MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01T
Fall Term 2006
Quiz 6: Angular Momentum
Section ______
Table and Group Number ______________________
Name ___________________________________
Problem 1: A uniform block of mass m and height h is sliding along a surface with constant
center of mass velocity vcm in the direction shown in the figure below. At the moment depicted
in the figure, the center of mass of the block is a (horizontal) distance d from the point S. What
is the magnitude and direction of the angular momentum of the block about the point S?
Remember to indicate your choice of coordinate system.
m
h
S
cm
vcm
d
First define a coordinate system -- the easiest one is probably with x to the right,
y to the top, and z out of the page. Then the vector from S to the center of mass of the block is:
!
h
r = di" + "j
2
and the momentum vector is:
#!
p = mvcm "i
#! ! #!
With angular momentum defined as L = r ! p, we obtain:
#! ! #! "
h %
h
L = r ! p = $ di" + "j ' ! mvcm "i = (mvcm k"
#
2 &
2
Problem 2: A hockey puck of mass m1 slides along ice with an initial velocity v0 and strikes one
end of a uniform stick of length l and mass m2 that is also lying on the ice. After the collision,
the center of mass of the stick moves in a direction parallel to the puck with an unknown final
velocity vrod. The stick also rotates about its center of mass with an unknown final angular
velocity ωrod. The puck continues to move in the same straight line as before it hit the stick with
(known) final velocity vf.
m2
l
cm
vrod
ωrod
m1
vf
v0
Before
After
(a) What is the magnitude of the final velocity, vrod, of the center of mass of the stick in
terms of the known quantities given above?
To calculate this velocity, we'll use conservation of (linear) momentum. It is not yet known if the
collision is elastic or not, so conserving energy would not be appropriate. The initial momentum
and final momentum are given by:
pi = m1v0 and p f = m1v f + m2 vrod
! vrod =
(
m1
v0 " v f
m2
)
(b) Is the angular momentum of the puck-stick system conserved in this collision? Why or
why not? If it is, write down the equation for conservation of angular momentum about
the stick’s center of mass [note (after the fact) – this was supposed to mean “about an
origin that lies at the initial position of the stick’s center of mass”] in terms only of
the known and unknown quantities.
!
#!
ext
"!
= 0 " L is conserved. Put simply, there are no external torques for the puck-stick system.
Given that, we can calculate initial and final angular momenta about an origin that lies at the initial
position of the center of mass of the stick. This was very poorly worded in the quiz, and you can
definitely blame your TA for that (and the incorrect explanation after the quiz).
With this origin, the puck has an angular momentum about the origin both before and after the
collision, while the stick has an angular momentum after the collision only due to rotation about its
"!
!
own center of mass. [Note: After the collision, L rod = r rod $ m v #i + I% k# = x #i $ m v #i
2 rod
rod
rod
2 rod
+ I% rod k$ = I% rod k# .] Let's now look at the distance between the puck and the origin:
!
l
& r puck = x puck #i ' #j
2
While we may never know what x puck is, it doesn't matter because we'll ultimately take the cross
!
product of r puck and the momentum (of the puck) to calculate its angular momentum. To state
this explicitly:
"!
!
l +
l
(
L puck = r puck $ m1v puck #i = * x puck #i ' #j - $ m1v puck #i = m1v puck k#
)
2 ,
2
In the above equation, v puck is either v0 or v f depending on whether we are looking before or
after the collision. With that information in hand, the initial and final angular momenta are:
"!
"!
l
l
l
1
L i = m1v0 k# and L f = m1v f k# + I% rod k# = m1v f k# + m2l 2% rod k#
2
2
2
12
Using conservation of angular momentum (and dropping the vector signs):
l
l 1
Li = L f & m1v0 = m1v f + m2l 2% rod
2
2 12
(c) What is the magnitude of the final angular velocity, ωrod, of the stick about its center of
mass in terms of the known (including ones you’ve now calculated) quantities?
l
l 1
= m1v f + m2l 2! rod
2
2 12
6 m1
6
Solving for ! rod , we obtain: ! rod =
v0 " v f = vrod
l m2
l
From the above, we have: m1v0
(
)