Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational.

RotationChapters 8 and 9

Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational motion, there exists a linear analog. Combined with a few "bridge relationships“ between rotational and linear motion, a table of analogies presents a powerful tool for conceptual understanding.

Axis of Rotation

Rigid Body RotationRIGID BODY is an extended object whose size and shape do NOT change as it moves.

Rotational Kinematics

Chapter 8

Kinematics is a complete description of how motion occurs without consideration of the causes of motion

Some terms to Describe Rotational Motion

Angular Position, q in radiansDirection is + when measured CCW from x-

axis - when measured CW from x-axis

Angular Displacement, D , q change in angular position

0

q(rad) = s/rArc length, s

Since for a full circle, s=2pr, 1 rev = 2p rad = 360o 1 rad =57.3o

ExampleWhich particle has angular position 5p/2?

y

A.

x

y

B.

x

y

C.

x

y

D.

x

A


Angular Velocity, w in rad/s, is the velocity of rotation. Direction is along axis of rotation

given by the right hand rule (+ is CCW, - is CW)

fTt

22

Angular

velocity

Tangential

velocity

w

wvT

vT

Only for rad

r

v

tr

s T

T

rvT

2

Using Equations as Guide to Thinking

How do the variables relate to each other?

T

2

A BC D If A is 1m from the center and

D is 4 m from the center and all take 10 sec to complete a revolution, what are the velocities at A and D

vA = ______m/s

vD = ______m/s

wA = ______rad/s

wD = ______rad/s

TANGENTIAL ANGULAR Speed Speed

0.63

2.5

0.63

0.63

rv

A bike wheel with diameter 80cm rotates at a rate of 45rpm a)What is the angular velocity of the bike tire (in rad/s)

? b)What is the tangential speed of the tire in m/s? c)At 90rpm, what is the tangential velocity of the tire?

sm

rvT/88.1

)71.4)(4.0(

vTwsradsrad

sx

rev

radxrpm

/71.4/5.160

min1

1

245

2 x 1.88 =3.76m/s


Angular Acceleration, a in rad/s2, is the rate of change of the angular velocity.Direction is along axis of rotation, parallel or anti-parallel to w

r

a

tr

v

tT

Angularaccelerati

on

Tangentialacceleratio

n

Only for rad


Rotational variables

Translational or Linear variables

ra

rv

rs

T

T

Only if angle unit

is in radians

wf wi

wiwf

SPEEDING UPw and a are in same

direction

wfwi

wiwf

rotating CCW (+): w +, a +

SLOWING DOWNw and a are in opposite

direction

rotating CW (-): w -, a -

rotating CCW (+): w +, a -

rotating CW (-): w -, a +

12

The Vector Nature of Angular Variables

Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector.

Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity.

Problem: Assume the particle is speeding up.a) What is the direction of the instantaneous velocity, v?

b) What is the direction of the angular velocity, w?c) What is the direction of the tangential acceleration, aT?

d) What is the direction of the angular acceleration a?

e) What is the direction of the centripetal acceleration, ac?f) What is the direction of the overall acceleration, a, of the

particle?g) What changes if the particle is slowing down?w

v

Tangent to circleOut of page

w

Same direction as v, tangent to circle

aT aC

Same direction as w, out of page

a

Into center of circle

a

aT and a would point anti- parallel to w and a

A jet awaiting clearance for takeoff is momentarily stopped on the runway. As seen from the front of one engine, the fan blades are rotating with an angular velocity of -110 rad/s, where the – sign indicates a CW rotation. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in 14 s. Find the angular acceleration of the blades assuming it to be constant.

-16 rad/s2

Linear and Rotational KinematicsEquations for Constant Angular Acceleration

2.5

.4

.3)(.2

.1

20

2

221

0

0

021

ttt

t

xavv

attvxatvv

vvvtvx

2.5

.4

.3)(.2

.1

20

2

221

0

0

021

Rotational Motion

(a = constant)

Linear Motion(a = constant)

Applying the Equations of Rotational Kinematics

1.Draw a picture to represent the system if necessary.

