Rotational Motion AP Physics Lyzinski, CRHS-South.

Rotational Motion

AP Physics

Lyzinski, CRHS-South

Day #1

Sections 10.1 & 10.2

rC

P

r s

Polar coordinates (r, )

CCW is positive, CW is negative

s = arc length =

= angular position (measured in radians, not degrees)

(deg)180

)( rad

= Angular displacement 12

Rs

R

s

radncecircumfere

so

2)(2360

Where does it come from?

Degree to radian conversion

= Angular velocity avg angular velocity = t

Instantaneous angular velocity = dt

d

tt

lim

= Angular acceleration avg angular acceleration = t

Instantaneous angular acceleration = 2

2

limdt

d

dt

d

tt

Units of radians per second (rad/s)

Units of radians per second squared (rad/s2)

Translational vs. Rotational Motion

dt

dxv

2

2

dt

d

dt

d

atvv 12

t 12

xavv 221

22 22

12

2

tvatx 12

21

tt 12

21

)( 2121 vvtx )( 212

1 t

CAUTION!!! These equations can only be used if a or are constant!!!!!

dt

d

2

2

dt

xd

dt

dva

In-Class Example Problem

A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98 rad/s. What is the constant angular acceleration of the wheel?

#6

267.133985.5698

985.56)98)(3(478.232)(

???

478.2321

2

1

3737

98sec3

12

1121

2121

2

srad

srad

srad

t

t

radrev

radrevrev

t

Class Practice

• A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98 rad/s. What is the constant angular acceleration of the wheel?

In-Class Example Problem#9 The tub of a washing machine goes into its spin cycle, starting from

rest and gaining angular speed steadily for 8.00 s, at which time it is turning at 5.00 rev/s. At this point, the person doing the laundry opens the lid and a safety switch turns off the machine. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn while its in motion?

totalrevs

revt

t

revt

t

srev

srev

50

30)05)(12()(

sec1205

20)50)(8()(

50sec8

21

2121

21

21

2121

21


A certain wheel begins to rotate. Its position varies with time according

to the equation radtt 563 2

A Mr. L original

Find the wheel’s average angular acceleration between 5 and 9 sec.

26sec4

24

sec59

3660

sec59

)5()9(

606)9(6)9(

366)5(6)5(

66

srads

radsrad

srad

srad

srad

srad

t

tdt

dt

Duh!!! The acceleration is

constant

Day #1 HW Assignment

pp. 299-300

Do problems 1-7 all (skip # 6 and #7b)

Day #2

Section 10.3

r s

v Relating Rotational Motion ( and ) to Translational Motion (x, v, and a)

Point of rotation

r

dt

dr

dt

rd

dt

dsv

)(

r

dt

dr

dt

rd

dt

dvat

)(

222 )(

rr

r

r

var

4222222 )()( rrraaa rt

Note: As r increases, v and a

get larger, while and

stay the same.


A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. The tires have diameter 58.0 cm and do not slip on the pavement. (a) Find the number of revolutions each tire makes during this motion. (b) What is the final angular speed of a tire in revolutions per second?

#16

srev

srad

srad

srad

srads

m

sm

sm

t

revradtt

mR

aRa

aatvvtvv

1.129.75)9)(429.8(0

33.5437.3410)9)(429.8(

429.829.

4.2

4.2sec9220

2

2

2

2

12

221

12

21

1221


The drive train of a bicycle is shown. The wheels have a diameter of 67.3 cm and the pedal cranks are 17.5 cm long. The cyclist pedals at a steady cadence of 76.0 rev/min. The chain engages with a front sprocket 15.2 cm in diameter and a rear sprocket 7.00 cm in diameter. (a) Calculate the speed of a link in the chain relative to the bicycle frame. (b) Calculate the angular speed of the bicycle wheels. (c) Calculate the speed of the bicycle relative to the road.

#12

sm

wheelwheelwheels

srad

wwsprocketrearwheelchain

sm

srad

sprocketfrontpedalschain

sprocketrear

sprocketfront

wheel

sradrev

pedals

rv

rv

mrv

mr

mr

mr

8.5)3365)(.257.17(

257.17)035(.604.

