9 Rotation Rotational Kinematics: Angular Velocity and Angular Acceleration Rotational Kinetic...

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Angular Velocity & Acceleration

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9 Rotation

• Rotational Kinematics: Angular Velocity and Angular Acceleration

• Rotational Kinetic Energy• Calculating the Moment of Inertia• Newton’s Second Law for Rotation• Applications of Newton’s Second Law for

Rotation• Rolling Objects• Hk: 29, 37, 41, 47, 67, 71, 79, 85, 105.

Rotation

Angular Velocity & Acceleration

[radians] Angle of Definition rdsd

[rad/s]Velocity Angular of Definition dtd

[rad/s/s]on AcceleratiAngular of Definition dtd

Ex. Angular Velocity & Acceleration

radiansin seconds,in t 532

:bygiven wheelngacceleration Point 2 tt

10t3)532( 2 ttdtd

1010t)3( dtd

Linear/Angular Relation for Rotation

rdds

rdtda

dtrd

dtdsv

rv

ra

Angular Kinematics

to to )(2

1 2

21 tto

222o

221 mvKE

]m[kg Inertia ofMoment

theof Definition 2

2

mrI

221 IKE

)( 2221 rm

2221 mr

Rotational Kinetic Energy

Ex. Rotational KE

rad/s 6s 1

rad 23dtd

second.per srevolution 3at 1m r circle ain uniformly rotates ball 2kgA

J 6)rad/s 6)(2(

2)1)(2(22

212

21

222

mkgIKE

mkgmkgmrI

Continuous Objects dmrI 2

dxLMdm

LM

dxdmdmrI

LLength RodThin UniformEx.

2

|03

31

0

2

0

2 LLLr

LMdrr

LMdr

LMrI

2313

313

31 0 MLL

LMI

Parallel Axis Theorem2

212

21 cmcm ImvKE

2212

21 )( cmIhm

2221 )( cmImh

Theorem Axis Parallel 2cmImhI

Ex. Parallel Axis Theorem

2121

22

2121

2

Rod. Uniform

MLLMIMhI

MLI

cmend

cm

2312

121

1232

1212

41 )( MLMLMLML

Tangential Acceleration

Newton’s Second Lawfor Rotation

Particlefor Law Second mrmaF tt

I) of (in terms Law Secondr 2

IrFmrrF

t

t

Rotationfor Law Second sNewton' Inet

N][m Torque of Definition trF

Ex. Newton’s 2nd Law for Rotation

RaIITRcw

maTmgFcw

2RaIT

maRaImg 2

aRImmg

2

2RImmga

Rolling Motion

Rotational Power

drdFsFdW tt

dtddtdWP

Power Rotational P

Summary

• Angular Velocity, Acceleration• Rotational Kinetic Energy• Calculating the Moment of Inertia• Newton’s Second Law for Rotation• Rotational Power• Rolling Objects

Problems

09-1. A compact disk rotates from rest to 400 rev/min in 4.0 seconds.

a) Convert 400 rev/min into radians/second.

b) Calculate the average angular acceleration in rad/s/s.

c) Calculate the number of radians and the number of revolutions subtended by the disk during this interval.

d) How far does a point on the edge of the disk, radius 6.0cm, travel during this interval?e) If the disk has mass of 15.5 grams, what is its kinetic energy when rotating at 400 rev/min?

f) What average power is needed to accelerate the disk to 400rev/min in 4.0 seconds? (this turns out to be a relatively small part of the total power consumed by a CD player)

a) Convert 400 rev/min into radians/second.

sradsrev

radrev /9.4160min12

min400

b) Calculate the average angular acceleration in rad/s/s.

ssrads

sradtav //5.10/

00.409.41

c) Calculate the number of radians and the number of revolutions subtended by the disk during this interval.

radtav 8.8342

09.41

revradrevrad 3.13

218.83

d) How far does a point on the edge of the disk, radius 6.0cm, travel during this interval?

mcmradcmrs 03.5503)8.83)(6(

e) If the disk has mass of 15.5 grams, what is its kinetic energy when rotating at 400 rev/min?

