Comparing rotational and linear motion

Angular VelocityAngular velocity,w, is the rate of change in angular displacement. (radians per second.)

=w 2pf Angular frequency f (rev/s). =w 2pf Angular frequency f

(rev/s).

Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm):

w = Angular velocity in rad/s.

q

t

Angular AccelerationAngular acceleration is the rate of

change in angular velocity. (Radians per sec per sec.)

The angular acceleration can also be found from the change in frequency, as follows:

2 ( ) 2

fSince f

t

2 Angular acceleration (rad/s )t

Torque and Angular Acceleration

When an object is subject to a net force, it undergoes an acceleration. (Newton’s 2nd)

When a rigid object is subject to a net torque, it undergoes an angular acceleration.

Force and Linear Acceleration

Inertia of RotationConsider Newton’s second law for the inertia of rotation

F = 20 N

a = 4 m/s2

Linear Inertia, m = F/a m = = 5 kg

20 N

4 m/s2

F = 20 NR = 0.5 ma = 2 rad/s2

Force does for translation what torque does for rotation:

Rotational Inertia, I

I = = = 2.5 kg m2(20 N)(0.5 m)

2 rad/s2

ta

ImaF ,

Moment of Inertia

This mass analog is called the moment of inertia, I, of the object

is defined relative to rotation axisSI units are kg m2

i

iirmI 2

More About Moment of Inertia

• I depends on both the mass and its distribution.• If an object’s mass is distributed further

from the axis of rotation, the moment of inertia will be larger.

Common Moments of Inertia

Common moments of inertia are on page 251.

Example 1 A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias.

2 21 12 2 (3 kg)(0.2 m)I mR

R

I = mR 2

Hoop

R

I = ½mR 2

Disk

2 2(3 kg)(0.2 m)I mR

I = 0.120 kg m2

I = 0.0600 kg m2

Important AnalogiesFor many problems involving rotation, there is an analogy to be drawn from linear motion.

xf

R

4 kg

wt wo = 50

rad/s t = 40 N m

A resultant force F produces negative acceleration a for a mass m.F ma

Im

A resultant torque tproduces angular acceleration a of disk with rotational inertia I.

I

Example 2Treat the spindle as a solid cylinder.

a) What is the moment of Inertia of the spindle?

b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel?

c) What is the acceleration of the bucket?

d) What is the mass of the bucket?

M

Solutiona) What is the moment of Inertia of the spindle?Given: M = 5 kg, R = 0.6 m

M

middleabout rod,,12

1

endabout rod,,3

1

shell spherical,3

2

sphere solid,5

2

cylinder solid,2

1shell lcylindrica,

2

2

2

2

2

2Inertia of Moments

ML

ML

MR

MR

MR

MR

2

2

1MRI = 0.9 kgm2

Solutionb) If the tension in the rope is 10 N, what is a?Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m

M

IrF

formula Basic IrT = a (0.6m)(10 N)/(0.9 kg∙m2)

a = 6.67 rad/s2

c) What is the acceleration of the bucket?Given: r=0.6 m, a = 6.67 rad/s

ra formula Basic

a=4 m/s2a = (6.67 rad/s2)(0.6 m)

Solutiond) What is the mass of the bucket?Given: T = 10 N, a = 4 m/s2

M

maF formula Basic

ag

TM

TMgMa

M = 1.72 kg

Comparing rotational and linear motion

Combined Rotation and Translation

vcm

vcm

vcm First consider a disk sliding without friction. The velocity of any part is equal to velocity vcm of the center of mass.

vR

P

Now consider a ball rolling without slipping. The angular velocity about the point P is same as for disk, so that we write:

Orv

R v R

Two Kinds of Kinetic Energy

vR

P

Kinetic Energy of Translation:

K = ½mv2

Kinetic Energy of Rotation:

K = ½I2

Total Kinetic Energy of a Rolling Object:

2 21 12 2TK mv I

KE of center-of-mass motionKE due to rotation

Example 3

What is the kinetic energy of the Earth due to the daily rotation?Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m.

T

2

formula Basic

First, find w

= 7.27 x10-5 rad/s

2

sphere Solid

5

2MRI

2

formula Basic

2

1 IKE

= 2.78 x1029 J

360024

2

22

5

1 MRKE

Summary – Rotational Analogies

Quantity Linear Rotational

Displacement

Displacement x

Radians

Inertia Mass (kg) I (kgm2)

Force Newtons N Torque N·m

Velocity v “ m/s ” Rad/s

Acceleration a “ m/s2 ” Rad/s2

Momentum mv (kg m/s) I (kgm2rad/s)

Analogous FormulasLinear Motion Rotational Motion

F = ma = IK = ½mv2 K = ½I2

Work = Fx Work = tq

Power = Fv Power = IFx = ½mvf

2 - ½mvo2 = ½If

2 - ½Io2

Example 4A solid sphere rolls down a hill of height 40 m.What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!)

22

formula Basic

2

1

2

1 Imvmgh

2

sphere solidFor

5

2mrI

rv formula Basic

v = 23.7 m/s

222

5

2

2

1 rvgh

5/7

2,

5

2

2

1 222 ghvvvgh

Angular Momentum

mvrL

IL

Analogy between L and p

Angular Momentum

Linear momentum

L = Iw p = mv

t = DL/Dt F = Dp/Dt

Conserved if no net

outside torques

Conserved if no net outside forces

Rigid body

Point particle

Example 5A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

SolutionKnown: M, R, m, v0

Find: wF

First, find L0

mvrL formula Basic

mvRL 0Next, find Itot

2

Particle

2

cylinder Solid

2

1

MRI

MRI

22

2

1MRmRITot

Now, given Itot and L0, find w

IL formula Basic

TotI

L0 = 2.71 rad/s

Summary of Formulas:I = SmR2I = SmR2

2 20½ ½fI I

mgho

½Iwo2

½mvo2

=mghf

½Iwf2

½mvf2

Height?

Rotation?

velocity?

Height?

Rotation?

velocity?

212K I

Powert

Work o o f fI I

Conservation: