Physics 207: Lecture 2 Notes
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Transcript Physics 207: Lecture 2 Notes
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
RL
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