604.)076(.959.7

07.

152.

3365.

959.776 min


pp. 300-301

Do problems 10, 13, 14, 15, 16, 19

Day #3

Section 10.4

Rotational Energy

A solid object is a collection of particles. When the object rotates, each of these particles moves, thus possessing kinetic energy. If we add up all these individual energies, we can find the energy associated with the rotating object.

However, as we have learned previously, the velocity of each particle depends on how far the particle is from the axis of rotation. Particles close the axis move slower than particles far from the axis (according to v = r). Therefore, it might be useful to express each individual kinetic energy in terms of w (which is the same for each particle) instead of v (which changes based on distance from the axis).

22i2

12i2

12i2

1

i

)(

K Energy Kinetic Rotational

ii

iii

ii

iR

rmrmvm

K

i

irm2

i

221 IKR

Notice that w was taken out of the summation because it is the same for every particle, no matter how far the particle is from the axis of rotation.

KE of each individual particle

The term has been given the name “the Moment of Inertia”, or “I” .

Therefore, “I” has units of 2mkg

What is “Inertia”?

Remember, in translational motion, an object’s inertia is its “tendency” to want to either remain at rest or moving at a constant velocity. An object’s mass is a direct measure of its inertia.

In rotational motion, the individual particles have masses at different distances from the axis of rotation. The Moment of inertia is the rotational analog of mass. It is a measure of how difficult it is to change an object’s motion about its axis of rotation. The closer the mass is to the axis, the easier it is to change its rotational motion. Thus, objects with more of their mass far from the axis rotation have a higher moment of inertia.

Translational vs. Rotational Motion (revised)

221 mvK

dt

dxv

2

2

dt

d

dt

d

atvv 12 t 12

xavv 221

22

221

22

tvatx 12

21 tt 1

221

)( 2121 vvtx )( 212

1 t

dt

d 2

2

dt

xd

dt

dva

221 IKR


A Penn State Baton Twirler is spinning her 2 ft long baton, which has identical end masses of 300 grams. Assuming the rod itself to be mass-less, find the moment of inertia of the baton if she rotates it about (a) line “a” (which is through the center of the rod) or (b) line “b” (which is 4 inches off-center).

22

22

22

22

0619.)2032)(.300(.

)4064)(.300(.

0557.)3048)(.300(.

)3048)(.300(.

2032.8

4064.16

3048.1

mkgmkg

mkgrmI

mkgmkg

mkgrmI

min

min

mft

iia

iia

A Mr. L original

a b

In-Class Example Problem#25

v

Before

After

0.12 kg 60.0 kg

14.0 cm

2.86 m Find the speed that the small mass leaves the Trebuchet. Assume the rod

to be mass-less.

sm

srad

ab

ii

rv

mgHIMghmgh

EE

mkgrmI

5.24)86.2)(55.8(

55.8

)3)(8.9)(12(.)158.2(

)14)(.8.9)(60()14)(.8.9)(12(.

158.2)14)(.60()86.2)(12(.

221

221

2222


pp. 301-302

Do problems 21, 22, 23

Day #4

Section 10.5

dmrmrrmIi i

iim

iii

22

0

2 lim

2222

12

1

5

2

2

1MRIMRIMRIMRI

2222

3

1

3

2

2

1MRIMRIMRIMRI

Calculating Moments of Inertia

Notice that the object with more of its mass further away from the axis of rotation has a

larger moment of inertia (and thus it will be harder to change it rotational motion)

dmrmrrmIi i

iim

iii

22

0

2 lim

2222

12

1

5

2

2

1MRIMRIMRIMRI

2222

3

1

3

2

2

1MRIMRIMRIMRI

Calculating Moments of Inertia

Again, notice that the object with

more of its mass further away from the axis of rotation

has a larger moment of inertia

Using Calculus to find Moments of Inertia

• First, make sure to figure out what your “tiny pieces” look like.

• Second, choose the appropriate density function for your “tiny piece” (for linear, for area, or for volume).

• Third, use the appropriate density function and solve for dm.

