As soon as the disc hits the rod's end, it simultaneously sets the rod into spin of angular velocity w about its CG and imparts the rod's CG a linear velocity of v2. In other words the rod not only spins but also moves to the left as a whole. The disc itself continues to move to the left with a velocity of v1. Let the mass of the disc be m.

Since there are no external forces to the (rod + disc) system their combined momentum will be conserved. Hence, we have,

Since there are no external torques acting on the (rod + disc) system, their combined angular momentum must be conserved as well. Considering an axis that passes through the rod's CG, perpendicular to plane of the rod and conserving the angular momentum of the (rod+ disc) system before and after collision we have,

Note that the moment of inertia of the rod about th axis passing through its CG perpendicular ot the plane of its motion is (see the list of moment of inertias).

From (1) and (2) we get,

Since the collision is elastic, the total kinetic energy of the (rod+disc) system before and after collision will be conserved and so we have,

From (7) it is clear that the disc will stop when and will reverse its direction when .

## Monday, February 16, 2009

Subscribe to:
Posts (Atom)