Rotational KinematicsRotational Kinematics

and Energyand Energy

Rotational MotionRotational Motion

Up until now we have been looking at the Up until now we have been looking at the kinematics and dynamics of kinematics and dynamics of translationaltranslational motion – motion – that is, motion without rotation. Now we will widen that is, motion without rotation. Now we will widen our view of the natural world to include objects that our view of the natural world to include objects that both rotate and translate.both rotate and translate.

We will develop descriptions (equations) that We will develop descriptions (equations) that describe rotational motiondescribe rotational motion

Now we can look at motion of bicycle wheels, Now we can look at motion of bicycle wheels, roundabouts and divers.roundabouts and divers.

II.II. Rotation with constant angular accelerationRotation with constant angular acceleration

III. Relation between linear and angular variablesIII. Relation between linear and angular variables- Position, speed, acceleration- Position, speed, acceleration

I.I. Rotational variablesRotational variables - Angular position, displacement, velocity, acceleration- Angular position, displacement, velocity, acceleration

IV. Kinetic energy of rotationKinetic energy of rotation

V. Rotational inertiaV. Rotational inertia

VI. TorqueVI. Torque

VII. Newton’s second law for rotationVII. Newton’s second law for rotation

VIII. Work and rotational kinetic energyVIII. Work and rotational kinetic energy

Rotational kinematicsRotational kinematics

• In the kinematics of rotation we encounter In the kinematics of rotation we encounter new kinematic quantitiesnew kinematic quantities– Angular displacementAngular displacement – Angular speedAngular speed – Angular accelerationAngular acceleration – Rotational InertiaRotational Inertia II– TorqueTorque

• All these quantities are defined All these quantities are defined relative to an relative to an axis of rotationaxis of rotation

Angular displacementAngular displacement

• Measured in radians or degreesMeasured in radians or degrees• There is no dimensionThere is no dimension

= = ff - - ii

Axis of rotationAxis of rotation

i

f

CWradif 3

2

Angular displacement and arc lengthAngular displacement and arc length

• Arc length depends Arc length depends on the distance it is on the distance it is measured away measured away from the axis of from the axis of rotationrotation

CWrif 3

2

Axis of rotationAxis of rotation

Qsp

sq

Pri

fr

s

qq

pp

rS

rS

Angular SpeedAngular Speed

• Angular speed is the rate of change of Angular speed is the rate of change of angular positionangular position

• We can also define the We can also define the instantaneous angular speedinstantaneous angular speed

t

t lim

0

t

Average angular velocity and Average angular velocity and tangential speedtangential speed

• Recall that speed is distance divided by Recall that speed is distance divided by time elapsedtime elapsed

• Tangential speed is arc length divided by Tangential speed is arc length divided by time elapsedtime elapsed

• And because we can writeAnd because we can write

t

svt

r

v

tr

s

tT

1

r

s

Average Angular AccelerationAverage Angular Acceleration

• Rate of change of angular velocityRate of change of angular velocity

• Instantaneous angular accelerationInstantaneous angular acceleration

ttif

tt

0

lim

Angular acceleration and tangential Angular acceleration and tangential accelerationacceleration

• We can find a link between tangential We can find a link between tangential acceleration acceleration aatt and angular acceleration and angular acceleration αα

• SoSo

r

a

trv

r

v

ttT

if

if

r

aT

Centripetal accelerationCentripetal acceleration

• We have thatWe have that

• But we also know thatBut we also know that

• So we can also say So we can also say

r

va Tc

2

222 rvT

rac2

Example: RotationExample: Rotation

• A dryer rotates at A dryer rotates at 120 rpm120 rpm. What distance do your . What distance do your clothes travel during one half hour of drying time in clothes travel during one half hour of drying time in a a 70 cm70 cm diameter dryer? What angle is swept out? diameter dryer? What angle is swept out?

