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. |
\[{{L}_{i}}={{L}_{f}}\] |
\[\therefore \] \[mv\left( \frac{l}{2} \right)=I\omega \] |
\[\Rightarrow \] \[mv\frac{1}{2}=\frac{2}{3}m{{l}^{2}}\cdot \omega \] |
\[\Rightarrow \] \[\omega =\frac{3mv}{4lm}\] |
\[=\frac{3\sqrt{2mK}}{4lm}=\frac{3}{2\sqrt{2}}\cdot \sqrt{\frac{K}{m{{l}^{2}}}}\] |
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