A thin uniform rod X of mass M and length L is pivoted at a point on the rod a height L/3 above its lower end, at the edge of a table. 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= gravitatio…

Question

A thin uniform rod X of mass M and length L is pivoted at a point on the rod a height L/3 above its lower end, at the edge of a table. 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)

Accepted Answer

(4) √(3g)/L. The pivot sits L/3 from the lower end, while the rod's centre is at L/2 from that end. So the centre is L/2-L/3=L/6 above the pivot, and that is how far it falls as the rod swings from vertical to horizontal. Moment of inertia about the pivot: I=(ML²)/(12)+M(L/6)²=(ML²)/(12)+(ML²)/(36)=(ML²)/9. Energy: MgL/6=1/2·(ML²)/9ω². (gL)/6=(L²ω²)/(18), so ω²=(3g)/L. ω=√(3g)/L.

Step-by-step solution

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