ROTATIONAL MOTIOND R . B E N J A M I N C H A N

A S S O C I AT E P R O F E S S O RP H Y S I C S D E PA R T M E N T

F E B R UA RY 2 0 1 4

PS 11 GeneralPhysics I for the Life Sciences

Questions and Problems for Contemplation

Chapter 8 Questions: 1, 4, 8, 10, 15, 18, 22, 24 Problems: 1, 2, 4, 8, 10, 15, 22, 27, 30, 38, 45, 52, 55,

61, 64 General Problems: 72, 80, 81

Describing Rotational Motion

Angular displacement q Let O be the axis of rotation How far the object has rotated

Only 2 directions possible: clockwise(-) and counter-clockwise(+)

Measured in radians 1 radian (rad) is the angle subtended

by an arc whose length is equal to the radius of motion

rl

θ

Distance traveled

Arc length traversed

For one complete revolution

q can be expressed in revolutions

rl 2

qrl

radrev 21

Example: Bike Wheel

A bike wheel rotates 4.5 revolutions. How many radians has it rotated?

If the wheel has a diameter of 45 cm, what is the distance traveled by a point on the rim of the wheel?

radrev

radrevs 28

25.4θ

cmradcmdl 630282

452

q

Example: Bird of Prey

A bird’s eye can distinguish objects that subtend an angle no smaller than 3 10-4 rad. How many degrees is this?

How small an object can the bird just distinguish when flying at a height of 100m? For small angles (<15), arc length

and chord length are nearly the same

017.02360103θ 4

radrad

cmradmrl 3)103)(100( 4 q

Angular Velocity w

Average w

Instantaneous w Dt must be very small

Velocity v of a point on a rotating wheel

Changes direction as vector turns Increases in proportion to distance from the axis

of rotation

tDD

qw

wrv

Angular Acceleration a

Average a

Instantaneous a Make Dt as small as possible

Tangential acceleration

Radial acceleration

ara tan

tDD

wa

222 )( ww r

rr

rvaR

Review of Linear and Angular Quantities

Frequency = number of complete revolutions per second = f w = 2f

Period = time required to complete one revolution = T = 1/f

Equations of Motion

Zero angular acceleration a = 0, w = constant Uniform circular motion q = wt + qo

Linear velocity is not constant Magnitude is constant: v = wr Direction is changing

Acceleration is not constant atan= 0 but aR= rw2 = constant (centripetal) Direction is changing

Example: Earth’s Rotation

How fast is the earth’s equator turning? w = 2/T = (2 rad)/84,600s = 7.27 x 10-5 rad/s v = rw = (6,380 km)(7.27 x 10-5 rad/s) = 464 m/s

How will your speed change as you go to the North or South pole? v = (r cos f)w = (464 cos f) m/s f = 14.5°, v = 449 m/s f = 30°, v = 402 m/s f = 60°, v = 232 m/s f = 90°, v = 0 m/s

The Coriolis Effect

As you go from the equator towards the N pole, you are moving faster than the ground you are moving into: veer to your right (earth rotates west to east)

As you go from the N pole towards the equator, you are moving slower than the ground you are moving into: veer to your right Clockwise flow!

To or from the S pole: veer to the left! Counterclockwise flow!

Example: Hard Drive

The platter of the hard drive of a computer rotates at 7200 rpm. What is the angular velocity of the platter?

If the reading head of the drive is 3.00 cm from the axis of rotation, how fast is the disk moving right under the head?

srad

srevrev

f 7541202sec60min1

min7200

22 w

smsradmrv /6.22)/754)(103( 2 w

Example (continued)

If a single bit requires 0.50 mm of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis? The number of bits passing the head per second is

or 45 megabits/s (Mbps)

sbitsbitm

sm /1045/1050.0

/6.22 66

Constant Angular Acceleration

a = constantw = wo + atq = qo + wot + ½ at2

Eliminate t between w and q w2 = wo

2 + 2aq

Total Acceleration

atotal = atan + aR

atan Constant magnitude,

changing directionaR

Variable magnitude, variable direction

Example: Centrifuge

A centrifuge motor is accelerated from rest to 20,000 rpm in 30s. Determine its angular acceleration and how many revolutions it makes while it is accelerating.