Rotational motion with constant Rotational motion with constant angular accelerationangular acceleration

• We will consider cases where We will consider cases where is constant is constant

• Definitions of rotational and translational Definitions of rotational and translational

quantities look similarquantities look similar

• The kinematic equations describing rotational The kinematic equations describing rotational

motion also look similarmotion also look similar

• Each of the translational kinematic equations Each of the translational kinematic equations

has a rotational analoguehas a rotational analogue

Rotational and Translational Rotational and Translational Kinematic EquationsKinematic Equations

2

22

22

1

fi

2f

if

vvv

xavv

avx

avv

i

i tt

t

Constant Constant motion motion

What is the angular acceleration of a car’s wheels What is the angular acceleration of a car’s wheels (radius (radius 25 cm25 cm) when a car accelerates from ) when a car accelerates from 2 m/s2 m/s to to 5 5 m/sm/s in in 8 8 seconds?seconds?

A rotating wheel requires A rotating wheel requires 3.00 s3.00 s to rotate through to rotate through 37.0 revolutions37.0 revolutions. Its angular speed at the end of the . Its angular speed at the end of the 3.00-s3.00-s interval is interval is 98.0 rad/s98.0 rad/s. What is the constant . What is the constant angular acceleration of the wheel?angular acceleration of the wheel?

Rotational DynamicsRotational Dynamics

• Easier to move door atEasier to move door at AA than atthan at BB using the using the same forcesame force FF

• MoreMore torquetorque is exerted atis exerted at AA than atthan at BB

A B

hinge

TorqueTorque

• Torque is the rotational analogue of ForceTorque is the rotational analogue of Force

• Torque, Torque, , is defined to be, is defined to be

WhereWhere FF is the force applied is the force applied tangent to the tangent to the rotationrotation and and rr is the distance from the axis of is the distance from the axis of rotationrotation

r

F

rF

TorqueTorque

• A general definition of torque isA general definition of torque is

• Units of torque are Units of torque are NmNm

• Sign convention used with torqueSign convention used with torque

– Torque is positive if object tends to rotate CCWTorque is positive if object tends to rotate CCW

– Torque is negative if object tends to rotate CWTorque is negative if object tends to rotate CW

r

F = Fsin= Fsin r r

Condition for EquilibriumCondition for Equilibrium

• We know that if an object is in (translational) We know that if an object is in (translational) equilibrium then it does not accelerate. We equilibrium then it does not accelerate. We can say thatcan say that F = 0F = 0

• An object in rotational equilibrium does not An object in rotational equilibrium does not change its rotational speed. In this case we change its rotational speed. In this case we can say that there is no can say that there is no net torquenet torque or in other or in other words that:words that:

= 0= 0

• An unbalanced torque (An unbalanced torque () gives rise to an angular ) gives rise to an angular acceleration (acceleration ())

• We can find an expression analogous to We can find an expression analogous to

F = maF = ma that relates that relates and and • We can see thatWe can see that

FFtt = ma = matt

• and and FFttr = mar = mattr = mrr = mr22

(since (since aatt = r = r• Therefore Therefore

Torque and angular accelerationTorque and angular acceleration

mrFt

= mr= mr22

Torque and Angular Acceleration

Angular acceleration is directly proportional to the net Angular acceleration is directly proportional to the net

torque, but the constant of proportionality has to do with torque, but the constant of proportionality has to do with

both the mass of the object and the distance of the object both the mass of the object and the distance of the object

from the axis of rotation – in this case the constant isfrom the axis of rotation – in this case the constant is mrmr22

This constant is called theThis constant is called the moment of inertiamoment of inertia. . Its Its

symbol issymbol is II, and its units are, and its units are kgmkgm22

II depends on the arrangement of the rotating system. depends on the arrangement of the rotating system.

