Lab VII Notes
Conservation of Angular Momentum
Week of 2008.10.27
Lab VII: Conservation of Angular Momentum
Conservation
You should have three objects: pucks of m1 and m2 and a rod of mr . Initially, only m1 is
moving. At some point in its pre-collision path, it is at r0 from the axis of rotation and has
a velocity v0 . Thus its initial angular momentum is
~ i = m1~r0 × ~v0 .
L
Since all the motion takes place in the x − y plane, we can just use
Li = m1 r0 v0 cos θ0 .
After the collision, m1 is at r1 and has a velocity v1 . Meanwhile, m1 and mr have ω about
the axis of rotation. Let R be distance from the axis of rotation to the center of m2 . The
total moment of inertia is
I = (m2 + 13 mr )R2 .
The final angular momentum is thus
Lf = m1 r1 v1 sin θ1 + (m2 + 31 mr )R2 ω.
You should compare the left- and right-hand sides in the equation,
m1 r0 v0 sin θ0 = m1 r1 v1 sin θ1 + (m2 + 13 mr )R2 ω.
Error Analysis
I instructed each class via e-mail to assume that the masses were known exactly. Thus for
the purposes of error propagation, Li = Li (r0 , v0 , θ0 ) and Lf = Lf (r1 , v1 , θ1 , R, ω). Here are
the correct expressions:
2
2
2
∂Li
∂Li
∂Li
2
2
2
δLi =
δr0 +
δv0 +
δθ02
∂r0
∂v0
∂θ0
2 2
= (m1 v0 sin θ0 ) δr0 + (m1 r0 sin θ0 )2 δv02 + (m1 r0 v0 cos θ0 )2 δθ02 .
1/2
δLi = (m1 v0 sin θ0 )2 δr02 + (m1 r0 sin θ0 )2 δv02 + (m1 r0 v0 cos θ0 )2 δθ02
.
Similarly,
δLf = (m1 v1 sin θ1 )2 δr12 + (m1 r1 sin θ1 )2 δv12 + (m1 r1 v1 cos θ1 )2 δθ12
1/2
.
+ [2(m2 + 13 mr )Rω]2 δR2 + [(m2 + 31 mr )R2 ]2 δω 2
You should report
Li = [value of Li ] ± δLi ,
Lf = [value of Lf ] ± δLf .
Lab VII Notes
Conservation of Angular Momentum
Week of 2008.10.27
Questions
You should cover these points in your discussion:
~ points into the page.
• Since ~p and ~r are in the plane of the drawing, L
~ does not change after the collision; if the direction of L
~ changes,
• The direction of L
angular momentum is not conserved.
• Linear momentum is not conserved during this collision because there is an external
force: m2 and mr experience a force towards the axis of rotation exerted by the hinge
on the side of the setup.
• If a different reference point was chosen, angular momentum would not be conserved
because there would be a net torque. Only if the hinge is the axis of rotation, we have
~r k F~ , and thus ~τ = ~r × F~ = 0, so angular momentum is conserved.
P
• If
F~ext 6= 0, angular momentum is still conserved as long as the force acts right
at the axis of rotation (~r = 0) or the force acts parallel to the radial vector (~r k F~ ;
~r × F~ = 0).
• If the original puck is traveling in a straight line, it still can be said to have angular
momentum. Angular momentum is conserved during the collision, and it is readily
~ 6= 0 after the collision, so we must have that L
~ 6= 0 before the collision
apparent that L
as well.
Point deductions
Here are the point deductions specific to this lab:
• Errors extracting r, v, or ω from the printout?
−1
• Li 6= Lf but not addressed?
−1
• δLi and δLf not computed?
−2
• δLi or δLf computed incorrectly?
−1
• Every two questions wrong or unanswered?
−1