2. Select a coordinate system to analyze your system with the x-axis along the initial angular position. Decide which direction of rotation will be + and which – (convention is + for CCW and – for CW). Keep directions consistent for the problem.

3. List the variables that are given

4. Write down the kinematic equation that will be used to solve the problem.

5. Isolate the unknown variable.

6. Solve for relevant unknowns by putting in numbers. Keep in mind that you may sometimes have more than one unknown and therefore need the same number of equations as unknowns.

17

tradrev

sradsrev

42

//5.000

Example: The blades of a ceiling fan start from rest and, after two revolutions, have an angular speed of 0.50 rev/s. The angular acceleration of the blades is constant. What is the angular acceleration of the fan?What is the angular speed after eight revolutions?

28

2

20

2

/39.0

)4(20

2

srad

t

radrev

168

?0

8

0

sradsrevsrad /28.6/1/2

)16(20

2

82

20

2

18

sradsradrev

radsrev /8.106/34

217/17

mrrrvT

30.0)8.106(32

The blade of a lawn mower is rotating at an angular speed of 17 rev/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade?

NONuniform Circular MotionMagnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle

v1v2

r

w

Dq a

aT = ra

ac

22

rr

va Tc

aT and a due to change of velocity magnitude or speed. They are due to forces acting in the line of motion.

ac is due to change in direction of velocity. It is due centripetal force.

rt

va TT

NONuniform Circular MotionMagnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle

v1

v2

r

w

Dq

a

aT = raac

22

rr

vac r

t

vaT

c

T

Tc

a

a

aaa

1

22

tanj

Acceleration and net force vectors do not point to the center in nonuniform circular motion. They have both radial (centripetal) and tangential components.

21

Example: A 220kg speedboat is making a circular turn (radius 32m) around a buoy. During the turn, the engine applies a net tangential force of 550N to the boat. The initial tangential speed of the boat is 5m/s. a) Find aT. b) After 2s into turn, find ac.

st

sradra

sradrv

T

T

0.2

/078.032/5.2/

?

/156.032/5/

2

0

22

rr

vac

v

v0

rDq

ac0

F

2/5.2

220550

sma

a

maF

T

T

TT

ac

aT

aT

ac=3.1 m/s2

Rotational Dynamics

Chapter 9

Dynamics is a complete description of WHY motion occurs

Rotation of a Rigid BodyRIGID BODY is an extended object whose size and shape do NOT change as it moves.

All points in object move along parallel paths. Object can be treated as a point.

All points in object move along different paths around rotation axis. One point – center of mass – follows the path of a point particle. Object must be treated as an EXTENDED object

Rotational DynamicsWhat causes rotational motion to change? What causes angular acceleration?

Linear MotionFORCE, F, causes accelerationF

rFrF sinAngular MotionTORQUE, t, causes angular acceleration

F F

No torque

No torqueF

2F

q

FF=Fsinq

Rotational DynamicsWhat causes rotational motion to change? What causes angular acceleration?

Linear Motion - FORCE, F, causes acceleration

rFrF sin

Angular Motion - TORQUE, t, causes angular acceleration

Distance to axis of rotation

Magnitude of

Force

Angle between r and

F

Units of Torque: N

m

TORQUE is a VECTOR+ when it produces CCW rotation about the axis- when it produces CW rotation about the axis

r

r/22F 2F

F F

F

AB

C D

E

A wheel turns freely on an axle at the center. Which one of the forces shown will provide the largest positive torque? Negative torque?E A

F

r=20cmq

F=Fsinq

F//

Problem: Luis uses a 20cm long wrench to tighten a nut. The wrench handle is tilted 30o above the horizontal and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut?

