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Physics I
95.141
LECTURE 21
11/24/10
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Exam Prep Question
•
The system to the right consists of a
cylinder (R=15cm, M=50kg) and 4 2kg
(point) masses attached to 30cm massless
rods. The system is free to rotate around
an axis through its center of mass.
•
•
•
•
a) (10 pts) What is the moment of inertia
of this system?
b) (10 pts) Assume a 10kg mass is
attached to a massless cord, wrapped
around the cylinder, and dropped from
rest. What is the acceleration of the
mass?
c) (5pts) What is the angular acceleration
of the cylinder/mass system?
d) (10pts) Determine the Kinetic Energy,
as a function of time, associated with i) the
rotating system and ii) the falling mass.
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
30cm
15cm
2kg
10kg
Exam Prep Question
•
The system below consists of a cylinder
(R=15cm, M=50kg) and 4 2kg (point)
masses attached to 30cm massless rods.
The system is free to rotate around an axis
through its center of mass.
•
30cm
15cm
a) (10 pts) What is the moment of inertia
of this system?
2kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Exam Prep Question
•
The system below consists of a cylinder (R=15cm,
M=50kg) and 4 2kg (point) masses attached to 30cm
massless rods. The system is free to rotate around an axis
through its center of mass.
•
30cm
15cm
b) (10 pts) Assume a 10kg mass is attached to a massless
cord, wrapped around the cylinder, and dropped from rest.
What is the acceleration of the mass?
2kg
10kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Exam Prep Question
•
The system below consists of a cylinder
(R=15cm, M=50kg) and 4 2kg (point)
masses attached to 30cm massless rods.
The system is free to rotate around an axis
through its center of mass.
•
30cm
15cm
c) (5pts) What is the angular acceleration
of the rotating cylinder/mass system?
2kg
10kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Exam Prep Question
•
The system below consists of a cylinder
(R=15cm, M=50kg) and 4 2kg (point) masses
attached to 30cm massless rods. The system is
free to rotate around an axis through its center
of mass.
•
30cm
15cm
d) (10pts) Determine the Kinetic Energy, as a
function of time, associated with i) the rotating
system and ii) the falling mass.
2kg
10kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Review Example
• What is the vector cross product of the two
vectors:

A  1iˆ  2 ˆj  4kˆ

B  2iˆ  3 ˆj  1kˆ
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Administrative Notes
• Exam III
– Wednesday 12/1
– In Class, 9am-9:50am
– Chapters 9-11
• Practice Exams posted
• Practice problems posted by end of day
• Exam Review Scheduled for 11/29….subject to
change. Will probably have to be shifted to
Tuesday.
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Outline
•
•
Vector Cross Products
Conservation of Angular Momentum
•
What do we know?
– Units
– Kinematic equations
– Freely falling objects
– Vectors
– Kinematics + Vectors = Vector
Kinematics
– Relative motion
– Projectile motion
– Uniform circular motion
– Newton’s Laws
– Force of Gravity/Normal Force
– Free Body Diagrams
– Problem solving
– Uniform Circular Motion
– Newton’s Law of Universal Gravitation
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
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Weightlessness
Kepler’s Laws
Work by Constant Force
Scalar Product of Vectors
Work done by varying Force
Work-Energy Theorem
Conservative, non-conservative Forces
Potential Energy
Mechanical Energy
Conservation of Energy
Dissipative Forces
Gravitational Potential Revisited
Power
Momentum and Force
Conservation of Momentum
Collisions
Impulse
Conservation of Momentum and Energy
Elastic and Inelastic Collisions2D, 3D Collisions
Center of Mass and translational motion
Angular quantities
Vector nature of angular quantities
Constant angular acceleration
Torque
Rotational Inertia
Moments of Inertia
Angular Momentum
Review of Lecture 20
• Introduced concept of Angular Momentum
L  I
• Conservation of Angular Momentum
– With no external torques acting on a system, the
angular momentum of the system is conserved.
• Vector Cross products
  
C  A B
95.141, F2010, Lecture 21
Department of Physics and Applied Physics

A  Ax iˆ  Ay ˆj  Az kˆ

B  B x iˆ  B y ˆj  Bz kˆ
iˆ
 
A  B  Ax
ˆj
Ay
kˆ
Az
Bx
By
Bz
Review of Angular Motion
• We know equations of motion for angular motion
 2  o2  2    o 
o   f

2
1 2
   o   o t  t
2
   o  t
• We know torques cause angular acceleration
  RF  I
• Objects can have rotational kinetic energy
KErot  I
1
2
2
• And angular momentum
L  I
• So why the cross product?
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Torque and the Cross Product
• When we first introduced torque as the product of the
radius and the perpendicular component of the Force,
we were only interested in the magnitude of the torque!
 