Solution Assuming constant angular acceleration

2/7030

0/2100 srads

sradt

o

wwa

Example continued

Where the final angular velocity w is

The angular displacement in 30s is then

We divide by 2 to convert to revolutions

sradms

revrevradf /2100

/60min/2000022

w

radssrad 422 1015.3)30)(/70)(2/1(0 q

revrevradrad 34

100.5/21015.3

q

Rolling Motion

Translational + rotational motion

No Slipping Static friction between object

and rolling surfacewrv

Example: Bicycle

A bicycle slows down uniformly from a velocity of 8.40 m/s to rest over a distance of 115 m. The overall diameter of the tire is 68.0 cm. Determine the initial angular velocity of the wheels.

sradmsm

rvo

o /7.24340.0

/40.8w

Example (continued)

Determine the number of revolutions each wheel undergoes before stopping. The rim of the wheel turns 115m before stopping.

Thus,

Determine the angular acceleration of the wheel

revm

mrm 8.53

)340.0(2115

2115

2222

/902.0)8.53)(/2(2

)/7.24(02

sradrevrevrad

srado

q

wwa

Example continued

Determine the time it took the bicycle to stop

Note: when the bike tire completes one revolution, the bike advances a distance equal to the outer circumference of the tire (no slipping or sliding).

ssradsradt o 4.27/902.0/7.2402

aww

Announcements

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Center of Mass

You can reduce an object to a point and describe its translational motion by considering the motion of this point (called its center of mass)

Determining Center of Mass

Consider masses m1, m2, m3, … with coordinates (x1, y1), (x2, y2), (x3, y3), …

ii

iii

cm m

xm

mmmmxmxmxx

321

332211

ii

iii

cm m

ym

mmmmymymyy

321

332211

CM for a Leg

Determine the center of mass of a leg when a) stretched out and b) bent at 90°. Assume the person is 1.70 m tall.

Solution a) Straight leg

Essentially 1-D Measure distance from hip joint

CM is 52.1-20.4 = 31.7 units from base of foot For a height of 172 cm, xcm = 54.5 cm above the bottom of

the foot

4.204.36.95.21

)3.50)(4.3()9.33)(6.9()6.9)(5.21(

cmx units

CM of Leg

b) Bent leg

For a height of 172 cmxcm = (172 cm)(0.149) = 25.6 cmycm = (172 cm)(0.23) = 39.6 cm Center of mass of bent leg is 39.6 cm

above the floor and 25.6 cm from thehip joint!

9.144.36.95.21

)6.23)(4.3()6.23)(6.9()6.9)(5.21(

cmx units

0.234.36.95.21

)5.28)(5.21()2.18)(6.9()8.1)(4.3(

cmy units

CM Trajectory

Center of mass of swimmer in flight follows projectile motion (parabolic) trajectory

Center of mass of wrench follows constant velocity trajectory

Torque

What causes an object to rotate?Torque = force x lever arm

q sinrFFr

More Torque

Units: Nm (Newton-meter) Reserve J for work and energy

Torque is a vector quantity Direction determined by the right hand rule

Newton’s First law

Translational EquilibriumAll forces cancel out: SF = 0

Rotational EquilibriumTorques must balance out: SG = 0

When is an object in equilibrium?

Newton’s Second Law

F = maG = Ia

I = moment of inertia a = angular acceleration Only two possible directions

Counter-clockwise rotation Clockwise rotation

Moment of Inertia of Particles

For a single moving object with mass m = rF = rma = rmra

=mr2a I = mr2

For several objects rigidly attached to each other S = (Smiri

2)a I = Smiri

2

Changing Moment of Inertia

Determine the change in the moment of inertia of a particle as the radius of its orbit doubles