30o

Nm

rFrF

3.17)60sin100)(2.0(

sin

Static Equilibrium(Translational AND

Rotational)

Torque and Static EquilibriumObject at rest is in static equilibrium

Linear Motion

0F

Rotational Motion 0

Static Equilibrium of a Rigid BodyA rigid object at rest is in static equilibrium – it has

zero translational acceleration and zero angular acceleration.

0

0

x

x

F

F

0The net torque about ANY AXIS is zero if a rigid object is in rotational equilibrium.

Applying Equilibrium Conditions to a Rigid Body

1.Select object.

2. Draw a Free Body Diagram. We are now dealing with extended objects and the position of the forces are important.

3. Choose a convenient x- and y- axis and resolve forces into their x- and y-components

4. Apply Newton’s 2nd Law for equilibrium to each direction: Fx = 0 and Fy = 0 .

5. Select a convenient axis of rotation or pivot point. The rotation axis is arbitrary since object is in equilibrium with respect to ANY axis

6. Identify the point where each force acts and calculate the torque produced about the axis of rotation. Set t = 0.

\

6. Solve equations in 4 and 6 to determine unknowns

32

SEESAW or PLANK PROBLEM: A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum. Find the forces F1 and F2 that the bolt and the fulcrum, respectively, exert on the board

F1

F2

W

1.40m

3.90m

530

0

0

21

21

FF

WFF

Fy

NF

F

mgrrF W

4.946

1325)5.2)(530(4.1

0

0

1

1

11

NF 14762 Choose Axis

(r=0)

Center of GravityThe center of mass of an object or system is the point at which all the WEIGHT can be assumed to reside.

Sometimes the system is an assortment of particles and sometimes it is a solid object.

Mathematically, you can think of the center of mass as a “weighted average”.

CMn

nnCG x

mggmgm

xgmxgmxgmx

...

)(...)()(

21

2211

If gravitational field is uniform throughout object, then CM and CG are the same

Center of Gravity, TORQUE and EquilibriumIn extended objects or a group of objects, force of gravity acts at the CG so the weight of an object can produce a torque

CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG is in line with a pivot point, then the object or system is in equilibrium.

STABLE

CG

W

FN

0yF 0

CG

W

FN

0yF 0UNSTABLE

Center of Gravity, TORQUE and EquilibriumIn extended objects, force of gravity acts at the CG so the weight of an object can produce a torque

CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG of a system is in line with a pivot point, then the system is in equilibrium.

STABLE

CG

W

FN

0yF 0

CG

W

FN

0yF 0STABLE

Wb

CGCG

CG of the system of boy and seesaw is supported at fulcrum

1. Plank or See Saw problems

2. Shelf or Crane problems: beam sticking out of the wall

3. Ladder Problems

Some Common Equilibrium Problems

15 m

60.0

4.0 m

37

LADDER PROBLEM: An 8.00-m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N, stands 6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

38

Pivotr=0

r F=6

.3m NF

WWF

F

N

FLN

y

1230875355

0

0

2

2

sN

sN

x

fF

fF

F

1

1 0

0

FN1

WFWL

fS

FN2

fsNF

F

rWrWrF

N

N

LLFFN

727

040sin)4)(355(40sin)3.6)(875(50sin8

040sin40sin50sin

0

1

1

11

r L=4.

0m

39

SHELF PROBLEM: A safe of mass 430kg hangs by a rope from the end of a boom that is 3.0 m long. The boom consists of a hinged beam and a horizontal cable that connects the beam to a wall. The beam is uniform and has a mass of 85 kg; the mass of the cable and rope are negligible. What is the tension in the cable?

