Lecture 16
Goals:
• Chapter 12
 Extend the particle model to
rigid-bodies
 Understand the equilibrium
of an extended object.
 Analyze rolling motion
 Understand rotation about a
fixed axis.
 Employ “conservation of
angular momentum” concept
Assignment:
 HW8 due March 17th
Physics 207: Lecture 16, Pg 1
System of Particles (Distributed Mass):
 Until now, we have considered the behavior of very simple
systems (one or two masses).
 But real objects have distributed mass !
 For example, consider a simple rotating disk and 2 equal mass
m plugs at distances r and 2r.
w
1
2
 Compare the vtangential and kinetic energies at these two points.
Physics 207: Lecture 16, Pg 5
System of Particles (Distributed Mass):
1 K= ½ m v2 = ½ m (w r)2
w
v
v
1
2
 The rotation axis matters too!
2 K= ½ m (2v)2 = ½ m (w 2r)2
 Twice the radius, four times the kinetic energy
K Rotational  mv  m(wr )
1
2
2
1
2
2
 KEY POINT: It matters where you put your mass!
Physics 207: Lecture 16, Pg 6
Exercise Rotational Kinetic Energy
 We have two balls of the same mass. Ball 1 is attached
to a 0.1 m long rope. It spins around at 2 revolutions per
second. Ball 2 is on a 0.3 m long rope. It spins around at
2 revolutions per second.
K  mr w
1
2
2
2
 What is the ratio of the kinetic energy
of Ball 2 to that of Ball 1 ?
A. 1/9
B. 1/3
C. 1
D. 3
E. 9
Ball 1
Ball 2
Physics 207: Lecture 16, Pg 7
Exercise Rotational Kinetic Energy
 K2/K1 = ½ m wr22 / ½ m wr12 = 0.22 / 0.12 = 4
 What is the ratio of the kinetic energy of Ball 2 to
that of Ball 1 ?
(A) 1/9 (B) 1/3
(C) 1
(D) 3
(E) 9
Ball 1
Ball 2
Physics 207: Lecture 16, Pg 8
A special point for rotation
System of Particles: Center of Mass (CM)
 A supported object will rotate about its center of
mass.
 Center of mass: Where the system is balanced !
 Building a mobile is an exercise in finding
centers of mass.
m1
+
m2
m1
+
m2
mobile
Physics 207: Lecture 16, Pg 9
System of Particles: Center of Mass
 How do we describe the “position” of a system made
up of many parts ?
 Define the Center of Mass (average position):
 For a collection of N individual point like particles
whose masses and positions we know:

 mi ri
N

i 1
RCM 
M
RCM
m2
m1
r1
r2
y
x
(In this case, N = 2)
Physics 207: Lecture 16, Pg 10
Sample calculation:
 Consider the following mass distribution: m at ( 0, 0)

 mi ri
N

RCM  i 1
M
 XCM ˆi  YCM ˆj  ZCM kˆ
2m at (12,12)
m at (24, 0)
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 207: Lecture 16, Pg 11
System of Particles: Center of Mass
 For a continuous solid, one can convert sums to an
N
integral.

 mi ri
dm
y
r
x

i 1
RCM 
M



 r dm  r dm
RCM 

M
 dm
where dm is an infinitesimal
mass element but there is
no new physics
Physics 207: Lecture 16, Pg 12
Connection with motion...
 So, for a rigid rotating object whose CM is moving,
N

