question_answer

A cubical block of side \[l\] is given a kinetic energy K. It moves over a smooth flat surface and strikes a ridge at some point P as shown below.

Moment of inertia of cube about its centre is \[\frac{m{{l}^{2}}}{6}.\] Angular speed of block just after it strikes P will be

A) \[\sqrt{2}\frac{Kl}{m}\]

B) \[\frac{\sqrt{2K}\cdot m}{{{l}^{2}}}\]

C) \[\frac{3}{2\sqrt{2}}\cdot \sqrt{\frac{K}{m{{l}^{2}}}}\]

D) \[\frac{2}{\sqrt{3}}\cdot \sqrt{\frac{Km}{{{l}^{2}}}}\]

Correct Answer: C

Solution :

Moment of inertia of cube about an edge is (by parallel axis theorem)

\[I={{I}_{CM}}+m{{\left( \frac{l}{\sqrt{2}} \right)}^{2}}\]\[=\frac{m{{l}^{2}}}{6}+\frac{m{{l}^{2}}}{2}=\frac{2}{3}m{{l}^{2}}\]

As there is no external torque, angular momentum about P is conserved.

i.e.

\[{{L}_{i}}={{L}_{f}}\]

\[\therefore \]      \[mv\left( \frac{l}{2} \right)=I\omega \]

\[\Rightarrow \]   \[mv\frac{1}{2}=\frac{2}{3}m{{l}^{2}}\cdot \omega \]

\[\Rightarrow \]   \[\omega =\frac{3mv}{4lm}\]

\[=\frac{3\sqrt{2mK}}{4lm}=\frac{3}{2\sqrt{2}}\cdot \sqrt{\frac{K}{m{{l}^{2}}}}\]