37o

FHx

FHy rB=1.5m

r=3.0m

WS

WBT1

T1

T2

////

//Pivotr=0

40

SHELF PROBLEM: What is the tension in the cable? Find the vertical and horizontal hinge forces

37o

FHx

FHy rB=1.5m

r=3.0m

WS

WBT1

T1

T2

////

//

Pivotr=0

NTT

TggTrTrWr TTBB

TTWB

61460805.135.100969.997

037sin)3(53sin)430)(3(53sin)85(5.1037sin53sin53sin

00

2

2

2

2211

21

Which of these objects is in static equilibrium? D

A. B.

C. D.

A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown at right. Where should a third force of magnitude 5 N be applied to put the piece of plywood into equilibrium? A

Problem: A cat walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one 0.440 m from the left end of the board and the other 1.50 m from its right end. When the cat reaches the right end, the plank just begins to tip. What is the mass of the cat?

Rolling MotionCombination of rotation and translation

vT = wR

vT

vCM

vCM

vCM

+

pure ROTATION

pure TRANSLATION

When an object rolls without slipping, there is a relationship between the rotational and translational motion.

When there is slidding or slipping, there is NO relationship between the rotational and translational motions.

vCM

R

Rolling Motion(without slipping)

rvCM

)/( sradin

Linear speed

Rotational speed

raCM

)/( 2sradin

Linear acceleratio

n

Rotational acceleration

DxCM = s = rqhttp://tube.geogebra.org/student/m23809

http://tube.geogebra.org/student/m23809


Rolling MotionCombination of rotation and translation

Rolling and Slipping

CM Travels more than 1 circumference for every full rotation

No relationship between the translational motion (vCM) and rotational motion (Rw) dCM > Rq vCM > Rw aCM > Ra

Rolling without Slipping

CM Travels 1 circumference for every full rotation

There is a relationship between the translational motion (vCM) and rotational motion (Rw) dCM = Rq vCM = Rw aCM = Ra

ROLLING WITHOUT SLIPPINGVelocity at any point on a rolling object

Velocity at point in

pure ROTATION

(in CM ref frame)

Rw=vT

vT

vCM= vT

vCM

vCM = -vT

2vCM

vCM

0

+ =

Velocity at pointin

pure TRANSLATION

(relative to ground)

Velocity at pointin

COMBINED ROLLING MOTION

(relative to ground)

A wheel rolls without slipping. Which is the correct velocity vector for point P on the wheel?

vCMP

A. B. C. D. E.

C

vCM

vT(In CMframe)

v=2vCM

vT(In CM frame)

v

vCM

49

Example An Accelerating Car: An automobile starts from rest and for 20.0 s has a constant linear acceleration of 0.800 m/s2 to the right. During this period, the tires do not slip. The radius of the tires is 0.330 m. At the end of the 20.0-s interval, what is the angle through which each wheel has rotated?

st

sradra

20?

/42.233.0/8.0/

0

2

0

rad

tt

484)20)(42.2(0 2

21

221

0

Rotational Dynamics and

Newtons 2nd Law

What is the RELATIONSHIP between TORQUE and ANGULAR ACCELERATION?

r

w

F FT

Fr

22

mrr

vmmaFF T

ccr

FT causes tangential acceleration; it causes a change in the speed or magnitude of the velocity.

Fr changes only the direction of velocity. It does not affect the speed; it does no work. It produces no torque.

tT maF

NONuniform Circular Motion and Newtons 2nd Law

m

r

w

F FT

Fr

2

90sinmr

rmarF

maF

TT

TT

NONuniform Circular Motion and Newtons 2nd Law

This is the relationship between torque and angular acceleration for a single particle moving in circular motion. To look at rotational dynamics, must expand idea to an extended rigid object.

(Only if angle unit is

radians)

w

2

2222

211

321

.....

