Chapter 9 Rotations of Rigid Bodies

Up to this point when studying the motion of objects we have made the (implicit) assumption that these are “point objects” i.e. all the mass is concentrated at one point.

In chapter 9 we will drop this assumption and study the rotation of rigid bodies. These are objects that do not change volume or shape. All that is needed to describe the rotations of rigid bodies is Newton’s laws of motion. From these we will develop the equations that describe rotational motion

The following new concepts will be introduced:

Moment of Inertia ( I ) Torque ( ) Angular Momentum ( L ) (9-1)

Cartesian coordinates: (x, y)

Polar coordinates: (r, )

How to convert Cartesian to polar coordinates and vice versa

(x, y) (r, ) O

P

Ax

y

y

x

r

2 2

From triangle OAP we have:

cos , sin

, tan

x r y r

yr x y

x

(9-2)

Consider the rotation of a flat rigid body (the hand of a clock in the picture) in the xy-plane about the origin O which is fixed. Any point P of the rigid point (the tip of the hand in the picture) remains at a constant distance from O.

All we need to describe the motion of P (and thus the motion of the rigid body) is the angle between the position vector r of point P and the x-axis. We express the angle as function of time (t)

Note: Point O is the only point of the rotating body that does not move

O

(9-3)

In a three dimensional rigid body that rotates about a fixed axis all points on the rotation axis do not move

Any point P on the rigid body rotates on a plane that is perpendicular to the rotation axis, on a circular orbit of fixed radius r

(9-4)

Rotation axis

r

Consider a point P on a pot that is rotating on a potter’s wheel (fig.a). The position of P is described by the angle

( ) ( )

lim

av

t t t

t td

t dtt o

Average angular velocity

Instantaneous angular velocity

Units: rad/s (9-5)

t

t + t

The angular velocity vector

The vector is defined as follows:

1. Magnitude

2. Direction and sense of

Direction: The axis of coincides with the rotation axis

Sense: Curl the fingers of the right hand in the direction of rotation. The thumb points along

(9-6)

d

dt

Special case: Rotation with constant angular velocity

This equation is the analog in rotational motion of the equation that describes motion along the x-axis with constant velocity:

x(t) = xo + vt

Note: In this chapter and the next we will take advantage of the similarity in form between equations that describe motion along a line and rotational motion.

(9-7)

o00 0 0

Integrate both sides from 0 to t

-

tt tt

o

dd dt

dt

d dt t t

t

(t) = o + t eqs.1

Rotational motion with constant angular velocity

Period T is the time required to complete one revolution ( = 2)

Frequency f is the number of revolutions per second

From the definition

Relationship between f and : In eqs. 1 we set o = 0 and = 2 2 = T = /f

1fT

(9-8)

2 f

2

2

( ) ( )

lim

0

av

t t t

t t

d d

t dt dtt

In general the angular velocity is not constant but it changes with time. For this reason we introduce the notion of angular acceleration to describe the rate at which changes with t

Average angular acceleration

Units: rad/s2

Instantaneous angular acceleration

(9-9)

t

t + t

Motion with constant angular acceleration (9-10)

t

0 00 0

Integ

rate both sides

Compar

from 0 to t

d

Integrate both s

e wi th

ides f o

r m

:

tt t

o

o

dd dt

dt

dt t t

dd

v v at

dtdt

t 2

0 00 0 0 0

2

0

2

Compar

0 to t

d

e with:

2

2

2

tt t tt t

o

o o

o o

o

tdt dt t dt t

t

av t

t

tx x

o t

2

2o o

tt

Non-uniform acceleration

In general angular acceleration is not constant

In this most general case we decompose the vector a into two components:

1. ac (centripetal acceleration) that points towards the center C 2. at (tangential acceleration) that points along the tangent

rC

Rotation axis

(9-11)

22 ( )

, c t

v dv d r da r a r r

r dt dt dt

ta r

Rotational kinetic energy of a rigid body that rotates about an axis with angular velocity . Divide the object into N elements with masses m1, m2, …, mN

mi

vi

Ri

O

x

y

Each mass element mi moves on a circle of radius Ri with speed vi = Ri. The kinetic energy of mi

is given by:

Ri

(9-12)

2 2 2

2 2i i i i

i

m v m RK

mi

vi

Ri

O

x

y

(9-13)

Ri

i

2 2

1 2

22 2 2

1 1 1 2 1

The kinetic energy of m is:

2The total rotational kinetic energy K

is the sum of all the kinetic energies

...

...2

The sum in the brackets is known a

i ii

N

N

m RK

K K K K

K m R m R m R

s

the "moment of inertia"

or "rotational inertia"

(symbol I) of the object2

2

IK

2

in linear motion2

mvK

Moment of Inertia I

Recipe for the determination of I for a given object

1. Divide the object into N elements with masses m1, m2, …, mN

2. The contribution of each element Ii = miRi2

3. Sum all the terms I = I1 + I2 + …. + IN

4. Take the limit as N

Ri

(9-14)

2I R dm

2 2 21 1 2 2 ... N NI m R m R m R

The moment of inertia of an object depends on:

a. The mass and shape of the object, and b. the position of the rotation axis

2I R dm

(9-15)

Moment of inertia of a rod of length L and mass M Linear mass density = M/L Divide the rod into elements of length dx and mass dm = dx

Axx’

(9-16)

/ 2/ 2 32 2 2

/ 2 / 2

3 2 2

2

0

Consider a rotation axis through the center of mass O

3

( )

12 12 12

For a rotation axis through the end point A we have:

