Inelastic Collision with Rotating Rod · CAP 2001 #25

Physics · mechanics

Problem

A mass m is moving at speed v perpendicular to a rod of length d and mass M = 6m which pivots around a frictionless axle running through its centre. It strikes and sticks to the end of the rod. The moment of inertia of the rod about its centre is Md2/12Md^2/12. Then the angular speed of the system right after the collision is (a) 2v/3d2v/3d. (b) 2v/d2v/d. (c) v/dv/d. (d) 3v/2d3v/2d.
Intuitive explanation

This is a collision problem where linear momentum becomes rotational momentum. The key is to use conservation of angular momentum about the pivot point: the incoming mass's linear momentum (relative to the axis) converts entirely into the spinning system's rotational momentum. Calculate the mass's angular momentum as L=mvdL = mvd (momentum times perpendicular distance from pivot), then set it equal to the final rotational inertia times angular speed, where the total inertia includes both the rod and the stuck mass at distance d/2d/2 from the axis.

Intuitive explanation

This is a collision problem where linear momentum becomes rotational momentum. The key is to use conservation of angular momentum about the pivot point: the incoming mass's linear momentum (relative to the axis) converts entirely into the spinning system's rotational momentum. Calculate the mass's angular momentum as L=mvdL = mvd (momentum times perpendicular distance from pivot), then set it equal to the final rotational inertia times angular speed, where the total inertia includes both the rod and the stuck mass at distance d/2d/2 from the axis.

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