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

Answer 1

Answer: (A)

$I_{0}$ is moment of inertia of block about an axis through $O$ and perpendicular to the plane of block.
If $I_{c}$ is moment of inertia about center of mass then $I_{c}=\frac{Ma^{2}}{6}$
hence from parallel axis theorem $I_{0}=I_{c}+Mr^{2}$ and $r^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}=\frac{a^{2}}{2}$
$I_{0}=\frac{Ma^{2}}{6}+\frac{Ma^{2}}{2}=\frac{2 Ma^{2}}{3}$
From conservation of angular momentum about $O$
$Mv\frac{a}{2}=\frac{2}{3}Ma^{2}\omega $
$\omega =\frac{3 v}{4 a}$