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. $m r^{2}\left(\frac{2 g h}{v^{2}}+1\right)$
• B. $m r^{2}\left(\frac{g h}{v^{2}}-1\right)$
• C. $m r^{2}\left(\frac{2 g h}{v^{2}}-1\right)$
• D. $m r^{2}\left(\frac{g h}{v^{2}}+1\right)$

Answer 1

Answer: (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:
$\left(K_{1}+K_{2}+U_{g}\right)_{i}+\mid W_{\text {other }}=\left(K_{1}+K_{2}+\dot{U}_{g}\right)_{f}$
Specializing, we have
$0+0+m g h+0=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}+0$
mgh $=\frac{1}{2} m v^{2}+\frac{1}{2} I \frac{v^{2}}{r^{2}} $
$2 mgh-m v^{2}=I \frac{v^{2}}{r^{2}} \text { and finally, } I=m r^{2}\left(\frac{2 g h}{v^{2}}-1\right)$