Question

The front solid cylinder has mass $\frac{M}{3} $ while the back one solid cylinder has mass $\frac{2M}{3}$. The centers of these cylinders are connected by massless rod as shown. Both the cylinders have same radii R. The system is released from rest on the inclined plane. The cylinders roll down. The speed of the rod after system descending a vertical distance h is

• A. $\sqrt{\frac{2gh}{3}}$
• B. $\sqrt{2gh}$
• C. $\sqrt{\frac{4gh}{3}}$
• D. $\sqrt{\frac{3gh}{7}}$

Answer 1

The Correct Option is C

Applying conservation of energy, $E _{1} = E _{2}$ or $\Delta U _{1}+\Delta KE _{1} =\Delta U _{2}+\Delta KE _{2}$ $\Rightarrow\left( M _{1}+ M _{2}\right) gh +0$ $=0+\frac{1}{2}\left( M _{1}+ M _{2}\right) v ^{2}+\frac{1}{2}\left( I _{1}+ I _{2}\right) \omega^{2}$ Given, $M _{1}=\frac{ M }{3}$ and $M _{2}=\frac{2 M }{3}$ $\because I =\frac{1}{2} MR ^{2}$ In case of pure rolling, $v _{ CM }= R \omega$ $\therefore\left(\frac{ M }{3}+\frac{2 M }{3}\right) gh$ $=\frac{1}{2}\left[\frac{ M }{3}+\frac{2 M }{3}\right] v ^{2}+\frac{1}{2} \cdot \frac{1}{2}\left[\frac{ M }{3}+\frac{2 M }{3}\right] R ^{2} \omega^{2}$ $\Rightarrow Mgh =\frac{1}{2} Mv ^{2}+\frac{1}{2} \cdot \frac{1}{2} Mv ^{2}$ $\left[\because V _{ CM }= R \omega\right]$ $\Rightarrow gh =\frac{1}{2} v ^{2}+\frac{1}{4} v ^{2}$ $\Rightarrow \frac{3 v ^{2}}{4}= gh$ $\therefore v =\sqrt{\frac{4 gh }{3}}$