A light rigid rod of length 2a carries two identical point masses m, one at each end. The rod is suspended horizontally at its centre by a vertical torsion wire of torsional constant C. If the rod is rotated through a small angle about the wire and released, what is the time period?
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Question

A light rigid rod of length 2a carries two identical point masses m, one at each end. The rod is suspended horizontally at its centre by a vertical torsion wire of torsional constant C. If the rod is rotated through a small angle about the wire and released, what is the time period?

Options

G1 Physics · Accepted Answer

The moment of inertia of the two masses about the vertical axis is I = ma²+ma² = 2ma². The restoring torque is τ = −Cθ. Hence, Iθ̈ = −Cθ. Therefore, ω² = C/I = C/(2ma²). Thus, T = 2π/ω = 2π√(2ma²/C). Correct option: B.