10. Rotational Motion

1. Angular Velocity & Acceleration

2. Torque

3. Rotational Inertia & the Analog of Newton’s Law

4. Rotational Energy

5. Rolling Motion

How should you engineer the blades

so it’s easiest for the wind to get the

turbine rotating?

Examples of rotating objects:

• Planet Earth.

• Wheels of your bike.

• DVD disc in the player.

• Circular saw.

• Pirouetting dancer.

• Spinning satellite.

Ans. blade mass toward axis

10.1. Angular Velocity & Acceleration

(Instantaneous) angular velocity

ˆt

ω ω

Average angular velocity

= angular displacement ( positive if CCW )

0ˆlim

t t

ω ω ˆ

d

d t

ω

Angular speed:d

d t

1 d s

r d t Circular motion: s r v

r

d sv

d tLinear speed: r

ω̂ // rotational axis

ω̂

in radians

ˆ ω

ˆ ω

ω̂ in radians

1 rad = 360 / 2 = 57.3

r̂θ̂

Polar coord ( r, )

Example 10.1. Wind Turbine

A wind turbine’s blades are 28 m long & rotate at 21 rpm.

Find the angular speed of the blades in rad / s,

& determine the linear speed at the tip of a blade.

21 rpm 21 / min 2 /

60 / min

rev rad rev

s

2.2 /rad s

v r 28 2.2 /m rad s 62 /m s

Angular Acceleration

(Instantaneous) angular acceleration0

limt t

d

d t

Trajectory of point on rotating rigid body is a circle,

i.e. r = const.

Its velocity v is always tangential:

d

dtv

a

Tangential component:

We shall restrict ourselves to rotations about a fixed axis.

2

2

d

d t

ˆα ω

Its acceleration is in the plane of rotation ( ) :

ˆt

drd t

a θ ˆr θ

ˆr v θ

ˆ/ / θ

Radial component:ˆ

r

dr

d t

θa 2 ˆr r

2

ˆv

r r

ˆˆ

d d

d t dt

θr

v

at

ar

a

t r a aˆ

ˆd dr rdt dt

θ

θ

ˆ r

ω̂

Angular vs Linear

Example 10.2. Spin Down

When wind dies, the wind turbine of Example 10.1 spins down with

constant acceleration of magnitude 0.12 rad / s2.

How many revolutions does the turbine make before coming to a stop?

2 20 2

2 20

2 4

# of rev.

2

2

0 2.2 /

4 3.14 0.12 /

rad s

rad s

3.2

10.2. Torque

sinr F

Torque : ˆτ ττ̂

plane of r & F

[ ] = N-m ( not J )

r sin

Rotational analog of force

Example 10.3. Changing a Tire

You’re tightening the wheel nuts after changing a flat tire of your car.

The manual specify a tightening torque of 95 N-m.

If your 45-cm-long wrench makes a 67 angle with the horizontal,

with what force must you pull horizontally to do the job?

sinr F

95 0.45 sin 180 67Nm m F

230F N

sin sin Note:

10.3. Rotational Inertia & the Analog of Newton’s Law

F m aLinear acceleration:

Rotating baton (massless rod of length R + ball of mass m at 1 end):

tR F

t tF m aTangential force on ball: m R2m R I

2I m R = moment of inertia

= rotational inertiaof the baton

Calculating the Rotational Inertia

Rotational inertia of discrete masses 2

i ii

m rI ri = perpendicular distance of mass i to the rotational axis.

Rotational inertia of continuous matter 2I r dm

r = perpendicular distance of point r to the rotational axis.

( r) = density at point r.

2r dV r

Example 10.4. Dumbbell

A dumbbell consists of 2 equal masses m = 0.64 kg

on the ends of a massless rod of length L = 85 cm.

Calculate its rotational inertia about an axis ¼ of the

way from one end & perpendicular to it.

2 23

4 4

L LI m

250.64 0.85

8kg m

25

8m L

20.29 kg m

GOT IT? 10.2

Would I

(a)increase

(b)decrease

(c)stay the same

if the rotational axis were

(1)at the center of the rod

(2) at one end?

(b)

(a)

Example 10.5. Rod

Find the rotational inertia of a uniform, narrow rod of mass M and length L

about an axis through its center & perpendicular to it.

2I r dm/2 2

/2

L

L

Mx dx

L

2r dV/2

3

/2

1

3

L

L

Mx

L

2

12

ML

/2 2

/2

L

Lx dx

Example 10.6. Ring

Find the rotational inertia of a thin ring of radius R and mass M about the ring’s axis.

2I R dm

2M R

2

M

R L

2 2

0R R d

22

02

M Rd

I = MR2 for any thin ring / pipe

2 2

0 0

LI R R d d z

2M R3 22

MR L

R L

2

M

R

Pipe of radius R & length L :

2R dm

2R dm

Example 10.7. Disk

Find the rotational inertia of a uniform disk of radius R & mass M

about an axis through its center & perpendicular to it.

