Question

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

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

Answer 1

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

After hitting the ridge at point $O$, block will rotate immediately with angular velocity $\omega$ about axis of rotation passing through $O$.


${\tau }_{\text{ext}}=0$ about point $O$.
(since torque due to $N$ and $mg$ cancel)
Hence, applying angular momentum conservation:
$L_i=L_f$
$Mv\left(\frac{a}{2}\right)=I_0\times \omega ...\left(i\right)$


$\text{MOI}$ about $O$ from parallel axis theorem,
$I_0=I_{\text{CM}}+M{d}^2$
$I_0=\frac{M{a}^2}{6}+M{\left(\frac{a}{\sqrt{2}}\right)}^2$
$\Rightarrow I_0=\frac{2M{a}^2}{3}$
Substituting in Eq. $\left(i\right)$,
$Mv\left(\frac{a}{2}\right)=\left(\frac{2M{a}^2}{3}\right)\omega$
$\therefore \omega =\frac{3v}{4a}$