 2

25

2sin

5

MR

MR mR

ωφ

θ=

+ (3)

Substituting vt Rθ = and integrating over the time interval during which the mass travels from one pole to the other, we have

( )

2

2 2 20

25

2sin

5

Rt

V

t

MRdt

MR mR vt R

πω

φ=

=

=+

∫ (4)

Making the substitutions,

( ),vt R u dt R v du≡ = (5)

we can rewrite (4) as

2

2 2 20

2

20

25

2sin

5

21 sin

MR Rdu

vMR mR u

R duv u

=+

=+

π

π

ωφ

ωβ

(6)

where 5 2m Mβ ≡ and where we have used the fact that the integrand is symmetric around

2u π= to write φ as twice the value of the integral over half the range. Using the identity

(2 1sin 1 cos 2

2u = − )u (7)

we express (6) as

2

0

21 1

1 c2 2

R duv u

πωφ

β β=

+ −

∫os 2

(8)

or, changing the variable to x = 2u,

0 1 1

1 c2 2

R dxv x

π

os

ωφ

β β=

+ −

∫ (9)

Now, we can use Eq. (E.15), Appendix E, to obtain

370 CHAPTER 11

( ) ( )1

0

1 tan 22tan

1 1

22 5

22 5

xRv

R Mv M m

MT

M m

− +

= + +

=+

=+

πβω

φβ β

π ω

ω (10)

where T R vπ= is the time required for the particle to move from one pole to the other.

If m = 0, (10) becomes

( )0m Tφ ω= = (11)

Therefore, the angle of retardation is

( ) ( )0m mα φ φ= = − (12)

or,

2

12 5

MT

M mα ω

= − +

(13)

11-11.

a) No sliding:

PP

22

From energy conservation, we have

2C.M.

1 12 2 22

mg mg mv I 2ω= + + (1)

where v is the velocity of the center of mass when one face strikes the plane; v is related to ω by

CM C.M.

CM 2v = ω (2)

I is the moment of inertia of the cube with respect to the axis which is perpendicular to one face and passes the center:

216

I m= (3)

Then, (1) becomes

DYNAMICS OF RIGID BODIES 371

( )2 2

2 21 1 12 1

2 2 2 6 32mg m

mω 2mω ω

− = + = (4)

from which, we have

( )2 32 1

2g

ω = − (5)

b) Sliding without friction:

In this case there is no external force along the horizontal direction; therefore, the cube slides so that the center of mass falls directly downward along a vertical line.

PθP

While the cube is falling, the distance between the center of mass and the plane is given by

cos2

y θ= (6)

Therefore, the velocity of center of mass when one face strikes the plane is

0 4 4

1 1sin

2 22= =

= − = − = −

π θ π

y θ θ θ ω (7)

From conservation of energy, we have

2

21 1 1 12 2 2 2 62

mg m m 2mg ω ω = + − + (8)

from which we have

( )2 122 1

5g

ω = − (9)

11-12. According to the definition of the principal moments of inertia,

( ) ( )( )

2 2 2 2

2 2 2

2

2

2

j k i k i j

j k i

i i

I I x x dv x x dv

x x dv x dv

I x dv

+ = + + +

= + +

= +

∫ ∫

∫ ∫

ρ ρ

ρ ρ

ρ (1)

since

2 0ix dvρ >∫we have

372 CHAPTER 11

j k iI I I+ ≥ (2)

11-13. We get the elements of the inertia tensor from Eq. 11.13a:

( )

( ) ( ) ( )

2 211 ,2 ,3

2 2 23 4 2 2 13

I m x x

m b m b m b mb

α α αα

= +

= + + =

2

b b

2

Likewise and 222 16I m= 2

33 15I m=

( ) ( )

12 21 ,1 ,2

2 24 2 2

I I m x x

m b m b mb

α α αα

= = −

= − − − = −

Likewise 213 31I I mb= =

and I I 223 32 4mb= =

Thus the inertia tensor is

2

13 2 12 16 4

1 4 15I mb

− = −

The principal moments of inertia are gotten by solving

2

13 2 12 16 4

1 4 15mb

λλ

λ

− − 0− − = −

Expanding the determinant gives a cubic equation in λ:

