Question

A cubical block of side ' a' is moving with velocity V on a horizontal smooth plane as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Applying conservation of angular momentum w.r.t ridge at 'O',
$MV\left(\frac{a}{2}\right)=I\omega$
$\Rightarrow$ where 'I' is the moment of inertia of the block about the edge,
$I=\frac{M}{12}(a^2+a^2)+\frac{M{a}^2}{(\sqrt{2}{)}^2}$
$=\frac{M}{12}\left(2a^2\right)+\frac{M{a}^2}{2}$
$=M{a}^2(\frac{1}{6}+\frac{1}{2})=\frac{2}{3}M{a}^2$
$\therefore MV\left(\frac{a}{2}\right)=\left(\frac{2}{3}M{a}^2\right)\omega$
$\Rightarrow \omega =\frac{3V}{4a}$