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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2016
  • Hard
  • MCQ

A cubical block of side \(30\,cm\)  is moving with  velocity \(2\,\,ms^{-1}\) on a smooth horizontal surface. The surface has a bump at a point \(O\) as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is

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Answer & Solution

Correct Answer

(B) \(5.0\)

Step-by-step Solution

Detailed explanation

Angular momentum, \(mvr = I\omega \)

momrnt of Inertia \((I)\) of cubical block is given by

\(\begin{array}{l}
I = m\left( {\frac{{{R^2}}}{6} + {{\left( {\frac{R}{{\sqrt 2 }}} \right)}^2}} \right)\\
\therefore \,\,\omega  = \frac{{m.2\frac{R}{2}}}{{m\left( {\frac{{{R^2}}}{6} + {{\left( {\frac{R}{{\sqrt 2 }}} \right)}^2}} \right)}}\\
 \Rightarrow \omega  = \frac{{12}}{{8R}} = \frac{3}{{2 \times 0.3}} = \frac{{10}}{2} = 5\,rad/s.
\end{array}\)

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