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 point A is L A which lies in the positive z direction and the angular momentum of M about point B is L B . The correct statement for this system is :

Solution

Physics sir · 3 steps

Measure the position vector from each point, and watch whether it stays in the plane of the circle.

1. About point A. A is the centre of the circle, so r⃗_A is horizontal with length r. The velocity v⃗ is tangent to the circle and also horizontal. So L⃗_A = M r⃗_A × v⃗ points straight along +z and has magnitude Mrv, which never changes.
2. About point B. From B, the position vector runs along the rod, tilted at α to the axis. It is still perpendicular to v⃗, so |L⃗_B| = Mlv stays fixed. Its direction is perpendicular to the rod, so it tilts away from z and swings round the axis as M goes round.

   τ⃗ = r⃗ × F⃗ₙₑₜ, F⃗ₙₑₜ = centripetal, pointing to A

   Point	Torque of net force	Magnitude of L⃗	Direction of L⃗
   A	zero, since r⃗_A and F⃗ₙₑₜ lie along the same line	constant, Mrv	fixed along +z
   B	nonzero and horizontal, rotating with M	constant, Mlv	rotates on a cone round the axis

3. So. L⃗_A is constant in both magnitude and direction, while L⃗_B keeps its magnitude but changes direction. The correct option is (a).

The idea to keep. Angular momentum depends on the reference point. It is conserved only about a point where the net torque is zero. For a conical pendulum that point is the centre of the circle, not the pivot.

Answer from the NTA final answer key. Solution by PrepWiser.