• Magnitude given by RFsinθ…same as cross product R, F
FROM LECTURE 19

R
  RF
  RF sin 
95.141, F2010, Lecture 21
Department of Physics and Applied Physics

F
Torque and the Cross Product




• However, we since learned that
I
• And we know that angular acceleration points in direction
of axis of rotation… so Torque must as
well!



• Torque is cross product of R,F   R  F
FROM LECTURE 19

R
  RF
  RF sin 
95.141, F2010, Lecture 21
Department of Physics and Applied Physics

F
Angular Momentum of a Particle
• We have already defined angular momentum as
L  I
• But this definition is for objects rotating with
some angular velocity and moment of inertia
around an axis of rotation.
• More general, alternate, definition:
  
Lrp
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Equivalence of our two definitions
• Suppose we have a mass rotating around an
axis   10 rad s
m=2kg
2m
• Use cross product
  
Lrp
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
y
x
Equivalence of our two definitions
• Suppose we have a mass rotating around an
axis   10 rad s
m=2kg
2m
• Use
L  I
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
y
x
So why use cross product?
• Cross products are messy…why would we ever
use them, instead of the simpler L  I   RF
• Because the cross product allows us to
determine the angular momentum of, or torque
on, objects which are not necessarily moving
with constant, or even circular motion!
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Example
• Calculate the angular momentum (about the origin) of
the rock of mass m dropped from rest off the cliff.
d
(0,0)
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Example
• What Torque is exerted (about the origin) on the
rock?
d
(0,0)
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Relationship of torque to angular momentum
• When we discussed linear momentum, we

nd

revised Newton’s 2 Law to state
dp
 F  dt
• Similarly, we can write an expression for net 
 dL
torque in terms of angular momentum
  dt
• Double check with our falling rock:

L  dmgt kˆ
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Newton’s 2nd Law: Angular Form
• The vector sum of all of the torques acting on a
particle, object, or system, is equal to the time
rate of change of the angular momentum of the
particle, object or system.

 dL
  dt
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Conservation of Angular Momentum
• We used the revised expression for conservation
of linear momentum to argue that if there is no
net external force on a system or object, then
the momentum of the system or object is
conserved.
• Similarly:
– If the net external torque acting on a system of object
is zero, then the angular momentum of that object will

remain constant.

dL
L constant!
  0  dt
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Conservation of Momentum
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Does this make sense?
• What is happening?
• What do we need to know?
– System= Container + Skinner

L final

Linitial  0



 LSkinner  Lcontainer  Linitial  0


L  I
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Moments of Inertia
• Skinner=point mass
• Shipping Container
2
I Skinner  M Skinner RSkinner
R  1.25m
M  70kg
2.5m
2.5m
12m
Mass=3,500kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
σ Container
• Determine surface mass density σ
• Divide into 6 slabs
2.5m
• Total Mass = 3500kgs
A  2 Abottom  2 Asides  2 Aends
2.5m
A  2(2.5  12)  2(2.5  12)  2(2.5  2.5)
A  132.5m 2
M
kg

 26.4 2
A
m
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
12m
I Container
• Divide into 6 slabs
Top & Bottom:
M topbot  Atopbot
M topbot  1584kg
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Ends
• Use parallel axis theorem
  2.5m
h=6m
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
I sides  121 M 2  Mh 2
M ends  330kg
Sides
• Use parallel axis theorem
I sides  121 M 2  Mh 2
M sides  1584kg
h  1.25m
  12m
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Conservation of Momentum
• Skinner seems to be making one rotation every
2 seconds.
95.141, F2010, Lecture 21
Department of Physics and Applied Physics
Finally
• In the clip, it takes about 20 seconds to turn the
container 90 degrees.
95.141, F2010, Lecture 21
Department of Physics and Applied Physics