It might be different when the same mass is rotating about It might be different when the same mass is rotating about

a different axisa different axis

= mr= mr22

Newton’s Second Law for RotationNewton’s Second Law for Rotation

• We now have thatWe now have that

• Where Where II is a constant related to the is a constant related to the distribution of mass in the rotating distribution of mass in the rotating systemsystem

• This is a new version of Newton’s This is a new version of Newton’s second law that applies to rotationsecond law that applies to rotation

= I= I

A fish takes a line and pulls it with a tension of A fish takes a line and pulls it with a tension of 15 N15 N for for 20 20 seconds. The spool has a radius of seconds. The spool has a radius of 7.5 cm7.5 cm. If the moment . If the moment of inertia of the reel is of inertia of the reel is 10 kgm10 kgm22, through how many rotations , through how many rotations does the reel spin? (Assume there is no friction)does the reel spin? (Assume there is no friction)

Angular Acceleration andAngular Acceleration and II

The angular acceleration reached by a The angular acceleration reached by a rotating object depends on, rotating object depends on, MM, , rr, (their , (their distribution) and distribution) and

When objects are rolling under the influence When objects are rolling under the influence of gravity, only the mass distribution and the of gravity, only the mass distribution and the radius are importantradius are important

••

Moments of Inertia for Rotating ObjectsMoments of Inertia for Rotating Objects

II for a small mass for a small mass mm rotating about a point a rotating about a point a distance distance rr away is away is mrmr22

What is the moment of inertia for an object that What is the moment of inertia for an object that is rotating – such as a rolling object?is rotating – such as a rolling object?

Disc?Disc?Sphere?Sphere?Hoop?Hoop?Cylinder?Cylinder?

Moments of Inertia for Rotating Moments of Inertia for Rotating ObjectsObjects

The total torque on a rotating system is the sum The total torque on a rotating system is the sum of the torques acting on all particles of the of the torques acting on all particles of the system about the axis of rotation – system about the axis of rotation –

and sinceand since is the same for all particles:is the same for all particles:

I mr2 = m1r12+ m2r2

2+ m3r32+…

Axis of rotationAxis of rotation

Continuous ObjectsContinuous Objects

To calculate the moment of inertia for To calculate the moment of inertia for continuous objects, we imagine the object to continuous objects, we imagine the object to consist of a continuum of very small mass consist of a continuum of very small mass elements elements dmdm. Thus the finite sum . Thus the finite sum ΣmΣmi i rr22

ii

becomes the integralbecomes the integral

dmrI 2

Moment of Inertia of a Uniform RodMoment of Inertia of a Uniform Rod

L

Find the moment of inertia of a uniform rod of Find the moment of inertia of a uniform rod of length length LL and mass and mass MM about an axis perpendicular about an axis perpendicular to the rod and through one end. Assume that the to the rod and through one end. Assume that the rod has negligible thickness. rod has negligible thickness.

Example:Example:Moment of Inertia of a DumbbellMoment of Inertia of a Dumbbell

A dumbbell consist of point masses A dumbbell consist of point masses 2kg2kg and and 1kg 1kg attached by a attached by a rigid massless rod of length rigid massless rod of length 0.6m0.6m. Calculate the rotational inertia . Calculate the rotational inertia of the dumbbell (a) about the axis going through the center of of the dumbbell (a) about the axis going through the center of

the mass and (b) going through the the mass and (b) going through the 2kg2kg mass. mass.

Example:Example:Moment of Inertia of a DumbbellMoment of Inertia of a Dumbbell

2222 4.0)6.0)(1()( kgmmkgLmIb

Moment of Inertia of a Uniform HoopMoment of Inertia of a Uniform Hoop

R

dmAll mass of the hoop All mass of the hoop MM is at distance is at distance r = Rr = R from from the axisthe axis

20

2

00

222 MRmRdmRdmRdmrIM

MM

Moment of Inertia of a Uniform DiscMoment of Inertia of a Uniform Disc

R

dr

Each mass element is a hoop of radius Each mass element is a hoop of radius rr and and thicknessthickness drdr. Mass per unit area. Mass per unit area

σ = M / A = M /πRσ = M / A = M /πR22

r

We expect that We expect that II will will be smaller than be smaller than MRMR22 since the mass is since the mass is uniformly distributed uniformly distributed from from r = 0r = 0 to to r = Rr = R rather than being rather than being concentrated atconcentrated at r = Rr = R as it is in the hoop.as it is in the hoop.