3 244 622 2820 0λ λ λ− + − =

Solving numerically gives

1

2

3

10.00

14.35

19.65

λ

λ

λ

=

=

=

21

22

23

Thus the principal moments of inertia are 10

14.35

19.65

I mb

I mb

I mb

=

=

=

To find the principal axes, we substitute into (see example 11.3):

DYNAMICS OF RIGID BODIES 373

( )

( )

( )

1 2 3

1 2

1 2 3

13 2 0

2 6 4

4 15 0

i i i i

i i i i

i i i i

− − + =

− + 1 − + =

+ + − =

λ ω ω ω

ω λ ω ω

ω ω λ ω

3 0

For i = 1, we have ( )1 10λ =

11 21 31

11 21 31

11 21 31

3 2 0

2 6 4

4 5 0

0

ω ω ω

ω ω ω

ω ω ω

− + =

− − + =

− + =

Solving the first for 31ω and substituting into the second gives

11 21ω ω=

Substituting into the third now gives

31 21ω ω= −

or

11 21 31: : 1 : 1 : 1ω ω ω = −

So, the principal axis associated with is 1I

( )13

x y z+ −

Proceeding in the same way gives the other two principal axes:

2 : .81 .29 .52

3 : .14 .77 .63

i

i

= − + −

= − + +

x y

x y

z

z

We note that the principal axes are mutually orthogonal, as they must be.

11-14.

z

y

x Let the surface of the hemisphere lie in the x-y plane as shown. The mass density is given by

33

32 23

M M MV bb

ρππ

= = =

First, we calculate the center of mass of the hemisphere. By symmetry

374 CHAPTER 11

CM CM

CM

0

1v

x y

z z dvM

ρ

= =

= ∫

Using spherical coordinates 2( cos , sin )z r dv r dr d dθ θ θ= = φ we have

( )

223

CM0 0 0

43

sin cos

3 1 1 32

2 2 4 8

b

r

z d dM

b bb

ππ

φ θ

ρφ θ θ θ

ππ

= = =

=

= =

∫ ∫ ∫ r dr

We now calculate the inertia tensor with respect to axes passing through the center of mass:

z′ = z

x′

y′

38

b

By symmetry, . Thus the axes shown are the principal axes. 12 21 13 31 23 32 0I I I I I I= = = = = =

Also, by symmetry I . We calculate I using Eq. 11.49: 11 22I= 11

2

11 1138

I J M v = − (1)

where 11J = the moment of inertia with respect to the original axes

( )

( )

( )

( )

2 211

2 2 2 2 2 2

2 24 2 2 2

30 0 0

223 2

0

2

sin sin cos sin

3sin sin cos sin

2

3sin 2 cos sin

10

25

v

v

b

r

J y z dv

r r r dr d d

Mr dr d d

b

Mbd

Mb

π π

θ φ

π

θ

ρ

θ φ θ θ θ φ

θ φ θ φ θπ

π θ π θ θ θπ

= = =

=

= +

= +

= +

=

∫ ∫ ∫

θ= +

Thus, from (1)

2 211 22

2 9 835 64 320

I I Mb Mb Mb= = − = 2

Also, from Eq. 11.49

DYNAMICS OF RIGID BODIES 375

( )33 33 330I J M J= − =

( should be obvious physically) 33 33I J=

So

( )2 233

4 3 2sin

5

v

v

I x y dv

r dr d d M

ρ

ρ θ θ φ

= +

= =

∫ 2b

211 22

233

Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are

83I

320

2I

5

I Mb

Mb

= =

=

11-15.

θ

P

g

We suspend the pendulum from a point P which is a distance from the center of mass. The rotational inertia with respect to an axis through P is

(1) 20I MR M= + 2

where is the radius of gyration about the center of mass. Then, the Lagrangian of the system is

0R

(2

1 cos2

IL T U Mg

θ )θ= − = − − (2)

Lagrange’s equation for θ gives

sin 0I Mgθ θ+ = (3)

For small oscillation, sin θ θ≅ . Then,

0Mg

Iθ θ+ = (4)

or,

2 20

0g

Rθ θ+ =

+ (5)

376 CHAPTER 11

from which the period of oscillation is

2 202

2R

τ πω

+= = (6)

If we locate another point P′ which is a distance ′ from the center of mass such that the period

of oscillation is also τ, we can write

2 2 20 0R Rg g+ + 2′

=′

(7)

from which . Then, the period must be 20R = ′

2

2g

τ π+′

= (8)

or,

2g

τ π+ ′

= (9)

This is the same as the period of a simple pendulum of the length + ′

1

. Using this method, one does not have to measure the rotational inertia of the pendulum used; nor is one faced with the problem of approximating a simple pendulum physically. On the other hand, it is necessary to locate the two points for which τ is the same.

