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Illustration 11.5: Conservation of Angular Momentum
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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 (rp 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·m2/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·m2/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: Lz = (xpy - ypx). At t = 16 seconds, we have (-5.06) (-1.73) - (-12.51) (-4.69) = -50 = 50 kg·m2/s (again, into the page).
Note that in general, A × B = (AyBz - AzBy) i + (AzBx - AxBz) j + (AxBy - AyBx) k.