...

ii

nn

n

rmrmrmrm

Rotational Motion and Newtons 2nd Law

m1

r1

m2r2

m3

r3

F2

F1

F3

t1 = r1FT1 = r1m1aT1 = m1r12a

t2 = r2FT2 = r2m2aT2 = m2r22a

t3 = r3FT3 = r3m3aT3 = m3r32a

If we did this for ALL particles in the object, the net torque on the entire object would be

ALL pts on rigid body rotate at same

a

Rotational Inertia, I

w 2

iirm

Rotational Motion and Newtons 2nd Law

m1

r1

m2r2

m3

r3

F2

F1

F3


a


I (angle unit must be in radians)

Newtons 2nd Law for Rotation of a Rigid Body

Rotational equivalent of

FORCE

Rotational equivalent of MASS

2iirmI

Rotational Inertia (Moment of Inertia)

Moment of inertia is the rotational equivalent of mass.Rotational Inertia of an object depends on the mass AND on how the mass is distributed about an axis of rotation. It is constant for a particular rigid body and a particular axis.

ri is the perpendicular distance between the mass and the rotation axis.

(Units are kg m2)

Both involve same mass.Left is easier to rotate since the mass is distributed closer to the rotation axis.

2iirmI

Rotational Inertia (Moment of Inertia)ri is the perpendicular distance between the mass and the rotation axis.

(Units are kg m2)

I depends on where the axis of rotations

isBoth involve same mass.Top is easier to rotate since the mass is distributed closer to the rotation axis.

Which rod has a greater Moment of Inertia (is harder to rotate)?

Larger I (harder to rotate)

Smaller I (easier to rotate)

Problem: Two 3kg masses are at the ends of a 4m bar with negligible mass. What is the moment of inertia of the object when

2

22

222

211

2

24)2(3)2(3

kgm

rmrmrmI ii

a) when the masses rotate around the center of the rod

b) when the masses rotate around one end of the rod

2

2

222

211

2

48)4(3)0(3

kgm

rmrmrmI ii

3kg

w

4m

3kg3kg

w

2m3kg

2m

22111 36kgmrmI

Rotational Inertia (Moment of Inertia)Three small spheres that rotate about a vertical axis are shown.

The perpendicular distance between the axis of rotation and center of mass of each sphere is given. Rank the 3 spheres from greatest to least rotational inertia.

1 m

2 m

3 m

wm1=36kg

m2=9kg

m3=4kg

2

22222

36)2)(9(

kgmrmI

2

22333

36)3)(4(

kgmrmI

2iirmI

Rotational Inertia (Moment of Inertia)

For a solid object

ri is the perpendicular distance between the mass and the rotation axis.

(Units are kg m2)

dmrI 2

Moments of Inertia of Common Shapes

Linear Dynamics Rotational Dynamics

Net Force (N):

Fnet

Net Torque (N m):

tnet

Mass (kg): mMoment of Inertia (kg m2):

I

Acceleration (m/s2): aAngularAcceleration (rad/s): a

Newtons 2nd Law

Newtons 2nd Law

m

Fa net

Inet

Acceleration is caused by forces.

The larger the mass, the smaller the acceleration.

Angular acceleration is caused by torques.

The larger the moment of inertia, the smaller the angular acceleration.

Using Newton’s 2nd Law for RotationIn the Scottish game of caber toss, contestants toss a heavy uniform pole, landing it on its end. A 5.9m pole with a mass of 79kg has just landed on its end. It is tipped 25o from the vertical and is starting to rotate about the end that touches the ground. A) Find the angular acceleration. B) What is the tangential acceleration of the left end of the log?

FgFN

Pivotr=02

231

21

231

/1.12

25sin325sin)(

sradL

gMLMgL

ML

I

W

231MLI

25o

gg

LraT 63.02

25sin3

Ropes and PulleysLinear Motion Angular

MotionFrictionless, massless pulley

Pulley does NOT rotate

Rope slides over frictionless pulley

Rope tension does NOT change when it passes over a frictionless, massless pulley

Pulley has mass and friction

Pulley rotates

If pulley TURNS WITHOUT THE ROPE SLIPPING ON IT, then vrope=vT pulley=Rw arope = aTpulley=Ra

Rope tension CHANGES when it passes over a rotating pulley with friction.