11-16. The rotation matrix is

( )cos sin 0sin cos 00 0

θ θθ θ

= −

λ (1)

The moment of inertia tensor transforms according to

( ) ( ) ( )( )t=′Ι λ Ι λ (2)

That is

( )

( ) ( )

( ) ( )

1 10

2 2cos sin 0 cos sin 01 1

sin cos 0 0 sin cos 02 2

0 0 1 0 00 0

A B A B

I A B A B

C

θ θ θ θθ θ θ θ

+ − −

= − − +′

1

DYNAMICS OF RIGID BODIES 377

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 1 1 1cos sin sin cos 0

2 2 2 2cos sin 01 1 1 1

sin cos 0 cos sin sin cos 02 2 2 2

0 0 1 0 0

A B A B A B A B

A B A B A B A B

C

θ θ θθ θθ θ θ θ θ θ

+ + − − + + − = − − + + − − + +

θ

( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

2 2

2 2

2 2

2 2

1 1cos cos sin sin

2 21 1

sin cos2 2

0

1 1cos sin 0

2 21 1

sin sin cos cos 02 2

0

A B A B A B

A B A B

A B A B

A B A B A B

C

+ + − + += − − + −

− − − + − − + +

θ θ θ θ

θ θ

θ θ

θ θ θ θ

or

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2

2 2

1 1 1cos sin cos sin 0

2 2 21 1 1

sin cos cos sin 02 2 2

0 0

A B A B A B A B

A B A B A B A B

C

θ θ θ θ

θ θ θ θ

+ + − − − − = − − + − + − −′

I (3)

If 4θ π= , sin cos 1 2θ θ= = . Then,

( )0 0

0 00 0

A

B

C

=′

I (4)

11-17.

x3

x2

x1

378 CHAPTER 11

The plate is assumed to have negligible thickness and the mass per unit area is sρ . Then, the inertia tensor elements are

( )2 211 1 1 2s x dxρ= −∫I r dx

( )2 2 22 3 1 2 2 1 2s sx x dx dx x dx dx Aρ ρ= ≡∫ ∫= + (1)

( )2 2 222 2 1 2 1 1 2s sx dx dx x dx dxρ ρ= − =∫ ∫I r (2) B≡

( ) ( )2 2 2 233 3 1 2 1 2 1s sr x dx dx x x dx dxρ ρ= − = +∫ ∫ 2

B

I (3)

Defining A and B as above, becomes 33I

33I A= + (4)

Also,

( )12 1 2 1 2sI x x dx dxρ C= −∫ ≡ − (5)

( )21 2 1 1 2sI x x dx dxρ C= −∫ = − (6)

( )13 1 3 1 2 310sI x x dx dxρ I= − =∫ = (7)

( )23 2 3 1 2 320sI x x dx dxρ I= − =∫ = (8)

Therefore, the inertia tensor has the form

00

0 0

A C

C B

A B

− = − +

I (9)

11-18. The new inertia tensor ′I is obtained from I by a similarity transformation [see Eq. (11.63)]. Since we are concerned only with a rotation around the , the transformation matrix is just

3 -axisx

φλ , as defined in Eq. (11.91). Then,

1φ φ

−=′I Iλ λ (1)

where

1 tφ φ− =λ λ (2)

Therefore, the similarity transformation is

= −I cos sin 0 0 cos sin 0sin cos 0 0 sin cos 00 0 1 0 0 0 0

A C

C B

A B

θ θ θ θθ θ θ θ

− − −′ + 1

Carrying out the operations and simplifying, we find

DYNAMICS OF RIGID BODIES 379

( )

( )

2 2

2 2

1cos sin 2 sin cos 2 sin 2 0

21

cos 2 sin 2 sin sin 2 cos 020 0

A C B C B A

C B A A C B

A B

θ θ θ θ θ

θ θ θ θ θ

− + − + − = − + − + +′ +

I (3)

Making the identifications stipulated in the statement of the problem, we see that

11 22

12 21

,I A I B

I I C

= =′ ′ ′ ′ = = −′ ′ ′

(4)

and

33I A B A B= + = +′ ′ ′ (5)

Therefore

00

0 0

A C

C B

A B

−′ ′ = −′ ′ ′ +′ ′

I (6)

In order that and be principal axes, we require C′ = 0: 1x 2x

( )1cos 2 sin 2 0

2C B Aθ θ− − = (7)

or,

2

tan 2C

B Aθ =

− (8)

from which

11 2tan

2C

B Aθ − = −

(9)

Notice that this result is still valid if A = B. Why? (What does A = B mean?)

