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A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

1. g/L
2. 2g/L
3. 2g/3L
4. 3g/2L

Correct answer: 3g/2L

Solution

The torque about the hinge due to the rod's weight is Mg(L/2). Using the rotational analog of Newton's second law, torque = Iα, where I = (1/3)ML² for a rod hinged at one end. Solving for angular acceleration, α = (3g/2L).

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