Please wait for the animation to completely load.

A red mass (1 kg) is incident on an identical black mass (1 kg) that is attached
to a massless rigid string so that it can rotate around the origin as shown
**(position is given in meters and time is given in seconds)**. At
t = 2.6 s the red mass undergoes a completely elastic collision with the black
mass. Restart.

Consider the beginning part of the animation in which a 1-kg red ball is incident
on a 1-kg black ball that is constrained to move in a circle. From which point
should we measure the red ball's angular momentum? The best place, given the collision
with the pendulum, is the point (0, 0), the pivot. This is because we can easily
measure the angular momentum of the pendulum about this point. What then is the
angular momentum of the red ball before the collision? Certainly it must be changing,
since **r** changes. No! The angular momentum for a particle is
given by the cross product:
**L** = **r** × **p,** which means
we can consider the part of
**r** that is perpendicular to
**p(**r p sin θ, where θ is the angle between
**r** and **p)**. Since **p** is in the
negative x direction, the part of
**r** that is perpendicular to **p** is y. Therefore,
|L| = 50 kg·m^{2}/s. Note that even though **r** changes,
y does not. The direction of the angular momentum is found with the RHR and is
into the page (which is the negative z direction).

Now what happens to the angular momentum after the first collision? Given that only
the black ball moves, we find that |L| = mvr = Iω = 50 kg·m^{2}/s
(again, into the page). The angular momentum is the same as before the collision.
Given that there are no external torques (The pendulum string does not create a
torque. Why?), angular momentum is conserved.

What about after the second collision? Well, this is a bit harder. The radius vector
**r** changes (before the first collision the radius changed, but
the part of the radius perpendicular to the momentum was constant). We must use
a better definition of the magnitude of
**r** × **p** than rp sin θ. In general we
get for the z component of the angular momentum: L
_{z} = (xp_{y} – yp_{x}). At t = 16 seconds, we have (-5.06)
(-1.73) – (-12.51) (-4.69) = -50 = 50 kg·m^{2}/s (again, into
the page).

Note that in general,
**A** × **B** = (A
_{y}B_{z} – A_{z}B_{y})
**i** + (A_{z}B_{x} – A_{x}B_{z})
**j** + (A_{x}B_{y} – A
_{y}B_{x})** k**.

Illustration authored by Mario Belloni and Wolfgang Christian.

HTML updated for JavaScript by Ricky Davidson.

Physlets were developed at Davidson College and converted from Java to JavaScript
using the
SwingJS
system developed at St. Olaf College.