Moment of Inertia of a Uniform DiscMoment of Inertia of a Uniform Disc

R

dr

r

rdrA

MdAdm 2

R R

drrrdrrdmrI0 0

322 22

242

4

2

1

24

2MRR

R

MR

A

MI

Moments of inertia Moments of inertia II for Different Mass Arrangements for Different Mass Arrangements

Moments of inertia Moments of inertia II for Different Mass Arrangementsfor Different Mass Arrangements

Which one will win?Which one will win?

A hoop, disc and sphere are all rolled A hoop, disc and sphere are all rolled down an inclined plane. Which one will win?down an inclined plane. Which one will win?

Which one will win?Which one will win?

A hoop, disc and sphere are all rolled A hoop, disc and sphere are all rolled down an inclined plane. Which one will win?down an inclined plane. Which one will win?

1. Hoop I = MR1. Hoop I = MR22

2. Disc I = ½MR2. Disc I = ½MR22

3. Sphere I = 2/5MR3. Sphere I = 2/5MR22

= / I

1.1. 11 = = / MR/ MR22

2.2. αα22= 2(= 2(/ MR/ MR22))

.. = 2.5(= 2.5(/ /

MRMR22))

Kinetic energy of rotationKinetic energy of rotation

Reminder:Reminder: Angular velocityAngular velocity,, ωω is the same for all is the same for all particles within the rotating body. particles within the rotating body.

Linear velocityLinear velocity,, vv of a particle within the rigid body of a particle within the rigid body depends on the particle’s distance to the rotation axis (depends on the particle’s distance to the rotation axis (rr).).

2222

233

222

211

2

1)(

2

1

2

1

...2

1

2

1

2

1

i

ii

i

ii

i

ii rmrmvm

vmvmvmK

Moment of InertiaMoment of Inertia

2

2

1 IK

Kinetic energy of a body in pure rotationKinetic energy of a body in pure rotation

Kinetic energy of a body in pure translationKinetic energy of a body in pure translation

2

2

1COMMvK

VII. Work and Rotational kinetic energyVII. Work and Rotational kinetic energy

TranslationTranslation RotRotation

WmvmvKKK ifif 22

2

1

2

1WIIKKK ifif 22

2

1

2

1

f

i

x

x

FdxW f

i

dW

Work-kinetic energy Work-kinetic energy TheoremTheorem

Work, rotation about fixed axisWork, rotation about fixed axis

dFW )( ifW Work, constant torque

vFdt

dWP

dt

dWP

Power, rotation about Power, rotation about fixed axisfixed axis

Proof:

2222222222

2

1

2

1)(

2

1)(

2

1)(

2

1)(

2

1

2

1

2

1ififififif IImrmrrmrmmvmvKKKW

f

i

dWddrFdsFdW tt

dt

d

dt

dWP

Rotational Kinetic EnergyRotational Kinetic Energy

We must rewrite our statements of conservation of We must rewrite our statements of conservation of mechanical energy to include mechanical energy to include KEKErr

Must now allow that (in general):Must now allow that (in general):

½ mv2+mgh+ ½ I2 = constant

Could also add in e.g. spring Could also add in e.g. spring PEPE

Example - Rotational KE

• What is the linear speed of a ball with radius What is the linear speed of a ball with radius 1 cm1 cm when it reaches the end of a when it reaches the end of a 2.0 m2.0 m high high 3030oo incline?incline?

mgh+ ½ mvmgh+ ½ mv22+ ½ I+ ½ I22 = constant = constant• Is there enough information?Is there enough information?

