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