Question

A cubical block of side length $a$ and mass $M$, 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 immediately after it hits the ridge is -

• A. $\frac{3v}{4a}$
• B. $\frac{3v}{2a}$
• C. $\frac{\sqrt{3}v}{2a}$
• D. $\frac{\sqrt{3}v}{4a}$

Answer 1

The correct option is A $\frac{3v}{4a}$


From the above diagram, we have,

$r^2=\frac{a^2}{2}$

Net torque about point $O$ will be zero.

Therefore, angular momentum $\left(L\right)$ about point $O$ will be conserved, $i.e.L_i=L_f$

$\Rightarrow Mv\left(\frac{a}{2}\right)=I_0\omega$

$\Rightarrow \frac{Mva}{2}=[\frac{M{a}^2}{6}+M\left(\frac{a^2}{2}\right)]\omega$

$\Rightarrow \frac{Mva}{2}=\frac{2M{a}^2\omega }{3}$

$\Rightarrow \omega =\frac{3v}{4a}$

Hence, option $\left(A\right)$ is the correct answer.