A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity ω. The angular momentum of M about the point A, th…

Question

A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity ω. The angular momentum of M about the point A, the centre of that circular path, is L_A, which lies in the positive z direction, and the angular momentum of M about the point B, the upper end of the rod, is L_B. The correct statement for this system is

Accepted Answer

(3) L_A is constant, both in magnitude and direction. A is the centre of the circle, in the plane of the motion, so from A the position vector has constant length r and is always perpendicular to ⃗v. ⃗L_A=M⃗r×⃗v has magnitude Mvr, constant, and points along the axis of the circle — the +z direction — which never moves. So ⃗L_A is constant in both magnitude and direction. B is the pivot at the top of the rod, off the plane of the circle. The vector from B to M has a vertical part as well as a horizontal one. Its cross product with ⃗v therefore has a component along the axis, which is steady, and a component perpendicular to the axis, which turns with the mass. |⃗L_B| stays the same, because the geometry repeats, but its direction sweeps round a cone once per revolution. Only L_A is constant in both magnitude and direction.

Step-by-step solution

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