Download Physics 2211: Lecture 28

Physics 2211: Lecture 37



Work and Kinetic Energy
Rotational Dynamics Examples
Atwood’s machine with massive pulley
Falling weight and pulley
Translational and Rotational Motion Combined
Rotation around a moving axis
Important kinetic energy theorem
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 1
Work

Consider the work done by a force F acting on an object
constrained to move around a fixed axis. For an
infinitesimal angular displacement d:
dW  F  ds  F rd cos 
 F rd cos  2   
 F rd sin
  F r sin  d
dW   d

F

r
d
ds  rd ˆ
axis

For constant  :

Analog of

W will be negative if  and D have opposite signs!
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
W =  D
W  F  Dr
Lecture 37, Page 2
Work & Kinetic Energy
WNET = DK

Recall the Work/Kinetic Energy Theorem:

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:
WNET  DK  12 I  2f  i2 
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 3
Example: 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?
R
M
F
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 4
Disk & String

The work done is W =  D
The torque is  = RF (since  = 90o)
The angular displacement D is
2 rad/rev x 10 rev.
R

So W = (.1 m)(10 N)(20 rad) = 62.8 J
 



D
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
M
F
Lecture 37, Page 5
Disk & String
WNET  62.8 J  DK  12 I 2
Recall that I for a disk about
its central axis is given by:
I  12 MR
So
R
2
DK  12  12 MR 2   2
4WNET


2
MR
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm

4  62.8 J 
.04 kg .1 m 
M
2
  792.5 rad/s
  7568 rpm
Lecture 37, Page 6
Example
Work & 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) disk 1
2
1
(2) disk 2
(3) same
F
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
F
Lecture 37, Page 7
Example
Solution

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 = DK.
But we know DK 
1 2
I
2
2
1
So since I1 = I2
1 = 2
F
F
d
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 8
Review: Torque and Angular
Acceleration
 NET  I



This is the rotational analog
of FNET = ma
Torque is the rotational analog of force:
The amount of “twist” provided by a force.
Moment of inertia I is the rotational analog of mass
If I is big, more torque is required to achieve a given
angular acceleration.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 9
Example
Rotations


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
F2
(b) 2
(c) 4
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
F1
Lecture 37, Page 10
Example
Solution
We know
but
  I
  FR
so FR  mR 2
F  mR
Since R2 = 2 R1
I  mR 2
and
F2 mR2 R2


F1 mR1 R1
F2
2
F1
F2
F1
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 11
Work & Power

The work done by a torque  acting through a
displacement D is given by:
W   D

The power provided by a constant torque is therefore given
by:
dW
d
P

 
dt
dt
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 12
Atwood’s Machine with Massive Pulley



y
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 SF = ma
 T1  m1g = m1(-a)
 T2  m2g = m2a
a
For the pulley use   I  I
R
a 1
 T1R - T2R  I  MRa
R 2
(Since I 
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
1
MR 2 for a disk)
2
x
M

R
T1
T1
T2
T2
m2
m1
a
m1g
m2g
Lecture 37, Page 13
a
Atwood’s Machine with Massive Pulley

y
We have three equations and three
unknowns (T1, T2, a). Solve for a.
x
T1  m1g =  m1a
(1)
T2  m2g = m2a
(2)
T1
(3)
T1
T1 - T2 = 1/2 Ma


m1  m 2
a
g
 m1  m 2  M 2 
M

R
T2
m2
m1
a
m1g
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
T2
m2g
Lecture 37, Page 14
a
Falling 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.

I
R
T
T
Starting at rest, what is the speed of the
mass after it has fallen a distance L.
m
a
mg
L
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 15
Falling weight & pulley

For the hanging mass use SF = ma
mg - T = ma
For the pulley + flywheel use  = I

 = TR = I
Realize that a = R


a
TR  I
R
Now solve for a using the above
equations.

Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
R
T
T
m
a
 mR 2 
a
g
2
 mR  I 
I
mg
L
Lecture 37, Page 16
Falling weight & pulley

Using 1-D kinematics we can solve for
the speed of the mass after it has
fallen a distance L:

I
R
T
v f2  v 02  2aDy
T
v f  2aL
where
mgR 2
a
mR 2  I
m
a
mg
L
2mgLR 2
vf 
mR 2  I
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 17
Falling weight & pulley

Conservation of Energy Solution:
E  mv  I  mgy
2
1
2
 v R
where
E
1
2

2
1
2
2
v
mR 2  I 2  mgy
R

Einitial  0  mgL
Efinal 
Einitial  Efinal
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
1
2

v f2
mR  I 2
R
2

2mgR 2L
vf 
mR 2  I

I
R
T
T
m
a
mg
L
y=0
Lecture 37, Page 18
Rotation 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?
M
R
F
Top view
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 19
Rotation around a moving axis
A

The CM moves according to F = MA

The distance moved by the CM is thus D 

The disk will rotate about
its CM according to  = I


So the angular displacement is  
1
I  MR 2
2
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm

M

I

F
M
1 2
F 2
At 
t
2
2M
RF
2F

2
1
MR
2 MR
1 2
F 2
t 
t
2
MR
A
R
F
Lecture 37, Page 20
Rotation 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 2
t (a)
2M
Divide (b) by (a):

D

2
R

F 2
t (b)
MR
R  2D
The length of string
pulled out is L = R :
L  2D

F
F
D
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
L
Lecture 37, Page 21
Comments 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! 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!

Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
M
A
R
F
Lecture 37, Page 22
Important kinetic energy theorem

Consider the total kinetic energy of a system of two masses:
K  21 m1v12  21 m2v 22
Now write the velocities as a sum of
the velocity of the center of mass and
a velocity relative to the center of mass
2
v12  v1  v1  vCM
 u12  2vCM  u1
2
v 22  v 2  v 2  vCM
 u22  2vCM  u2
so
K
1
2
2
 21 m1u12  21 m2u22  vCM   m1u1  m2u2 
 m1  m2  vCM
= KCM
*m u
1 1
v1  vCM  u1
v 2  vCM  u2
= KREL
=0*
 m v  m2v 2 
 m2u2  m1 v1  vCM   m2 v 2  v CM    m1v1  m2v 2   Mv CM  M  1 1
  Mv CM  0
m

m

1
2

Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 23
Important kinetic energy theorem

Thus
K
1
2
2
 21 m1u12  21 m2u22
 m1  m2  vCM
= KCM
So
1
2
= KREL
2
K  21 MvCM
 K REL
2
is the kinetic energy of the center of mass (M is total mass).
MvCM
KREL is the kinetic energy due to motion relative to the center of mass.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 24
Connection with rotational motion
2
K  21 MvCM
  21 mi ui2
i
KCM

KREL
For a solid object rotating about its center of mass:
K REL   21 mi ui
i
KREL  
i
1
2
2
where ui   ri

2 2
mi  r    mi ri  
 i

2 2
i
1
2
but I CM 
2
m
r
 ii
i
2
KREL  12 I CM
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 25
Translational & rotational motion
combined

For a solid object which rotates about its center or mass
and whose CM is moving:
2
KNET  21 ICM 2  21 MVCM
VCM

Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 37, Page 26