Solution

It increases by

41

4)2( 2

2

2

2

mRmR

RmmR

II

f

i

%300%100114%100

i

if

III

Changing Your I

Vertical axis of rotationArms on the side

R = 25 cm, M = 9.6 kg

Raise your arms in a crucifixion pose R = 57.5 cm, M = 9.6 kg

433% increase

222 60.0)25.0)(6.9( mkgmkgMRIside

222 20.3)575.0)(6.9( mkgmkgMRIcross

Moments of Inertia for Various Objects

Rotational Kinetic Energy

22

2122

212

21 )()()( ww iiiiii rmrmvmKE

221 wIKErotational

2221

21

CMCM MvIKETotal w

Example: Ball Rolling Down an Inclined Plane

Determine the speed of a solid sphere of mass M and radius R when it reaches the bottom of an inclined plane if it starts from rest at a height H and rolls without slipping. Assume no slipping occurs. Compare the result to an object of the same mass sliding down a frictionless inclined plane.

Solution

Initial mechanical energy PE = MgH KEtrans = 0 KErot = 0

Final mechanical energy PE = 0 KEtrans = ½ Mv2

KErot = ½ Iw2

Conservation of EnergyMgH = ½ (Mv2 + Iw2)

Solution (continued)

I = (2/5)MR2 for a solid sphere rotating about an axis through its center of mass

w = v/RThus

MgH = ½ Mv2 + ½(2/5)MR2(v/R)2

(1/2 + 1/5) v2 = gHv = [(10/7)gH]1/2

v does not depend on the mass and radius of the sphere!!

Frictionless Incline

Ball slides down the incline and does not rollThus,

½ Mv2 = MgHv = (2gH)1/2

The speed is greater! None of the original PE is converted into rotational

energy.

Work Done on a Rotating Body

W = FDl = F rDqW = Dq

PowerP = W/DtP = Dq/Dt = w

Angular Momentum L

L = IwNewton’s second law becomes

Thus, tII

tII o

D

DD

wwwa

tL

DD

Conservation of Angular Momentum

If the net torque acting on a rotating object is zero, then its angular momentum remains constant.

fi

if

LLLL

tL

DD

0

0

ffii II ww

The Ice Skater

How can the ice skater spin so fast?

if

if I

I ww

The Diver

How can the diver make somersaults?Does she have to rotate initially?What trajectory does she follow?

The Hanging Wheel

Why is the wheel standing up?Why does it turn around about the point of

support?

Rotating Disk Demo

What happens when you tilt the rotating disk?

dtdL

HINT:

Drunk Driver Test/Tightrope Artist

Follow the line walkIncrease your moment of inertia to minimize

rotations

Quiz 8

1. A 4 kg mass sits at the origin, and a 10 kg mass sits at x = + 21 m. Where is the center of mass on the x-axis?

(a) + 7 m (b) + 10.5 m (c) + 14 m (d) + 15 m2. An object moving in a circular path experiences

(a) free fall. (b) constant acceleration. (c) linear acceleration. (d) centripetal acceleration.3. A boy and a girl are riding on a merry-go-round which is turning

at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular velocity?

a) The boy b) The girl c) Both have the same non-zero angular velocity. d) Both have zero angular velocity.

Quiz 8

4. A wheel starts at rest, and has an angular acceleration of 4 rad/s2. Through what angle does it turn in 3 s?

a) 36 rad b) 18 rad c) 12 rad d) 9 rad 5. A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on

the outer rim move in 2 s? a) 314 cm b) 4084 cm c) 8995.5 cm d) 17990.8 cm

6. What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70 cm when the bike is moving 8 m/s?

a) 91 m/s2 b) 183 m/s2 c) 206 m/s2 d) 266 m/s2

7. A bicycle is moving 4 m/s. What is the angular speed of a wheel if its radius is 30 cm?

a) 0.36 rad/s b) 1.2 rad/s c) 4.8 rad/s d) 13.3 rad/s

Quiz 8

8. An ice skater is in a spin with his arms outstretched. If he pulls in his arms, what happens to his kinetic energy?

a) It increases. b) It decreases. c) It remains constant but non-zero. d) It remains zero. 9. What is the quantity used to measure an object's resistance to

changes in rotation? a) mass b) moment of inertia c) linear momentum d) angular momentum

10. A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

a) 57.6 J b) 64.0 J c) 78.8 J d) 122 J