11-19.

θ = 0

x2

xη θ

θ = π

θ = π/2

1

The boundary of the plate is given by r keαθ= . Any point (η,θ ) has the components

1

2

cos

sin

x

x

η θ

η θ

= =

(1)

380 CHAPTER 11

The moments of inertia are

21 20 0

2 3

0 0sin

ke

ke

I A x d d

d d

αθ

αθ

π

π

ρ η η θ

ρ θ θ η η

= =

=

∫ ∫

∫ ∫

The integral over θ can be performed by using Eq. (E.18a), Appendix E, with the result

4

1 2k

I A P= =ρα

(2)

where

( )4

2

116 1 4

eP

πα

α−

=+

(3)

In the same way,

22 10 0

2

0 0cos

ke

ke

I B x d d

d

= =

=

∫ ∫

∫ ∫

αθ

αθ

π

π 3 d

ρ η η θ

ρ θ θ η η (4)

Again, we use Eq. (E.18a) by writing co 2s 1 sin2θ θ= − , and we find

( )4

22 1 8

2k

I B Pρ

αα

= = + (5)

Also

12 1 20 0

3

0 0cos sin

ke

ke

I C x x d d

d d

αθ

αθ

π

π

ρ η η θ

ρ θ θ θ η

= − = −

= −

∫ ∫

∫ ∫ η

(6)

In order to evaluate the integral over θ in this case we write ( )cos sin 1 2 sin 2θ θ θ= and use

Eq. (E.18), Appendix E. We find

(7) 412I C kρ= − = P

Using the results of problem 11-17, the entire inertia tensor is now known.

According to the result of Problem 11-18, the angle through which the coordinates must be rotated in order to make I diagonal is

11 2tan

2C

B Aθ − = −

(8)

Using Eqs. (2), (5), and (7) for A, B, C, we find

2 1

2C

B A α=

− (9)

DYNAMICS OF RIGID BODIES 381

so that

1

tan 22

θα

= (10)

Therefore, we also have

2θ2α

11 4 2+ α

2

2

1sin 2

1 4

2cos 2

1 4

θα

αθ

α

= + =+

(11)

Then, according to the relations specified in Problem 11-18,

21 cos sin 2 sinI A A C B 2θ θ= = − +′ ′ θ (12)

Using ( ) ( )2cos 1 2 1 cos 2θ θ= + and ( ) ( )2sin 1 2 1 cos 2θ θ= − , we have

( ) ( )11 1

cos 2 sin 22 2

A B A B CI A θ θ= = + + − −′ ′ (13)

Now,

( )

42

4

1 4

4

k PA B

A B k P

ρα

α

αρ

+ = +

− = −

(14)

Thus,

( )4

2 4 41 2 2

2 11 4 2

2 1 4 1 4

k Pk P k P

ρ αα αρ ρ

αI A

α α= = + − × − ×′ ′

+ + (15)

or,

( )41I A k P Q Rρ= = −′ ′ (16)

where

2

2

1 42

1 4

Q

R a

αα

+=

= +

(17)

Similarly,

( )42I B k P Q Rρ= = +′ ′ (18)

and, of course,

382 CHAPTER 11

3I A B I I1 2= + = +′ ′ ′ ′ ′ (19)

We can also easily verify, for example, that I12 0C= − =′ ′ .

11-20. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy

1 2b

E U mg= =

(2b

is the height of the center of mass above the floor.)

