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Lecture 37, Page 1 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Physics 2211: Lecture 37Physics 2211: Lecture 37
Work and Kinetic Energy Rotational Dynamics Examples
Atwood’s machine with massive pulleyFalling weight and pulley
Translational and Rotational Motion CombinedRotation around a moving axisImportant kinetic energy theorem
Lecture 37, Page 2 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
WorkWork
daxis
r
F
ˆ ds rd
Consider the work done by a force acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:
F
cos F F rdd dsW
2co s rdF in sF rd
dW d
W will be negative if and have opposite signs!
W F r
Analog of
For constant : W =
sin rF d
Lecture 37, Page 3 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Work & Kinetic EnergyWork & Kinetic Energy
Recall the Work/Kinetic Energy Theorem: WNET = K
This is true in general, and hence applies to rotational motion as well as linear motion.
So for an object that rotates about a fixed axis:
2 212 NET f iW K I
Lecture 37, Page 4 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Disk & StringExample: Disk & String
A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning).
How fast is the disk spinning after the string has unwound?
F
RM
Lecture 37, Page 5 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Disk & StringDisk & String
The work done is W = The torque is = RF (since = 90o)The angular displacement is
2 rad/rev x 10 rev.
F
RM
So W = (.1 m)(10 N)(20rad) = 62.8 J
Lecture 37, Page 6 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Disk & StringDisk & String
RM
Recall that I for a disk about
its central axis is given by:
212I MR
So 2 21 12 2 K MR
NET22
4 62.8 J4W
.04 kg .1 m
MR792.5 rad/s
7568 rpm
21262.8 NETW J K I
Lecture 37, Page 7 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExampleWork & EnergyWork & Energy
Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest.Which disk has the biggest angular velocity after the pull ?
(1)(1) disk 1
(2)(2) disk 2
(3)(3) same FF
1 2
Lecture 37, Page 8 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample SolutionSolution
FF
1 2
d
The work done on both disks is the same!W = Fd
The change in kinetic energy of each will therefore also be the same since W = K.
So since I1 = I2
1 = 2
But we know 21
2K I
Lecture 37, Page 9 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Review: Torque and Angular Review: Torque and Angular AccelerationAcceleration
This is the rotational analog of FNET = ma
Torque is the rotational analog of force:Torque is the rotational analog of force:The amount of “twist” provided by a force.
Moment of inertiaMoment of inertia I I is the rotational analog of massis the rotational analog of massIf I is big, more torque is required to achieve a given
angular acceleration.
NET I
Lecture 37, Page 10 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample RotationsRotations
Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular acceleration of the wheels is the same?
(a) 1
(b) 2
(c) 4
F1
F2
Lecture 37, Page 11 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample SolutionSolution
We know I
so
mRF
mRFR 2
1
2
1
2
1
2
RR
mRmR
FF
F1
F2
Since R2 = 2 R12
FF
1
2
I mR2but andFR
Lecture 37, Page 12 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Work & PowerWork & Power
The work done by a torque acting through a displacement is given by:
The power provided by a constant torque is therefore given by:
dW dP
dt dt
W
Lecture 37, Page 13 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Atwood’s Machine with Massive PulleyAtwood’s Machine with Massive Pulley
A pair of masses are hung over a massive disk-shaped pulley as shown.Find the acceleration of the blocks.
For the hanging masses use F = ma T1 m1g = m1(-a)
T2 m2g = m2a
I 1
22MR(Since for a disk)
Ia
RMRa
1
2
I Ia
R For the pulley use
T1R - T2Rm2m1
R
M
y
x
m2m1
m2g
aT1
m1g
a
T2
T1 T2
Lecture 37, Page 14 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Atwood’s Machine with Massive PulleyAtwood’s Machine with Massive Pulley
m2m1
R
M
y
x
m2m1
m2g
aT1
m1g
a
T2
T1 T2
am m
m m Mg
1 2
1 2 2
We have three equations and three unknowns (T1, T2, a). Solve for a.
T1 m1g = m1a (1)
T2 m2g = m2a (2)
T1 - T2 = 1/2 Ma (3)
Lecture 37, Page 15 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Falling weight & pulleyFalling weight & pulley
A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley.