T1/T1 / T3

T2

w

R

a Pulley RotationAt rim, vT = Rw aT = Ra

Rope Translationvrope = Rwarope = Ra

Ropes and Pulleys

Motion constraints for an object connected to a pulley of radius R by a nonslipping rope

T3T2

w

R

a

vobject = Rw

aobject = Ra

Note that these are magnitudes only

maFmg T

A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s?

MR

m

mg

FT

FT

Mg

F1

//

////

MaF

R

aMRRF

MR

I

T

TT

FT

21

221

221

Block Pulley

2

21

21

/8.42

2

)(

smMm

mga

mgMma

maMamg

a+

a+t


MR

m

mg

FT

FT

Mg

F1

//

//// Pulley

2/24

2.0

8.4

)2(

2

srad

MmR

mgRa

NMm

mMgMm

mgMMaFT

22.12

2

2

2

121


MR

m

mg

FT

FT

Mg

F1

//

//// Pulley

2/242.0

8.4

)2(

2

sradMmR

mgRaT

NMm

mMgMm

mgMMaFT

22.12

2

2

2

121

Rotational Work and Energy

WorkWhen a force acts on an object over a distance, it is said that work was done upon the object. Work tells us how much a force transfers energy to a system.

sF

FsW

//

cos

Work done by a constant force that points in the same direction as the displacement

FrWFsW

R

Constant force pulls rope out a distance, s

Wheel rotates through the angle q = s/r

Rotational Work

Unit: J must be in rad

r

FT

sr

FT

q

s

Work done by a force

Work done by a torque

w

2

21

2221

222122

222122

1121

321

.....

...

IK

rmKrmrmrm

KKKKK

R

iiR

nn

nRRRRR

Rotational Kinetic Energy

m1

r1

m2r2

m3

r3

K1 = ½m1v1T2 = ½m1r1

2w2

K2 = ½m2v2T2 = ½m2r2

2w2

K3 = ½m3v3T2 = ½m3r3

2w2

If we did this for ALL particles in the object, the total rotational kinetic energy would be


w


vT2

vT1

vT3

Kinetic Energy

(angle unit must be in radians)Rotational

Kinetic Energy

Rotational equivalent of MASS

221 IKR

221 mvKT

Translational Kinetic Energy

Rotational Kinetic Energy

Tnet KW RnetR KW

PowerTranslational

Rotational

vFt

sF

t

W

t

EP //

//

tt

W

t

EP RR

A solid disk and a hoop with mass M and radius R start from rest and roll without slipping down an incline. What is the velocity at the bottom in terms of h?

fNCi EWE

s

q

w Disk

Hoop

Sphere

SphericalShell

221 mrI

2mrI

232 mrI

252 mrI

h

ghv

MvMvMgh

MRMvMgh

IMvMgh

UKKUKKEE

B

BB

Rv

B

BB

gBRBTBgTRTTT

BT

B

34

2412

21

2221

212

21

2212

21

))((

0

Disk

Disk Hoop ghvB

Roll without slipvCM=Rw

System is isolated, WNC=0 since FN and fS do no work on object

Find the acceleration of an object rolling down an inclined plane. The object rolls without slipping. It has a mass of M and radius R.

MafMgMaF

s

x

sin

FN

Fg

Fg sinq

Fgcosq

fS

//

//

q

q

2R

Iaf

IIRfI

s

Ra

s

+ ,a t Torque around Rotation axis at CG

22

2

2

1

sinsin

sin)(

sin

MRI

RI

RI

RIa

g

M

Mga

MgMa

MaMg

Disk

Hoop

Sphere

SphericalShell

221 mrI

2mrI

232 mrI

252 mrI

+a

Which wins the RACE? 4 different spherical objects with the same mass and radius are released from rest and roll down an inclined plane. The objects roll without slipping. Rank the objects according to the order that they reach the finish line

AB

CD

Spherical Shell

DiskHoop

Sphere

C < A < D < B

Disk

Hoop

Sphere

SphericalShell

221 mrI

2mrI

232 mrI

252 mrI



NONuniform Circular MotionA single particle moving in circular motion has rotational kinetic energy and angular momentum and can experience a torque if a force acts tangentially.

