A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) ml²/12ω ²sin θ cos θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F H and F V about the CM. The value of θ is then such that :

Solution

Physics sir · 4 steps

Newton's laws give F_V and F_H; their torques about the CM must supply the given rate of change of L.

1. Find the forces. The CM moves in a horizontal circle of radius l/2 sin θ and does not move vertically. So F_V = mg, and F_H supplies the centripetal force: F_H = mω² l/2 sin θ, pointing towards the shaft.
2. Torques about CM. The pivot sits l/2 cos θ above the CM and l/2 sin θ to its side. So F_V has lever arm l/2 sin θ and F_H has lever arm l/2 cos θ. The two torques turn the rod in opposite senses, so the net torque is their difference.

   mg l/2 sin θ - mω²l/2 sin θ·l/2 cos θ = ml²/12 ω²sin θ cos θ

3. Solve. Cancel m l sin θ from every term: g/2 = lω²cos θ(1/4+1/12) = (lω²cos θ)/3.
4. So. cos θ = 3g/2lω², which is option (b).

The idea to keep. For a rotating rigid body, first get the contact forces from the motion of the CM. Then take torques of those forces about the CM and equate them to dL/dt about the CM.

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