it rotates about its center of mass!
 mi ri

RCM  i 1
M
K TOTAL  K Rotation  K Translatio n
K TOTAL  K Rotation  MV
2
CM
1
2
For a point p rotating:
K R  m p v p  m p (wrp )
1
2
p
p
p
p
2
1
2
VCM
p
w
p
p
p
Physics 207: Lecture 16, Pg 13
2
Rotational Kinetic Energy
 Consider the simple rotating system shown below.
(Assume the masses are attached to the rotation
axis by massless rigid rods).
 The kinetic energy of this system will be the sum
4
of the kinetic energy of each piece:
2
1
K  2  mi vi
 K = ½ m1v12 + ½ m2v22 + ½ m3v32 + ½ m4v42
i 1
m4
w
r4
m3
r3
r1
m1
r2
m2
Physics 207: Lecture 16, Pg 14
Rotation & Kinetic Energy
 Notice that v1 = w r1 , v2 = w r2 , v3 = w r3 , v4 = w r4
 So we can rewrite the summation:
4
4
K   mi v   miw r 
1
2
2
i
i 1
1
2
2 2
i
i 1
1
2
4
[  m r ]w
i 1
2
i i
2
 We recognize & define a new quantity, moment of inertia
or I, and write:
m4
K Rotational  I w
1
2
N
I   mi ri
i 1
2
2
w
r4
m3
r3
r1
r2
m1
m2
Physics 207: Lecture 16, Pg 15
Calculating Moment of Inertia
N
I   mi ri
2
where r is the distance from
the mass to the axis of rotation.
i 1
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
m
m
m
m
L
Physics 207: Lecture 16, Pg 16
Calculating Moment of Inertia...
 For a single object, I depends on the rotation axis!
 Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
I1 = 2mL2
m
m
m
m
I2 = mL2
I = 2mL2
R  L/2
RL
L
R  2L / 2
Physics 207: Lecture 16, Pg 17
Moments of Inertia
 For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
dm
 An integral is required to find I :
I   r dm
2
r
 Some examples of I for solid objects:
dr
r
L
R
Solid disk or cylinder of mass M
and radius R, about
perpendicular axis through its
center.
I = ½ M R2
Use the table…
Physics 207: Lecture 16, Pg 20
Exercise: Work & Energy
 Strings are wrapped around the circumference of two solid disks
and pulled with a force, F, for the same linear distance, d.
 Disk 1 has a bigger radius, but both are of identical material
(i.e., their density is r = M / V ) and have the same thickness.
 Both disks rotate freely around axes though their centers, and
start at rest.
 Which disk has the biggest angular velocity, w, at the end ?
Recall W = F d = DK ( =½ I w2)
w2
w1
(A) Disk 1
(B) Disk 2
(C) Same
start
finish
F
F
d
Physics 207: Lecture 16, Pg 24
Exercise Work & Energy
 Strings are wrapped around the circumference of two solid
disks and pulled with identical forces for the same linear
distance.
Disk 1 has a bigger radius, but both are identical material (i.e.
their density r = M/V is the same). Both disks rotate freely
around axes though their centers, and start at rest.
 Which disk has the biggest angular velocity after the end ?
W = F d = ½ I1 w12 = ½ I2 w22
w2
w1
w1 = (I2 / I1)½ w2 and I2 < I1
(A) Disk 1
(B) Disk 2
(C) Same
start
finish
F
F
d
Physics 207: Lecture 16, Pg 25
Work & Kinetic Energy:
 Recall the Work Kinetic-Energy Theorem: DK = WNET
 This applies to both rotational as well as linear motion.
 So for an object that rotates about a fixed axis
DK  I w  w   WNET
1
2
2
f
2
i
 For an object which is rotating and translating
K  I CMw  MV
1
2
2
1
2
2
CM
Physics 207: Lecture 16, Pg 26
Demo Example : A race rolling down an incline
 Two cylinders with identical radii and total masses roll down
an inclined plane.
 The 1st has more of the mass concentrated at the center
while the 2nd has more mass concentrated at the rim.
 Which gets down first?
M
Two cylinders with radius R and mass m
h
q
A) Mass 1
B) Mass 2
C) They both arrive at same time
M
who is 1st ?
Physics 207: Lecture 16, Pg 27
Same Example : Rolling, without slipping, Motion
 A cylinder is about to roll down an inclined plane.
 What is its speed at the bottom of the plane ?
M
h
q
M
v?
Physics 207: Lecture 16, Pg 28
Rolling without slipping motion
 Again consider a cylinder rolling at a constant speed.
2VCM
VCM
CM
Physics 207: Lecture 16, Pg 29
Motion
 Again consider a cylinder rolling at a constant speed.
Rotation only
VTang = wR
CM
Both with
|VTang| = |VCM |
2VCM
VCM
CM
Sliding only
VCM
CM
Physics 207: Lecture 16, Pg 30
Example : Rolling Motion
 A solid cylinder is about to roll down an inclined plane.
What is its speed at the bottom of the plane ?
 Use Work-Energy theorem
Ball has radius R
M
h
q
Mgh = ½ Mv2 + ½ ICM w2
and
M
v?
v =wR
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
v = 2(gh/3)½
Physics 207: Lecture 16, Pg 31
Example: The Frictionless Loop-the-Loop … last time
 To complete the loop the loop, how high do we
have to release a ball with radius r (r <<R) ?
 Condition for completing the loop the loop:
Circular motion at the top of the loop (ac = v2 / R)
 Use fact that E = U + K = constant !
Ub=mgh
ball has mass m &
r <<R
U=mg2R
h?
R
v
Tangential
Recall that “g” is the source of
the centripetal acceleration
and N just goes to zero is
the limiting case.
Also recall the minimum speed
at the top is
 gR
Physics 207: Lecture 16, Pg 32
Example: The Loop-the-Loop … last time
 If rolling then ball has both rotational and CM motion!
 E= U + KCM + KRot = constant = mgh (at top)
 E= mg2R + ½ mv2 + ½ 2/5 mr2 w2 = mgh
& v=wr
 E= mg2R + ½ mgR + 1/5 m v2 = mgh  h = 5/2R+1/5R
Ub=mgh
ball has mass m &
r <<R
U=mg2R
v
Tangential
 gR
h?
R
Just a little bit more….
Physics 207: Lecture 16, Pg 33
Exercise: Work Energy Example, Rotating Rod
 A uniform rod of length L=0.5 m and mass m=1 kg
is free to rotate on a frictionless pin passing through
one end as in the Figure. The rod is released from
rest in the horizontal position.
 What is its angular speed when it reaches the
lowest point ?
L
m
Physics 207: Lecture 16, Pg 34
Example: Rotating Rod
 A uniform rod of length L=0.5 m and mass m=1 kg is free to
rotate on a frictionless hinge passing through one end as
shown. The rod is released from rest in the horizontal position.
 What is its angular speed when it reaches the lowest point ?
The hinge changes everything!
W = m g h = ½ IHinge w2
L
m
mg
L/2
W = mgL/2 = ½ (m L2/3) w2
mg
and solve for w
Physics 207: Lecture 16, Pg 35
Lecture 16
K TOTAL  K Rotational  K Translatio nal
K TOTAL  K Rotational  MV
2
CM
1
2
K Rotational  I w
1
2
Assignment:
 HW7 due March 25th
 For the next Tuesday:
Catch up
2
I   mi ri
2
i
Physics 207: Lecture 16, Pg 36