(angle unit is radians)

r

w

F

FT

Fr

vT

Moment of inertia of single particle moving in a circle of radius r is mr2

(all mass is traveling at r)

Imr

rmarFrF TT

2

sin

22122

212

21 ImrmvK TR

Angular Momentum

Momentum

(angle unit must be in radians)

IL mvp Linear momentum

Angular Momentum

t

pFnet

t

Lnet

If there are no external forces on a system of particles, momentum of the system is conserved

If there are no external torques on a system of particles, angular momentum of the system is conserved

Conservation of Angular Momentum

Conservation of Angular Momentum

Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min.

Simultaneously, four students walk up and place four 5-kg boxes (I = mr2) symmetrically along its outer edge so that each box's center of mass is located 1.50 m from the axis of the carousel. What will become the carousel's new angular velocity at the instant the boxes are in place?

before after

rpmrpm

IIILL

BCCC

64.9'))25.11(481(')15)(81(

'4''

22

2221

25.1181

mkgmrImkgMRI

B

C

Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min.

Suppose instead that they had placed the boxes so that each box's center of mass was located 50 cm from the axis of the merry-go-round, what would have been its new angular velocity?

before after

rpmrpm

IIILL

BCCC

1.14'))25.1(481(')15)(81(

'4''

22

2221

25.181

mkgmrImkgMRI

B

C

Linear and angular momentum are SEPARATELY CONSERVED

One form of momentum does NOT transform into the other form

w

p=mv

L = 0 p = 0L ≠ 0

Linear momentum of the system is NOT conserved since there is an external force on the system (friction between turntable and surface). Linear momentum gets transferred to the EarthAngular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the bullet transfers to the turntable.

r

v

Angular Momentum of a Point ParticleIn general, a single particle moving in any way has angular momentum

relative to an axis point, O

q

r

O 2

mrIL

sinrpmvrL

Radius is closest distance to rotation

axis, r

Ballistic Pendulum

Linear momentum of the system is NOT conserved since the pendulum is fixed to the earth at the axis of rotation and this provides and external force. Linear momentum gets transferred to the Earth

Angular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the ball gets transferred to the pendulum.

w’

Ballistic Pendulum

ddM

dm, v

w’

dMm

mv

Mdmdmvd

IIIImvd

LLLL

LL

RbRb

RbRb

)('

)('

)('''

''

'

31

2312

Angular Momentum in Astronomy

L=mvR

L=mvR

The Angular momentum of the orbiting object is the same at every point on the orbit since the force of gravity produces no external torque. Therefore the orbiting object travels faster when it is closer to the rotation axis and slower when it is further away. This is Kepler’s 3rd Law

Linear Dynamics

Rotational Dynamics

Displacement s, d qVelocity v wAcceleration a aCause of acceleration

Fnet, net force tnet, net torque

Inertia m, Mass I, Moment of Inertia

Newton’s 2nd Law F = ma = t Ia

Work W = F//s WR = tq

Kinetic Energy K = ½mv2 K R= ½Iw2

Momentum p = mv L = Iw

A baseball catcher sits on a rotating stool. He reaches out 85.0cm tocatch a 40.0m/s fastball. After catching the ball he spins at a rate of60.0rpm. His mass is 80.0kg and the mass of the ball is 150g. Find therotational inertia of the catcher and the stool.

85 cm

0.81 kg m2

Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational.

Documents

angular velocity vector

velocity of rotation

angular velocity changes

rad slide

angular position rad

ms slide

causes of motion slide

o slide