Figure shows a counterweight of mass m suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. The turntable can rotate without friction. When the counterweight is released from rest, it descends through a distance h, acquiring a speed v. The moment of inertia I of the rotating apparatus is

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${mr}^{2} (\frac{2 gh}{v^{2}} + 1)$

${mr}^{2} (\frac{gh}{v^{2}} + 1)$

${mr}^{2} (\frac{gh}{v^{2}} - 1)$

${mr}^{2} (\frac{2 gh}{v^{2}} - 1)$

answer is C.

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Detailed Solution

Each point on the cord moves at a linear speed of v = $ωr$ where r is the radius of the spool. The energy conservation equation for the counter weight turn table-Earth system is:
${(K_{1} + K_{2} + U_{g})}_{i} + W_{other} = {(K_{1} + K_{2} + U_{g})}_{f}$

Specializing, we have
$0 + 0 + mgh + 0 = \frac{1}{2} {mv}^{2} + \frac{1}{2} {Iω}^{2} + 0$

mgh = $\frac{1}{2} {mv}^{2} + \frac{1}{2} I \frac{v^{2}}{r^{2}}$

2mgh - ${mv}^{2} = I \frac{v^{2}}{r^{2}} and finally, I = {mr}^{2} (\frac{2 gh}{v^{2}} - 1)$

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