LL

CM

L L

CM

L

xdI x dm x dx I x dx

L L L MLI

I x dx

I

3 3 2 2

0

( )

3 3 3 3

Lx L L L ML

Parallel axis theorem

Moment of inertia I about any axis of an object of mass m

Moment of Inertia ICM

about a parallel axis that passes through the center of mass

= + md2

I = ICM + md2

d is the distance between the rotation axis and the parallel axis that passes through the center of mass

(9-17)

Torque of a force acting on a point

Associated with any force F we can define a new vector , known as “the torque of F” , which plays an important role in rotational dynamics

Magnitude of = Fr r = “arm of the moment” r = rsin = Frsin

r

Direction of If the force tends to rotate the object on which it acts the clockwise (CW) direction (as in the picture) the torque points into the plane of rotation and has a negative sign. If on the other hand the rotation is counterclockwise (CCW) the torque points out of the plane and is positive

(9-18)

Fr

O

P

The equivalent of the “second law” in rotational dynamics

Consider the object shown in the figure which can rotate about a vertical axis. We divide the object into element of masses m1, m2, … mN On each of these elements we apply a force F1, F2, ... FN

For simplicity we assume that these forces are perpendicular to the object. Consider one element of mass mi The force Fi

results in a torque i = Firi , Fi = miai = miri i = miri2

Total torque = 1+ 2 + …+ N = (m1r12 +…+ mNrN

2) = I

Thus: Compare this with: F = ma

(9-19)

I

Example (9-6) page 250. A bucket of mass m = 12 kg is connected via a rope to a cylindrical flywheel of mass M = 88 kg and radius R = 0.5 m. The bucket is dropped and the flywheel is allowed to spin. Determine the angular velocity of the flywheel after the bucket has fallen for 5 s.

(9-20)

a

R

(9-21)

2

We substiture T from

System bucket

eqs.1 into eq

System b

(eq

ucke

s.2

( )

2 2

s.1)

(eqs.2)

(eqs.3)

(eqs.4

t

)

ynetF T mg ma

T mg ma

TR

mg ma R

a MR a MRaI I

R R

2

Compare eqs.3 and eqs.4 ( - ) 2

129.8 2.1 m/s

12 88 / 22

MRamg ma R

ma g

Mm

R

a

The acceleration of the bucket is also the acceleration of the rope and therefore the tangential acceleration at the rim of the flywheel

(9-22)

2

2

2.1 m/s

2.14.2 rad/s

0.54.2 5 21 rad/so

a

a

Rt

Torque of the gravitational force

The torque of the gravitational force on a rigid body Fg is equal to the torque of Fg acting at the center of mass of the object

Thus = mgR

R = Rsin

= mgRsin

R

(9-23)

Analogies between linear and rotational motion

Linear Motion Rotational Motion

x (distance) (rotation angle)

v (velocity) (angular velocity)

a (acceleration) (angular acceleration)

m (mass) I (moment of inertia)

F (force) (torque)

p = (linear momentum) L (angular momentum) p = mv L = I

(9-24)

Angular momentum L

By analogy to the definition of the linear momentum

we define the vector of angular momentum L as follows:

units: kg.m2/s

Since I > 0 L is parallel to

L

L I����������������������������

p mv����������������������������

(9-25)

( )

dL d I dI I

dt dt dt

������������������������������������������

���������������������������� dL

dt

��������������

for linear motiond p

Fdt

��������������

��������������

L

Conservation of angular momentum

(9-26)

If the torque of all external forces on a rigid body is zero then the angular momentum L does not change (it is conserved

If 0 0

L is a constant vector

dL

dt

d L

dt

��������������

��������������

��������������

0L ��������������

Rolling

When a wheel rolls on flat ground it executes two types of motion

a. All points on the rim rotate about the center of mass with angular velocity

b. The center of mass and all other points on the wheel move with velocity vcm

Note 1: The total velocity of any point on the wheel is the vector some of the velocities due to these two motions Note 2: and vcm are connected (9-27)

Rolling

vcmvcm

L1

(9-28)

1

1 1

1

L is the distance traveled in one revolution of the wheel

Thus: L is given by the equation:

=2 R = circumference of the rim of the wheel

2 2 R

cm

cm cm

L v T

L

v T v R RT

cmv R

Kinetic energy of a rolling object

vcmvcm

P

(9-29)

2

P

2

2 2 22 22

I is the wheel's moment of inertia about point P2

(parallel axis theorem)

( )( )

2 2 2 2 2The first term in the expression for K is the rotatio

P

P cm

cm cm cmcm

IK

I I mR

I I mvm RK I mR

nal energy

about the center of mass. The second term is the kinetic energy

due to the translation of the center of mass.

Find the acceleration of the center of mass of the rolling object (moment of inertia about the center of mass Icm , mass m, radius R) as it rolls down an inclined plane of angle

(9-30)2

sin

/

mga

m I R

(9-31)

2

Center of mass motion: sin -

Rotation about the center of mass: also

(eqs.1)

(eqs.2)

Substitu

Fxnet mg f ma

fR I

Ia IafR I f

R R

2

2 2

te f from eqs.1 in eqs.2 sin -

sin sin Note: a does not depend on m

/ 1 /

Iamg ma

Rmg g

am I R I mR

Rolling object 1 = cylinderRolling object 2 = hoop

(9-32)

2

sin

/

mga

m I R

22

1 2

1 22 21 2

1

Cylinder Hoop

2

sin sin

/ /

sin

/ 2

mRI I mR

mg mga a

m I R m I R

mga

m m

2

1 2

sin

2 sin sin(0.67) sin (0.5) sin

3 2

mga

m mg g

a g a g