2 m

Example - Rotational KE

So we

have that:

The velocity of the centre of mass and the

tangential speed of the sphere are the same,

so we can say that:

Rearranging for vf:

2

5

2MRISphere

222

222

5

1

2

15

2

2

1

2

1

ffi

ffi

Rvgh

MRMvMgh

R

v

R

vt

smgh

v

vvvgh

if

fffi

/3.57.0

28.9

7.0

7.05

1

2

1 222

mgh+ ½ mvmgh+ ½ mv22+ ½ I+ ½ Iωω22 = constant = constant

Example: Conservation of KEExample: Conservation of KErr

A boy of mass A boy of mass 30 kg30 kg is flung off the edge of a roundabout is flung off the edge of a roundabout ((m = 400 kgm = 400 kg,, r = 1 mr = 1 m) that is travelling at ) that is travelling at 2 rpm2 rpm. What is . What is the speed of the roundabout after he falls off?the speed of the roundabout after he falls off?

During a certain period of time, the angular position of a During a certain period of time, the angular position of a swinging door is described by swinging door is described by

θθ= 5.00 + 10.0= 5.00 + 10.0tt + 2.00 + 2.00tt22

wherewhere θθ is in radians andis in radians and tt is in seconds. Determine the is in seconds. Determine the angular position, angular speed, and angular angular position, angular speed, and angular acceleration of the door (a) atacceleration of the door (a) at tt = 0 = 0 and (b) atand (b) at tt = 3.00 s = 3.00 s..

The four particles are connected by The four particles are connected by rigid rods of negligible mass. The rigid rods of negligible mass. The origin is at the center of the origin is at the center of the rectangle. If the system rotates in rectangle. If the system rotates in the the xyxy plane about the plane about the zz axis with axis with an angular speed of an angular speed of 6.00 rad/s6.00 rad/s, , calculate (a) the moment of inertia calculate (a) the moment of inertia of the system about the of the system about the zz axis and axis and (b) the rotational kinetic energy of (b) the rotational kinetic energy of the system.the system.

Parallel axis theoremParallel axis theorem

2MhII COM

Proof:Proof:

Rotational inertia about a given axisRotational inertia about a given axis = = Rotational Inertia about a parallel axis Rotational Inertia about a parallel axis that extends trough body’s Center of that extends trough body’s Center of Mass + Mass + MhMh22

h h = perpendicular distance between the given axis and axis = perpendicular distance between the given axis and axis through COM.through COM.

dmbaydmbxdmadmyx

dmbyaxdmrI

)(22)(

)()(

2222

222

R

222 22 MhIMhbMyaMxdmRI COMCOMCOM

Many machines employ cams for various purposes, such as Many machines employ cams for various purposes, such as opening and closing valves. In Figure, the cam is a circular opening and closing valves. In Figure, the cam is a circular disk rotating on a shaft that does not pass through the center disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid of the disk. In the manufacture of the cam, a uniform solid cylinder of radius cylinder of radius RR is first machined. Then an off-center hole is first machined. Then an off-center hole of radius of radius RR/2/2 is drilled, parallel to the axis of the cylinder, and is drilled, parallel to the axis of the cylinder, and centered at a point a distance centered at a point a distance RR/2/2 from the center of the from the center of the cylinder. The cam, of mass cylinder. The cam, of mass MM, is then slipped onto the circular , is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed cam when it is rotating with angular speed ωω about the axis of about the axis of the shaft?the shaft?

Find the net torque on the wheel in Figure about the axle Find the net torque on the wheel in Figure about the axle through through OO if if aa = 10.0 cm = 10.0 cm and and bb = 25.0 cm = 25.0 cm..

A block of mass A block of mass mm11 = 2.00 kg = 2.00 kg

and a block of mass and a block of mass mm2 2 = =

6.00 kg6.00 kg are connected by a are connected by a massless string over a pulley massless string over a pulley in the shape of a solid disk in the shape of a solid disk having radius having radius RR = 0.250m = 0.250m and mass and mass M M = 10.0 kg= 10.0 kg. . These blocks are allowed to move on a fixed block-wedge These blocks are allowed to move on a fixed block-wedge of angle = of angle = 30.030.0 as in Figure. The coefficient of kinetic as in Figure. The coefficient of kinetic friction is friction is 0.3600.360 for both blocks. Draw free-body diagrams for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley. string on both sides of the pulley.

