A thin uniform rod (X) of mass M and length L is pivoted at a height (L/3) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is ____ .
( g= gravitational acceleration)

Solution

Physics sir · 4 steps

Track the centre of mass: it drops only L/6, and the rod turns about the pivot, not about its end.

1. Drop of centre. The centre of mass is at L/2 from the bottom, so it sits L/2-L/3=L/6 above the pivot. When the rod lies flat on the table, the centre of mass is at the pivot's height, so it falls by L/6.
2. Inertia about pivot. Use the parallel axis theorem with distance L/6 from the centre: I=ML²/12+M(L/6)²=(3ML²+ML²)/36=ML²/9.
3. Energy conservation. The pivot does no work, so the potential energy lost turns into rotational kinetic energy about the pivot.

   MgL/6=1/2(ML²/9)ω² ⇒ ω²=9g/3L=3g/L

4. So. ω=√(3g/L), which is option (b).

The idea to keep. For a body swinging about a fixed pivot, use Mg Δ h_(cm)=1/2Iₚᵢᵥₒₜω². Measure both the drop and the moment of inertia from the actual pivot.

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