Question

A cubical block of side a is moving with velocity v on a horizontal smooth plane as shown in figure. 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. $\sqrt{\frac{3}{2}}a$
• D. zero

Answer 1

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

When the block hits the ridge O, it will start rotating about an axis passing through O and perpendicular to the plane of the paper. Net torque about O is zero, therefore, angular momentum L about O will be conserved.
$L_i=L_f$
$\therefore Mv\left(\frac{a}{2}\right)=I_O\omega =(I_{cm}+M{r}^2)\omega =(\frac{M{a}^2}{6}+M{\left(\frac{a}{\sqrt{2}}\right)}^2)\omega =\frac{2}{3}M{a}^2\omega$

where $I_O$ is the MI of the cube about point O and $\omega$ is the angular speed of rotation of the block about O.
$\therefore \omega =\frac{3v}{4a}$
Note: The MI of a square plate is $\frac{M{a}^2}{6}$. Hence, MI of a cube will be also $\frac{M{a}^2}{6}$.