When the rod hits the floor, the potential energy is zero. Thus

22

12

E T Iω= =

where I is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length σ:

2 3end

0

1 13 3

b

I x dx bσ σ= = =∫ 2mb

Thus

2 22

16

T mb ω=

By conservation of energy

1 2U T=

2 212 6b

mg mb ω=

3gb

ω =

11-21. Using I to denote the matrix whose elements are those of I , we can write

=L Iω (11.54)

=′ ′ ′L I ω (11.54a)

We also have and and therefore we can express L and ω as =′x λ x ′xt=′x λ

t= ′L Lλ (11.55a)

t= ′ω λ ω (11.55b)

substituting these expressions into Eq. (11.54), we have

DYNAMICS OF RIGID BODIES 383

t t=′ ′L Iλ λ ω

and multiplying on the left by λ,

t t=′ ′L Iλλ λ λ ω

or

( )t=′ ′L Iλ λ ω

by virture of Eq. (11.54a), we identify

t=′I Iλ λ (11.61)

11-22. According to Eq. (11.61),

1

,ij ik k j

k l

I Iλ λ−=′ ∑ (1)

Then,

1

,

1

,

,

ii ik k ii i k

k i ikk i

k k kkk k

tr I I

I

I I

= =′ ′

=

= =

∑ ∑∑

∑ ∑

∑ ∑

I λ λ

λ λ

δ (2)

so that

tr tr=′I I (3)

This relation can be verified for the examples in the text by straightforward calculations.

Note: A translational transformation is not a similarity transformation and, in general, tr is not invariant under translation. (For example,

I

tr I will be different for inertia tensors expressed in coordinate system with different origins.)

11-23. We have

1−=′I Iλ λ (1)

Then,

384 CHAPTER 11

1

1

1

=′

= × ×

= ×

I I

I

I

λ λ

λ λ

λλ

so that,

=′I I (2)

This result is easy to verify for the various examples involving the cube.

11-24.

–a/2 a/2x2

x1

−3

2a

The area of the triangle is 23 4A a= , so that the density is

2

43

M MA a

ρ = = (1)

a) The rotational inertia with respect to an axis through the point of suspension (the origin) is

x2

x1

θ

( )

( )( )1

2 23 1 2 1 2

2 02 2

1 1 20 3

22

4 2

2

3 124 6

a

a x

I x x dx dx

dx x x dx

a Ma

− −

= +

= +

= =

∫ ∫

ρ

ρ

ρ

2

(2)

When the triangle is suspended as shown and when θ = 0, the coordinates of the center of mass are 2(0, ,0)x , where

DYNAMICS OF RIGID BODIES 385

( )1

2 2 1 2

2 0

1 20 3 2

1

2

2 3

a

a x

x x dx dxM

dx x dxM

a

− −

=

=

= −

∫ ∫

ρ

ρ2

(3)

The kinetic energy is

23

1 12 12

T I Ma2 2θ θ= = (4)

and the potential energy is

(1 cos2 3Mga

U )θ= − (5)

Therefore,

2 21cos

12 2 3Mga

L Ma θ θ= + (6)

where the constant term has been suppressed. The Lagrange equation for θ is

3 sin 0ga

θ θ+ = (7)

and for oscillations with small amplitude, the frequency is

3ga

ω = (8)

b) The rotational inertia for an axis through the point of suspension for this case is

x2

x1

−3

2a

( )2 30

2 23 2 1 2

032

2

2

512

x

a

I dx x x

Ma

= +′

=

∫ ∫ρ 1dx

(9)

The Lagrangian is now

2 25cos

24 3Mga

L Ma θ θ= + (10)

386 CHAPTER 11

and the equation of motion is

12

sin 05 3

ga

θ θ+ = (11)

so that the frequency of small oscillations is

12

5 3ga

ω = (12)

which is slightly smaller than the previous result.

11-25.

x2

x1

R

r

ρθ

The center of mass of the disk is 2(0, )x , where

( ) ( )

2 2 1 2 2 1 2lower upper

semicircle semicircle

2

0 0 0

3

2

sin 2 sin

23

R R

x x dx dx x dx dxM

r rdrd r rdrdM

RM

= +

= ⋅ +

∫ ∫

∫ ∫ ∫ ∫π π

π

ρ

ρ⋅

= −

θ θ θ

ρ

θ

(1)

Now, the mass of the disk is

2 2

2

1 12

2 2

32

M R

R

= ⋅ + ⋅

=

Rρ π ρ π

ρπ (2)

so that

24

9x R

π= − (3)

The direct calculation of the rotational inertia with respect to an axis through the center of mass is tedious, so we first compute I with respect to the and then use Steiner’s theorem. 3 -axisx