• Fourth, make sure that dm only has one variable in it, and then plug it into

dmrI 2

23

33

03

31

0

2

0

2

0

2

0

2

3

1

333

)()(

mLL

mLL

Lm

Lxdxx

dxxdxrdmrI

dmdxdmdLmLL

m

LL

LLL

Example: Find the moment of inertia of a

thin rod that rotates about its end.

L

y

xdm with

thickness dL

2214

41

24

414

41

04

41

0

3

0

2

0

2

222

222

222

)2(

)(2)(

MRRLLR

mRL

V

mRL

rLdrrLLrdrrdmrI

Lrdrdm

drrdrLrdrrdV

dmdVmVV

m

RRLL

heightringofarea


solid cylinder that rotates about its

central axis.

dr

r

zero, b/c tiny times tiny equals super tiny

dm with thickness dr

y

r

R

r

Top view of the “thin” cylindrical slices

2525

51

3343

8551

385

51

38

05

51

38

0

438

0

2342

0

2

234

3222234

3222233

3343

34

)2(

)2(

))()()((

)()()()(

)(

mRRR

mR

V

mR

rdrrdrrrdmrI

drr

drdrrdrrdrrdrrdm

drdrrdrrdrrdrrrdrr

rdrrdV

dmdVmVV

m

RRRR


sphere about its central axis.

zero, b/c tiny times tiny equals

super tiny

dm with thickness dr

R

Volume of outer sphere

Volume of inner sphere

Using the parallel axis theorem to calculate moments of inertia

2MDII CM

2

12

1MLI

If you know the moment of inertia of an object about a given axis, you can use the equation

to find the moment of inertia of this object about any axis parallel to the given axis. The distance between these axis is “D”.

D

Known: Unknown:

2312

412

121

2

22

121

2

MLMLML

MML

MDII

L

CM

Another example of using the parallel axis theorem

D = R/2

2

2

1MRI Known: Unknown:

2

432

412

21

2

22

21

2

MRMRMR

MMR

MDII

R

CM

R

Sections 10.6 & 10.7

Torque (): The tendency of a force to rotate

an object about a given axis.

r

F

F cos

F sin

FdrF sin

d

“d” is known as the “moment arm” or “lever arm”

***Notice that only forces perpendicular

to the lever arm cause a torque

Some notes about TORQUE• The units of torque () are N-m

• The direction of a torque is found using the Right-Hand-Rule– Place your fingers in the direction of the lever arm.– “Slap” in the direction of the force.– Your thumb points in the direction of the torque.– The direction of a torque is found using the Right-Hand-Rule

• Positive torques are CCW and negative torques are CW.

• Torque is NOT a force!!!

• Torque is NOT the same as work. They have the same units, but are VERY different.

• The net torque on an object is the vector sum of the individual torques. Therefore, i

How are torques and forces different?

• Forces can cause a change in motion in translational motion.

• Forces can cause a change motion in rotational motion. HOWEVER, the further the force is from the axis of rotation, the more “effective” it will be in changing motion. Thus, the force as well as the length of the “lever arm” are important in rotational motion. Therefore, instead of speaking only of a “force”, we speak of a “torque”.

Newton’s 2nd Law (for a particle)

I

mr

rrmrF

rmF

maF

t

t

tt

2

)(

)(

The net torque on a particle is proportional to its ANGULAR acceleration.

The net force on a particle is proportional to its TANGENTIAL acceleration.

Newton’s 2nd Law (for a rigid body)

Idmrdmrd

dmrd

mr

rrmrF

rmF

maF

t

t

tt

22

2

2

)(

)(

Every “tiny little” mass (dm) in the rigid body is located at a different distance from the axis of rotation, and this needs to be taken into account. Also, each of these masses is subjected to its own individual “tiny little” torque (d). To get the total torque, we need to sum up ALL of the “tiny little” ones (by integrating).

Mass-less Pulleys

M1

M=0

M2

R

R

T1 T2

21

21

221

21

221

21

0

)0(

TT

RTRT

RRTRT

MRRTRT

Inet

All a mass-less pulley does is change the direction of a force.

Mass-less pulleys don’t really exist (but make calculations easy )

Mass-ful Pulleys

M1

M

M2

RR

T1 T2

21

22

21

1

221

21

TT

RTMRRT

MRRTRT

Inet

The difference in the tensions causes the net torque which forces the pulley to rotate

The pulley in this example is modeled as a solid disk (and thus

I = ½ MR2)


The system below is at rest when the 10 kg mass is released. The pulley is not mass-less, but rather has a mass of 6 kg and a radius of 20 cm. If the surface has a coefficient of friction of 0.2, find the acceleration of the system.

A Mr. L original

20 kg

10 kg

20 kg 10 kgFf

T1

m1g

FN1

gmamT

amgmT

111

111

amgmT

amTgm

222

222

m2g

T2

T2

T1

278.1338.58

)6)(5(.)8.9)(20)(2(.2010)8.9(1021

1122

21

12

221

12

12

smaa

aaa

Magmamamgm

MaTT

R

aMRRTRT

IRTRT

a

a

Translational vs. Rotational Motion (revised)

221 mvK

dt

dxv

2

2

dt

d

dt

d

atvv 12 t 12

xavv 221

22

221

22

tvatx 12

21 tt 1

221

)( 2121 vvtx )( 212

1 t

dt

d 2

2

dt

xd

dt

dva

221 IKR

maFnet Inet

Section 10.8

Work & energy in Rotational Motion

) equals nowK (where

) o(similar t

)F o(similar t

x)F Wo(similar t

2212

21

RT

TR

KKUKW

mvKIK

vdt

d

dt

d

dt

dW

W

If no external torques or forces are present, then Eb = Ea.

Situations with ONLY KT

212

1222

1 mvmvxF

KW

Situations with ONLY KR

Rotating Wheel (where the axis of rotation is fixed) 2

1212

221 II

KW

Situation with BOTH KT & KR

A Rolling Object

(it is rotating and translating at the same time)

2212

21 vMIK CMCM

K due to rotation

K due to translation

Situation with BOTH KT & KR & U

M1

M2

M3

R

2Rv2

21

212

1212

121

31

2

Rv

212

1212

121

3

31

221

Rv

2212

1212

121

3

31

)(

)2(

)(

I and 2h,H , since

0

MRvMvMghMM

IvMvMhgM

ghMM

MR

IvMvMgHM

ghMghM

KUKU

EE ab

h

H

Zero level

Solid Disk

Situation with BOTH KT & KR & U

h

)(

0

0

hoop) thin a(for I and , since

)(0

2

Rv2

212

21

2

Rv2

212

21

2Rv

2212

21

2

2

RhgvMvMgRMgh

MRMvMgRMgh

MRMvMgRMgh

MR

IMvRMgMgh

KUKU

EE ab

Find the velocity of the thin hoop (with radius “R”) at the bottom.

R

Review Day

Section 10.9

Rolling without slipping

• In order to roll, an object needs to encounter friction, which applies a torque to the object and causes a rotation about its center of mass.

• If an object does not slip at all while rolling, it is said to undergo PURE rolling motion.

• For pure rolling motion,

rdt

dr

dt

rd

dt

dva

rdt

dr

dt

rd

dt

dsv

CMCM

CM

)(

)(

These conditions must hold for non-slip rolling.

A closer look at an object that rolls but doesn’t slip (part 1)

When an object undergoes PURE rotation, every point on the object has the same angular velocity, Therefore, all points that are equidistant from the axis of rotation have the same tangential velocity.

vt = R

vt = R

v = 0


When an object undergoes PURE translation (which is the equivalent of ALL slip and NO roll), every point on the object has the same velocity, namely the velocity of the center of mass.

vCM

vCM

vCM


When an object undergoes PURE Rolling, this is a combination of both ROTATION and TRANSLATION. While every point on the object has the same angular velocity, the contact point with the floor acts as a pivot point. Thus, points on the rotating object that are furthest from the floor have the largest tangential velocity.

.

vt = vCM+ RvCM

vCM

v = 0

Review for Test

Days 9 thru 11 HW Assignment

Day 12

Test Day !!!!

Rotational Motion AP Physics Lyzinski, CRHS-South.

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Transcript of Rotational Motion AP Physics Lyzinski, CRHS-South.