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. $\frac{\sqrt{3}v}{\sqrt{2}a}$
• 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 $I_c=\frac{M{a}^2}{6}$
hence from parallel axis theorem $I_0=I_c+M{r}^2$ and $r^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2=\frac{a^2}{2}$
$I_0=\frac{M{a}^2}{6}+\frac{M{a}^2}{2}=\frac{2M{a}^2}{3}$
From conservation of angular momentum about $O$
$Mv\frac{a}{2}=\frac{2}{3}M{a}^2\omega$
$\omega =\frac{3v}{4a}$