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.
.

Options

A. `mr^2((2gh)/(v^2) +1)`
B. `mr^2((g h)/(v^2) +1)`
C. `mr^2((2 g h)/(v^2)-1)`
D. `mr^2((g h)/(v^2)+1)`

Answer

C

Explanation

( c) Each point on the cord moves at a linear speed of `v = omega r` where `r` is the radius of the spool.The energy conservation equation for the counterweight-turntable-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 = (1)/(2)mv^2 + (1)/(2) I omega^2 + 0`
`mgh = (1)/(2) = (1)/(2) mv^2 + (1)/(2) I (v^2)/(r^2)`
`2mgh - mv^2 = I (v^2)/(r^2)` and finally, `I = mr^2 ((2gh)/(v^2) - 1)`.