DYNAMICS OF RIGID BODIES 387

22 23 0 0 0

4 2

2

3 14 2

R RI r rdr d r rd

R MR

= ⋅ + ⋅

= =

∫ ∫ ∫ ∫π π

πr dρ θ θ

πρ (4)

Then,

20 3 2

2 22

22

1 12 81

1 321

2 81

I I Mx

6MR M R

MR

= −

= − ⋅

= −

π

π (5)

When the disk rolls without slipping, the velocity of the center of mass can be obtained as follows:

Thus

RCM

θ

θ

CM 2

CM 2

sin

cos

x R x

y R x

θ θ

θ

= −

= −

ρ

x2

CM 2

CM 2

cos

sin

x R x

y x

θ θ θ

θ θ

= −

=

( )2 2 2 2 2 2 2 2CM CM 2 22 cox y V R x R x sθ θ θ+ = = + − θ

2

2 2V a θ= (6)

where

2 22 22 cosa R x R x θ= + − (7)

Using (3), a can be written as

2

16 81

81 9a R cosθ

π π= + − (8)

The kinetic energy is

388 CHAPTER 11

trans rot

20

1 12 2

T T T

Mv I 2

= +

= + θ (9)

Substituting and simplifying yields

2 21 3 8cos

2 2 9T MR θ θ

π = −

(10)

The potential energy is

21

cos2

1 81 cos

2 9

U Mg R x

MgR

= +

= −

θ

θπ

(11)

Thus the Lagrangian is

21 3 8 8cos 1 cos

2 2 9 9R R gL M θ θ

π π θ = − − −

(12)

11-26. Since =φ φω lies along the fixed 3 -axisx′ , the components of φω along the body axes

are given by the application of the transformation matrix λ [Eqs. (11.98) and (11.99)]: ( )ix

( )( )( )

1 1

22

33

00

φ

φ

φ

ω φ

ω φ

φ φω

= =

λ (1)

Carrying out the matrix multiplication, we find

( )( )( )

1

2

3

sin sincos sin

cos

φ

φ

φ

ω ψ θω φ ψ

θω

θ =

(2)

which is just Eq. (11.101a).

The direction of =θ θω coincides with the line of nodes and lies along the axis. The components of

1x′′′

θω along the body axes are therefore obtained by the application of the transformation matrix ψλ which carries the xi′′′ system into the system: ix

( )( )( )

1

2

3

cos0 sin0 0

θ

θ ψ

θ

ω θ ψω θω

ψ

= = −

λ (3)

DYNAMICS OF RIGID BODIES 389

which is just Eq. (11.101b).

Finally, since ψω lies along the body , no transformation is required: 3 -axisx

( )( )( )

1

2

3

001

ψ

ψ

ψ

ω

ω ψ

ω

=

(4)

which is just Eq. (11.101c).

Combining these results, we obtain

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

11 1

22 2

33 3

sin sin cos

cos sin sin

cos

+ + + + + +

+

= − +

φ θ ψ

φ θ ψ

φ θ ψ

ω ω ω

ω ω ω

ω ω ω

φ ψ θ θ ψ

φ φ ψ θ θ ψ

θ θ ψ

ω =

(5)

which is just Eq. (11.02).

11-27.

L

α

θ

ωx3x2

x3′

Initially:

1 1 1

2 1 2 1

3 3 3 3

0

sin sin

cos

L I

L L I I

L L xos I I

ω

θ ω ω α

θ ω ω α

= =

= = =

= = =

Thus

2 1

3 3

tan tanL IL I

θ α= = (1)

From Eq. (11.102)

3 cosω φ θ ψ= +

Since 3 cosω ω= α , we have

390 CHAPTER 11

cos cosφ θ ω α ψ= − (2)

From Eq. (11.131)

3 13

1

I II

ψ ω−

= −Ω = −

(2) becomes

3

1

cos cosII

=φ θ ω α (3)

From (1), we may construct the following triangle

I3

I3 tan α

θ

from which 31 22 2 2

3 1

stan

I

I Iθ

α= +

co

Substituting into (3) gives

2 2 2 21 3

1

sin cosI IIω

φ α α= +

11-28. From Fig. 11-7c we see that =φ φω is along the 3 -axisx′ , =θ θω is along the line of

nodes, and =ψ ψω is along the . Then, 3 -axx is

3φ φ=′ ′eω (1)

whee is the unit vector in the direction. 3′e 3x′

Projecting the lines of nodes into the 1 -x′ and 2 -axesx′ , we obtain

( )1 2cos sinθ θ φ= +′ ′ ′e eω φ (2)