Starting at rest, what is the speed of the mass after it has fallen a distance L.
I
m
R
T
mg
a
L
T
Lecture 37, Page 16 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Falling weight & pulleyFalling weight & pulley
For the hanging mass use F = mamg - T = ma
For the pulley + flywheel use = I = TR = I
Realize that a = R
amR
mRg
2
2 I
TRa
RI
Now solve for a using the above equations.
I
m
R
T
mg
a
L
T
Lecture 37, Page 17 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Falling weight & pulleyFalling weight & pulley
Using 1-D kinematics we can solve for the speed of the mass after it has fallen a distance L:
2 20 2fv v a y
2fv aL
where2
2
mgRa
mR
I
2
2
2f
mgLRv
mR
I
I
m
R
T
mg
a
L
T
Lecture 37, Page 18 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Falling weight & pulleyFalling weight & pulley
Conservation of Energy Solution:
2 21 12 2E mv mgy I
0initialE mgL
where v R
2
212 2
vE mR mgy
R I
2
212 2
ffinal
vE mR
R I
initial finalE E y = 0
2
2
2f
mgR Lv
mR
I
I
m
R
T
mg
a
L
T
Lecture 37, Page 19 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rotation around a moving axisRotation around a moving axis
A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds.
What length of string L has unwound after the puck has moved a distance D?
F
RM
Top view
Lecture 37, Page 20 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rotation around a moving axisRotation around a moving axis
The CM moves according to F = MA AF
M
D AtF
Mt
1
2 22 2 The distance moved by the CM is thus
F
M A
RI 1
22MR
The disk will rotate about its CM according to = I 21
2
2RF F
MR MR
I
So the angular displacement is 2 21
2
Ft t
MR
Lecture 37, Page 21 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rotation around a moving axisRotation around a moving axis
So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time:
D
F F
L
The length of string
pulled out is L = R:
L D2
(b)2
Ft
MR
Divide (b) by (a):2
D R
2 R D
(a)2
2
FD t
M
Lecture 37, Page 22 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Comments on CM accelerationComments on CM acceleration
We just used = I for rotation about an axis through the CM even though the CM was accelerating! The CM is not an inertial reference frame! Is this OK??
(After all, we can only use F = ma in an inertial reference frame).
YES!YES! We can always write = I for an axis through the CM.This is true even if the CM is accelerating.We will prove this when we discuss angular momentum!
F
R
M A
Lecture 37, Page 23 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Important kinetic energy theoremImportant kinetic energy theorem
Consider the total kinetic energy of a system of two masses:
2 21 11 1 2 22 2K m v m v
Now write the velocities as a sum of
the velocity of the center of mass and
a velocity relative to the center of mass
1 1CMv v u
2 2CMv v u
so
2 2 21 1 11 2 1 1 2 2 1 1 2 22 2 2CM CMK m m v m u m u v m u m u
2 2 21 1 1 1 12CM CMv v v v u v u
2 2 22 2 2 2 22CM CMv v v v u v u
= KCM = 0 *= KREL
1 1 2 21 1 2 2 1 1 2 2 1 1 2 2
1 2
0CM CM CM CM
m v m vm u m u m v v m v v m v m v Mv M Mv
m m
*
Lecture 37, Page 24 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Thus
KREL is the kinetic energy due to motion relative to the center of mass.
So 212 CM RELK Mv K
is the kinetic energy of the center of mass (M is total mass).212 CMMv
2 2 21 1 11 2 1 1 2 22 2 2CMK m m v m u m u
= KCM = KREL
Important kinetic energy theoremImportant kinetic energy theorem
Lecture 37, Page 25 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Connection with rotational motionConnection with rotational motion
For a solid object rotating about its center of mass:
212REL i i
i
K m u
2 21 12 2CM i i
i
K Mv m u
KCM KREL
where i iu r
2 2 2 21 12 2REL i i i i
i i
K m r m r
but2CM i i
i
I m r
21CM2 I
RELK
Lecture 37, Page 26 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
For a solid object which rotates about its center or mass and whose CM is moving:
VCM
Translational & rotational motion Translational & rotational motion combinedcombined
2 21 12 2NET CM CMK MV I