A uniform rod A uniform rod 1.1 m1.1 m long with mass long with mass 0.7 kg0.7 kg is pivoted at is pivoted at one end, as shown in Fig., and released from a horizontal one end, as shown in Fig., and released from a horizontal position. Find the torque about the pivot exerted by the position. Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod force of gravity as a function of the angle that the rod makes with the horizontal direction.makes with the horizontal direction.

A seesaw pivots as shown in Fig. (a) What is the net torque A seesaw pivots as shown in Fig. (a) What is the net torque about the pivot point? (b) Give an example for which the about the pivot point? (b) Give an example for which the application of three different forces and their points of application of three different forces and their points of application will balance the seesaw. Two of the forces must application will balance the seesaw. Two of the forces must point down and the other one up.point down and the other one up.

Four small spheres are fastened to the corners of a frame of Four small spheres are fastened to the corners of a frame of negligible mass lying in the negligible mass lying in the xyxy plane (Fig. 10.7). Two of the plane (Fig. 10.7). Two of the spheres have massspheres have mass mm = 3.1kg = 3.1kg and are a distanceand are a distance aa = 1.7 m = 1.7 m from the origin and the other two have massfrom the origin and the other two have mass MM = 1.4 kg = 1.4 kg and are and are a distancea distance aa = 1.5 = 1.5 mm from the origin. from the origin. (a) If the rotation of the system occurs about the (a) If the rotation of the system occurs about the yy axis, as in axis, as in Figure a, with an angular speedFigure a, with an angular speed ωω = 5.1rad/s = 5.1rad/s, find the moment , find the moment of inertiaof inertia IIyy about the about the yy axis and the rotational kinetic energy axis and the rotational kinetic energy about this axis. about this axis. Suppose the system rotates in the Suppose the system rotates in the xyxy plane about an axis (the plane about an axis (the zz axis) through axis) through OO (Fig. b). Calculate the moment of inertia about (Fig. b). Calculate the moment of inertia about the the zz axis and the rotational energy about this axis. axis and the rotational energy about this axis.

The reel shown in Figure has The reel shown in Figure has radiusradius RR and moment of and moment of inertiainertia II.. One end of the block One end of the block of massof mass mm is connected to a is connected to a spring of force constantspring of force constant kk, and , and the other end is fastened to a the other end is fastened to a cord wrapped around the reel. cord wrapped around the reel. The reel axle and the incline The reel axle and the incline are frictionless. The reel is are frictionless. The reel is wound counterclockwise so wound counterclockwise so that the spring stretches athat the spring stretches adistancedistance dd from its unstretched position and is then released from its unstretched position and is then released from rest. (a) Find the angular speed of the reel when the from rest. (a) Find the angular speed of the reel when the spring is again unstretched. (b) Evaluate the angular speed spring is again unstretched. (b) Evaluate the angular speed numerically at this point ifnumerically at this point if II = 1.00 kg·m = 1.00 kg·m22,, RR = 0.300 m = 0.300 m, , kk = 50.0 N/m = 50.0 N/m,, mm = 0.500 kg = 0.500 kg, , dd = 0.200 m = 0.200 m, and, and θθ= 37.0°= 37.0°..

A tennis ball is a hollow sphere with a thin wall. It is set A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at rolling without slipping at 4.03 m/s4.03 m/s on a horizontal section of on a horizontal section of a track, as shown in Figure. It rolls around the inside of a a track, as shown in Figure. It rolls around the inside of a vertical circular loop vertical circular loop 90.0 cm90.0 cm in diameter, and finally leaves in diameter, and finally leaves the track at a point the track at a point 20.0 cm20.0 cm below the horizontal section. below the horizontal section. (a) Find the speed of the ball at the top of the loop. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall from the track. (b) Find its Demonstrate that it will not fall from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction speed as it leaves the track. (c) Suppose that static friction between ball and track were negligible, so that the ball slid between ball and track were negligible, so that the ball slid

instead of rolling. Would instead of rolling. Would its its speed then be higher, speed then be higher, lower, lower, or the same at the top of or the same at the top of the the loop? Explain. loop? Explain.