ψω′ has components along all three of the xi′ axes. First, we write ψ′ω in terms of a component

along the x and a component normal to this axis: 3 -axis′

( )12 3sin cosψ ψ θ= +′ ′ ′e eω θ (3)

where

12 1 2sin cosφ φ= −′ ′ ′e e e (4)

Then,

( )1 2 3sin sin sin cos cosψ ψ θ φ θ φ θ= − +′ ′ ′ ′e e eω (5)

DYNAMICS OF RIGID BODIES 391

Collecting the various components, we have

1

2

3

cos sin sin

sin sin cos

cos

ω θ φ ψ θ φ

ω θ φ ψ θ

ω ψ θ φ

= +′

= −′

= +′

φ (6)

11-29. When the motion is vertical θ = 0. Then, according to Eqs. (11.153) and (11.154),

( )3P I Pφ ψφ ψ= + = (1)

and using Eq. (11.159), we see that

3 3P P Iψ φ ω= = (2)

Also, when θ = 0 (and ), the energy is [see Eq. (11.158)] 0θ =

23 3

12

E I Mgω= + h (3)

Furthermore, referring to Eq. (11.160),

23 3

12

E E I Mghω= − =′ (3)

If we wish to examine the behavior of the system near θ = 0 in order to determine the conditions for stability, we can use the values of Pψ , Pφ , and E′ for θ = 0 in Eq. (11.161). Thus,

( )22 2

3 3212 2

12

1 cos1cos

2 2 sin

II Mgh

I

ω θMgh θ θ

θ−

= + + (5)

Changing the variable to z = cos θ and rearranging, Eq. (5) becomes

( ) ( )

22

12 3 3212

12 1

z 2 2z Mgh I z I wI−

= + − (6)

The questions concerning stability can be answered by examining this expression. First, we note that for physically real motion we must have 2 0z ≥ . Now, suppose that the top is spinning very rapidly, i.e., that 3ω is large. Then, the term in the square brackets will be negative. In such a

case, the only way to maintain the condition 2 0z ≥ is to have z = 1, i.e., θ = 0. Thus, the motion at θ = 0 will be stable as long as

2 212 3 34Mgh I I ω 0− < (7)

or,

122 23 3

41

Mgh II ω

< (8)

392 CHAPTER 11

Suppose now that the top is set spinning with θ = 0 but with 3ω sufficiently small that the condition in Eq. (8) is not met. Any small disturbance away from θ = 0 will then give z a negative value and θ will continue to increase; i.e., the motion is unstable. In fact, θ will continue until z reaches a value 0z that again makes the square brackets equal to zero. This is a turning point for the motion and nutation between z = 1 and 0z z= will result.

From this discussion it is evident that there exists a critical value for the angular velocity, cω , such that for 3 cω ω> the motion is stable and for cω ω< there is nutation:

12

3

2c

Mgh I

Iω = (9)

If the top is set spinning with 3 cω ω> and θ = 0, the motion will be stable. But as friction slows the top, the critical angular velocity will eventually be reached and nutation will set in. This is the case of the “sleeping top.”

11-30. If we set , Eq. (1.162) becomes 0θ =

( )( )

( )

2

212

coscos

2 1 cos

P PMgh

Iφ ψ θ

E V θ θθ

−= = +′

− (1)

Re-arranging, this equation can be written as

( ) ( ) ( ) ( )3 2 212 12 12 122 cos 2 cos 2 cos 2 0Mgh I E I P P P Mgh I E I Pψ φ ψθ θ θ− + + − + −′ ′ 2

φ = (2)

which is cubic in cos θ.

V(θ ) has the form shown in the diagram. Two of the roots occur in the region 1 cos 1θ− ≤ , and one root lies outside this range and is therefore imaginary.

V(θ)

cos θ–1

+1

DYNAMICS OF RIGID BODIES 393

11-31. The moments of inertia of the plate are

( )

1 2

2

3 1 2

2

22

cos 2

1 cos 2

2 cos

I I

I

I I I

I

I

α

α

α

=

= + = + =

=

(1)

We also note that

( )1 2 2

22

1 cos 2

2 sin

I I I

I

α

α

− = − −

= − (2)

Since the plate moves in a force-free manner, the Euler equations are [see Eq. (11.114)]

( )

( )

( )

1 2 1 2 3 3

2 3 2 3 1 1

3 1 3 1 2 2

0

0

0

I I I

I I I

I I I

ω ω ω

ω ω ω

ω ω ω

− − =− − = − − =

(3)

Substituting (1) and (2) into (3), we find

( ) ( )

( ) ( )

2 22 1 2 2 3

2 2 3 2 1

2 3 1 2 2

2 sin 2 cos 0

cos 2 cos 2 0

0

I I

I I

I I

α ω ω α ω

α ω ω α ω

ω ω ω

− − =

− − = − =

(4)

These equations simplify to

23 1 2

1 2 3

2 3 1

tan = −= −=

ω ω ω α

ω ω ω

ω ω ω

(5)

From which we can write

21 2 3 2 2 1 1 3 3 cotω ω ω ω ω ω ω ω ω α= = − = − (6)

Integrating, we find

( ) ( ) ( )2 2 2 2 2 2 2 22 2 1 1 3 30 0 cot 0 cotω ω ω ω ω α ω− = − + = − + α (7)

Now, the initial conditions are

394 CHAPTER 11

( )

( )

( )

1

2

3

0 cos

0 0

0 sin

ω α

ω

ω α

= Ω =

= Ω

(8)

Therefore, the equations in (7) become

22 2 2 2 2 2 22 1 3cos cot cos αω ω (9) α ω α= − + Ω = − + Ω

From (5), we can write

2 22 3

21ω ω ω= (10)

and from (9), we have 2 2 21 3 cotω ω= α . Therefore, (10) becomes

22 3 cotω ω= α (11)

and using 2 2 2 2 23 2sin tanω α ω α= Ω − from (9), we can write (11) as

22 2 2 22

cottan sin

ωα

ω α α= −

−Ω (12)

Since 2 2d dtω ω= , we can express this equation in terms of integrals as

22 2 2 22

cottan sin

ddt

ωα

ω α α= −

−Ω∫ ∫ (13)

Using Eq. (E.4c), Appendix E, we find

( ) ( )1 2 tan1

tanh cottan sin sin

tω α

αα α α

− Ω Ω

− = − (14)

Solving for 2ω ,

( ) ( )2 cos tanh sint tω α= Ω Ω α (15)

11-32.

a) The exact equation of motion of the physical pendulum is

sin 0I MgL+ =θ θ

where , so we have 2I Mk=

2 singLk

= −θ θ

or

( ) ( )2

d cosdd d

gLt k

θθ

or

DYNAMICS OF RIGID BODIES 395

( ) ( )2d d cogLk

= sθ θ θ

so

22

2cos

gLa

k= +θ θ

where a is a constant determined by the initial conditions. Suppose that at t , 0= 0=θ θ and at

that initial position the angular velocity of the pendulum is zero, we find 02

2a

k−

= cosgl

θ . So

finally

( )02

2cos cos

glk

= −θ θ θ

b) One could use the conservation of energy to find the angular velocity of the pendulum at any angle θ , but it is exactly the result we obtained in a), so at 01=θ , we have

( ) 102

2cos cos 53.7 s

gLk

−= = − =ω θ θ θ

11-33. Cats are known to have a very flexible body that they can manage to twist around to a feet-first descent while falling with conserved zero angular momentum. First they thrust their back legs straight out behind their body and at the same time they tuck their front legs in. Extending their back legs helps to resist spinning, since rotation velocity evidently is inversely proportional to inertia momentum. This allows the cat to twist their body differently to preserve zero angular momentum: the front part of the body twisting more than the back. Tucking the front legs encourages spinning to a downward direction preparing for touchdown and as this happens, cats can easily twist the rear half of their body around to catch up with the front.

However, whether or not cats land on their feet depends on several factors, notably the distance they fall, because the twist maneuver takes a certain time, apparently around 0.3 sec. Thus the minimum height required for cats falling is about 0.5m.

11-34. The Euler equation, which describes the rotation of an object about its symmetry axis, say 0x, is

( )x x y z y z xI I I− − =ω ω ω N

x

where xN b= − ω is the component of torque along Ox. Because the object is symmetric about Ox, we have y zII = , and the above equation becomes

0

dd

xxx x x

b tII b e

t

−= − ⇒ =

ωxω ω ω

396 CHAPTER 11