JIK f 025.0)9.41)(1079.2( 25212

21

252212

21 1079.2)06.0)(0155.0( kgmmkgMRI

f) What average power is needed to accelerate the disk to 400rev/min in 4.0 seconds? (this turns out to be a relatively small part of the total power consumed by a CD player)

mWs

JtK

tWP 12.6

4025.0

09-2. System = 85 gram meter-stick & mass 65 grams at end.

a) Rod is vertical with 65 gram end up. What is system PE-g wrt to the axis?

b) System PE-g at bottom of swing?

c) Calculate the moment of inertia of the system about the axis of rotation in SI units.

d) If 90% of the gravitational potential energy lost in the swing of part (b) is converted to kinetic energy of the mass system, what is the rotational rate of the mass system in rad/s at the bottom of its swing?

e) What is the speed of the 65 gram point mass at swing bottom?

a) PE-g wrt to the axis?JgggymgymU g 0535.1)0.1()065.0()5.0()085.0(2211

b) System PE-g at bottom of swing?

JgggymgymU g 0535.1)0.1()065.0()5.0()085.0(2211

c) Moment of inertia?

2

223122

31

09333.0

)0.1)(065.0()0.1)(085.0(

kgm

mkgmkgMRMLI

d) 90% PE-g is converted to kinetic energy. What is the rotational rate of the system at the bottom of its swing?

JUK 8963.1)107.2(9.09.0

221 IK srad

IK /375.62

e) Speed of the 65 gram point mass at swing bottom?

smsradmrv /375.6)/375.6)(0.1(

09-3. Thin-walled circular cylinder mass 160 kg, radius 0.34 m.

(a) Calculate the moment of inertia of the rotor about its own axis. (b) Calculate the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev/min.

c) A 1400kg car slows from a speed of 20 m/s to 10 m/s and the rotational speed of the flywheel is increased from 10 000 rpm to 18 500 rpm. What percentage of the car’s kinetic energy change is funneled into the flywheel?

d) If the flywheel above were accelerated in 10s, what minimum force applied at the edge of the disk would be required?

(a) Calculate the moment of inertia of the rotor about its own axis.222 5.18)34.0)(160( kgmmkgMRI

(b) Calculate the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev/min.

JIK 82212

21 1096.1)4608(5.18

sradrpm /4608)30/(44000

c) A 1400kg car slows from a speed of 20 m/s to 10 m/s and the rotational speed of the flywheel is increased from 10 000 rpm to 18 500 rpm. What percentage of the car’s kinetic energy change is funneled into the flywheel?

JkgKcar 000,210)2010)(1400( 2221

JKdisk 940,11)10471936)(5.18( 2221

%7.5%100)000,210/940,11(% JJdtransferre

d) If the flywheel above were accelerated in 10s, what minimum force applied at the edge of the disk would be required?

ssradt

//9.8810

10471936

NrIF 480034.0/)9.88)(5.18(/

2222

12112

1 vmvmK sys

ii rv

2222

12112

1 )()( rmrmK sys

2222

2112

1 rmrmK sys

2iirmI2

21 IK sys

222

211 rmrmI

Ex. m1=m2=m3=m4 = m

r1=r2=r3=r4 = a

24

23

22

21 amamamamI

22222 4mamamamamaI

222

211 rmrmI

Ex. m1=m2=m3=m4 = m

r1=r2 = 0 r3=r4 = 2a

2222 )2()2(00 amammmI 222 84400 mamamaI

Example

to

A car wheel angularly accelerates uniformly from 1.5rad/s with rate 3.0rad/s2 for 5.0s. What is the final angular velocity?

sradssradsrad /5.16)0.5)(2/0.3(/5.1

What angle is subtended during this time?

to 21 rad45)5(5.165.12

1

22

1 tto rad455355.1 22

1

222o 25.27245325.1 2

srad /5.1625.272

Ex: Changing Units

srevradrev

sradsrad /1592.0

211/1

srevsrad

srevsradsrad /5.11/1

/1592.0/72/72

rpmsradrev

sradsrad 688

min160

2172/72

srevsT 0869.

5.111

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