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

Question

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 (A) m r2((2 g h/v2)+1) (B) m r2((g h/v2)-1) (C) m r2((2 g h/v2)-1) (D) m r2((g h/v2)+1)

Tardigrade Admin · Accepted Answer

Each point on the cord moves at a linear speed of

v = ω 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} + \dot{U}_{g})_{f}

Specializing, we have

0 + 0 + m g h + 0 = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} + 0

mgh

= \frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v ^{2}}{r ^{2}}

2 m g h - m v^{2} = I \frac{v ^{2}}{r ^{2}} and finally, I = m r^{2} (\frac{2 g h}{v ^{2}} - 1)

$m r^{2} (\frac{2 g h}{v ^{2}} + 1)$

$m r^{2} (\frac{g h}{v ^{2}} - 1)$

$m r^{2} (\frac{2 g h}{v ^{2}} - 1)$

$m r^{2} (\frac{g h}{v ^{2}} + 1)$

mr2(v22gh​+1)

mr2(v2gh​−1)

mr2(v22gh​−1)

mr2(v2gh​+1)

$m r^{2} (\frac{2 g h}{v ^{2}} + 1)$

$m r^{2} (\frac{g h}{v ^{2}} - 1)$

$m r^{2} (\frac{2 g h}{v ^{2}} - 1)$

$m r^{2} (\frac{g h}{v ^{2}} + 1)$