2I r dm 2

M

R

2dm r dr

21

2M R

2

2Mr dr

R

32 0

2 RMI r dr

R

Table 10.2. Rotational Inertia

Parallel - Axis Theorem

Parallel - Axis Theorem: 2cmI I M d

Ex. Prove the theorem for a set of particles.

GOT IT? 10.3.

Explain why the rotational inertia for a solid sphere is

less than that of a spherical shell of the same M & R.

22

5sphereI M R 22

3shellI M R

Mass of shell is further away from the axis.

Example 10.8. De-Spinning a Satellite

A cylindrical satellite is 1.4 m in diameter, with its 940-kg mass distributed uniformly.

The satellite is spinning at 10 rpm but must be stopped for repair.

Two small gas jets, each with 20-N thrust, are mounted on opposite sides of it & fire

tangent to its rim.

How long must the jets be fired in order to stop the satellite’s rotation?

sinr F I

1

10 2 / min /60

940 0.74 20

rpm rad rev skg m

N

0 To stop the spin:

Time required for a const ang accel t

212

2R F M R

4t M R

F

8.6 s

Example 10.9. Into the Well

A solid cylinder of mass M & radius R is mounted on a frictionless horizontal axle over a well.

A rope of negligible mass is wrapped around the cylinder & supports a bucket of mass m.

Find the bucket’s acceleration as it falls into the well.

I

netF mg T Bucket:

Cylinder: T R 2

aT I

R

Let downward direction be positive.

m a

aIR

2

amg I m a

R

21

ga

ImR

1

2

gMm

GOT IT? 10.4.

(a): There must be a net torque to increase the pulley’s clockwise angular velocity.

Two masses m is connected by a string that passes over a frictionless pulley of mass M.

One mass rests on a frictionless table; the other vertically.

Is the magnitude of the tension force in the vertical section of the string

(a) greater than, (b) equal to, or (c) less than

in the horizontal? Explain.

10.4. Rotational Energy

Rotational kinetic energy = sum of kinetic energies of all mass elements,

taken w.r.t the rotational axis.

21

2dK dm v 21

2dm r

21

2rotK dK r dm 2 21

2r dm

21

2rotK I

21

2 i ii

K m v 2 21

2 i ii

m r 21

2I Set of particles:

Example 10.10. Flywheel Storage

A flywheel has a 135-kg solid cylindrical rotor

with radius 30 cm and spins at 31,000 rpm.

How much energy does it store?

2

21 1135 0.30 31,000 2 / min /

4 60kg m rpm rad rev s

2 21 1

2 2M R

21

2rotK I

32 MJ

~ energy in 1 liter of gasolineModern flywheels 10s of kW of power for up to a

min.

Carbon composite to withstand strain of 30,000 rpm.

Magnetic bearings to reduce friction.

supercondutor to reduce electrical losses.

Flywheel for hybrid bus (30% fuel saving).

Energy & Work in Rotational Motion

Work-energy theorem for rotations:

2

1

W d

rotK 2 21 1

2 2f iI I

Example 10.11. Balancing a Tire

An automobile wheel with tire has rotational inertia 2.7 kg m2.

What constant torque does a tire-balancing machine need to apply in order to

spin this tire up from rest to 700 rpm in 25 revolutions?

2

2 12.7 700 2 / min /

60

2 25 2 /

kg m rpm rad rev s

rev rad rev

21

2 fI W

46 N m2

2fI

10.5. Rolling Motion

21

2total i ii

K m v

Composite object:

2 21 1

2 2total cmK M I V

Moving wheel:

is w.r.t. axis thru

cm

total cm internalK K K

21

2 i ii

m V u 2 212

2 i i ii

m V u V u

2 21 1

2 2 i ii

M m V u 0i i i i cmi i

dm m

dt u r R

2 21 1

2 2totalK M dm V u 2 2 21 1

2 2M r dm V

V = velocity of CM. ui = velocity relative to CM.

2 21 1

2 2total cmK M I VMoving wheel:V = velocity of CM

is w.r.t. axis thru CM

Rolling wheel: X R V R

Example 10.12. Rolling Downhill

A solid ball of mass M and radius R starts from rest & rolls down a hill.

Its center of mass drops a total distance h.

Find the ball’s speed at the bottom of the hill.

0 0 0 0trans rotE K K U M g hInitially:

trans rotE K K U

2 21 1

2 2M v I

Finally:

22 21 1 2

2 2 5

vM v M R

R

27

10M v

0E E10

7v g h 2g h

sliding ballNote: v is independent of M & R

GOT IT? 10.5.

Solid ball.

Smaller I smaller Krot larger v.

A solid ball & a hollow ball roll without slipping down a ramp.

Which reaches the bottom first? Explain.