# Dynamics of Rigid Bodies - · PDF fileCHAPTER 11 Dynamics of Rigid Bodies 11-1. The...

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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

= ==

(1)

z

y

x Using the definition of the moment of inertia,

( ) 2ij ij k i jk

I r x x x dv

= (2)

we have

( )

( ) ( )

2 233

2 2 2 2cos cos

I r z dv

r r r dr d d

=

=

(3)

or,

( ) ( )1 2

4 233

0 1 0

5

1 cos cos

42

5 3

R

I r dr d

R

+

=

=

d

(4)

353

354 CHAPTER 11

The mass of the sphere is

343

M = R (5)

Therefore,

233 5I MR

2= (6)

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

211 22 3325

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

d

=

=

(8)

This vanishes because the integral with respect to 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 2 325

I I I MR= = = (10)

11-2.

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

x3 = x3

R hCM

x1

x1x2

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 1 222 2I I

x x x dI I v+

= = = + + (1)

DYNAMICS OF RIGID BODIES 355

which, in cylindrical coordinates, can be written as

( )2 2 21 2 0 0 0 22h Rz h

d dz r z rd

I I r

= = + (2)

where

23M M

V R

= =

h (3)

Performing the integration and substituting for , we find

( )21 2 3 420I I M R h= = +2 (4)

3I is given by

( )2 2 23 1 2I x x dv r rdr d dz = + = (5) from which

233

10I 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 a = (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 c = = = (2)

Then, Eq. (1) reduces to

2 2 2 1 + + = (3)

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

If we denote by dv the volume element in the system and by d the volume element in the (,, ) 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

= = =

= =

(4)

because d 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

= +

= +

) (5)

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

( )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

=

=

=

=

=

4 r

(6)

Therefore,

( ) ( )2 22 2 2415

a b d a b2

+ = + (7)

and

( )2 23 15I 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

= (1)

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

3

2

0 3 3m m

I x dx= = = 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 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 J = = = = = V J L r Jp M

(6)

so that

CM =J

VM

(7)

and

( )0

Jz a

I =

(8)

where ( ) 20 2 5I M= a . The velocity of any point a on the ball is given by Eq. (11.1):

CM = + v V r (9)

For the point of contact Q, this becomes

( )

CM

51

2

Q a J

J z aM a

=

=

Jv V

(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 a = = = (14)

so that

( )0 CM2 2I MV z a = (15)

from which

2 2

0

CM CM CM

2 25 5

I Ma az a

MV MV V

= = = (16)

If we assume that the ball rolls without slipping before it contacts the rail, then VCM a= , 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)5s

I MR= (1)

360 CHAPTER 11

For the hollow sphere,

R sin

( )2

2 2

0 0

4 3

0

4

sin sin

2 sin

83

hI d R R

R d

R

d

=

=

=

or, using , we have 24 R M =

223h

I 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 2= + (3)

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

( ) sinU Mg y = (4)

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

2 221 1

sin2 2

IL M y y Mgy

R= + + (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 I

=+

(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

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 kx= = .

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 Mx= + = 2 (2)

where we use the result 2 2I MR= for a disk and dx = R d. Lagranges equations give us

3

02

Mx kx+ = (3)

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

23

km

= (4)

11-8.

x3

x1x2

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