Comparing rotational and linear motion Angular Velocity Angular velocity, is the rate of change in...

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Transcript of Comparing rotational and linear motion Angular Velocity Angular velocity, is the rate of change in...
Comparing rotational and linear motion
Angular VelocityAngular velocity,w, is the rate of change in angular displacement. (radians per second.)
=w 2pf Angular frequency f (rev/s). =w 2pf Angular frequency f
(rev/s).
Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm):
w = Angular velocity in rad/s.
q
t
Angular AccelerationAngular acceleration is the rate of
change in angular velocity. (Radians per sec per sec.)
The angular acceleration can also be found from the change in frequency, as follows:
2 ( ) 2
fSince f
t
2 Angular acceleration (rad/s )t
Torque and Angular Acceleration
When an object is subject to a net force, it undergoes an acceleration. (Newton’s 2nd)
When a rigid object is subject to a net torque, it undergoes an angular acceleration.
Force and Linear Acceleration
Inertia of RotationConsider Newton’s second law for the inertia of rotation
F = 20 N
a = 4 m/s2
Linear Inertia, m = F/a m = = 5 kg
20 N
4 m/s2
F = 20 NR = 0.5 ma = 2 rad/s2
Force does for translation what torque does for rotation:
Rotational Inertia, I
I = = = 2.5 kg m2(20 N)(0.5 m)
2 rad/s2
ta
ImaF ,
Moment of Inertia
This mass analog is called the moment of inertia, I, of the object
is defined relative to rotation axisSI units are kg m2
i
iirmI 2
More About Moment of Inertia
• I depends on both the mass and its distribution.• If an object’s mass is distributed further
from the axis of rotation, the moment of inertia will be larger.
Common Moments of Inertia
Common moments of inertia are on page 251.
Example 1 A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias.
2 21 12 2 (3 kg)(0.2 m)I mR
R
I = mR 2
Hoop
R
I = ½mR 2
Disk
2 2(3 kg)(0.2 m)I mR
I = 0.120 kg m2
I = 0.0600 kg m2
Important AnalogiesFor many problems involving rotation, there is an analogy to be drawn from linear motion.
xf
R
4 kg
wt wo = 50
rad/s t = 40 N m
A resultant force F produces negative acceleration a for a mass m.F ma
Im
A resultant torque tproduces angular acceleration a of disk with rotational inertia I.
I
Example 2Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of the spindle?
b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel?
c) What is the acceleration of the bucket?
d) What is the mass of the bucket?
M
Solutiona) What is the moment of Inertia of the spindle?Given: M = 5 kg, R = 0.6 m
M
middleabout rod,,12
1
endabout rod,,3
1
shell spherical,3
2
sphere solid,5
2
cylinder solid,2
1shell lcylindrica,
2
2
2
2
2
2Inertia of Moments
ML
ML
MR
MR
MR
MR
2
2
1MRI = 0.9 kgm2
Solutionb) If the tension in the rope is 10 N, what is a?Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m
M
IrF
formula Basic IrT = a (0.6m)(10 N)/(0.9 kg∙m2)
a = 6.67 rad/s2
c) What is the acceleration of the bucket?Given: r=0.6 m, a = 6.67 rad/s
ra formula Basic
a=4 m/s2a = (6.67 rad/s2)(0.6 m)
Solutiond) What is the mass of the bucket?Given: T = 10 N, a = 4 m/s2
M
maF formula Basic
ag
TM
TMgMa
M = 1.72 kg
Comparing rotational and linear motion
Combined Rotation and Translation
vcm
vcm
vcm First consider a disk sliding without friction. The velocity of any part is equal to velocity vcm of the center of mass.
vR
P
Now consider a ball rolling without slipping. The angular velocity about the point P is same as for disk, so that we write:
Orv
R v R
Two Kinds of Kinetic Energy
vR
P
Kinetic Energy of Translation:
K = ½mv2
Kinetic Energy of Rotation:
K = ½I2
Total Kinetic Energy of a Rolling Object:
2 21 12 2TK mv I
KE of centerofmass motionKE due to rotation
Example 3
What is the kinetic energy of the Earth due to the daily rotation?Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m.
T
2
formula Basic
First, find w
= 7.27 x105 rad/s
2
sphere Solid
5
2MRI
2
formula Basic
2
1 IKE
= 2.78 x1029 J
360024
2
22
5
1 MRKE
Summary – Rotational Analogies
Quantity Linear Rotational
Displacement
Displacement x
Radians
Inertia Mass (kg) I (kgm2)
Force Newtons N Torque N·m
Velocity v “ m/s ” Rad/s
Acceleration a “ m/s2 ” Rad/s2
Momentum mv (kg m/s) I (kgm2rad/s)
Analogous FormulasLinear Motion Rotational Motion
F = ma = IK = ½mv2 K = ½I2
Work = Fx Work = tq
Power = Fv Power = IFx = ½mvf
2  ½mvo2 = ½If
2  ½Io2
Example 4A solid sphere rolls down a hill of height 40 m.What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!)
22
formula Basic
2
1
2
1 Imvmgh
2
sphere solidFor
5
2mrI
rv formula Basic
v = 23.7 m/s
222
5
2
2
1 rvgh
5/7
2,
5
2
2
1 222 ghvvvgh
Angular Momentum
mvrL
IL
Analogy between L and p
Angular Momentum
Linear momentum
L = Iw p = mv
t = DL/Dt F = Dp/Dt
Conserved if no net
outside torques
Conserved if no net outside forces
Rigid body
Point particle
Example 5A 65kg student sprints at 8.0 m/s and leaps onto a 110kg merrygoround of radius 1.6 m. Treating the merrygoround as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merrygoround while moving tangentially.
SolutionKnown: M, R, m, v0
Find: wF
First, find L0
mvrL formula Basic
mvRL 0Next, find Itot
2
Particle
2
cylinder Solid
2
1
MRI
MRI
22
2
1MRmRITot
Now, given Itot and L0, find w
IL formula Basic
TotI
L0 = 2.71 rad/s
Summary of Formulas:I = SmR2I = SmR2
2 20½ ½fI I
mgho
½Iwo2
½mvo2
=mghf
½Iwf2
½mvf2
Height?
Rotation?
velocity?
Height?
Rotation?
velocity?
212K I
Powert
Work o o f fI I
Conservation: