Question

A cubical block is moving on a horizontal smooth plane. It hits a ridge at point $\text{O}$. The angular speed of the block after it hits $\text{O}$ is

• A. $10\text{rad/s}$
• B. $18\text{rad/s}$
• C. $12\text{rad/s}$
• D. $15\text{rad/s}$

Answer 1

The correct option is D $15\text{rad/s}$

Let us suppose angular speed of the block after it hits $\text{O}$ is $\omega$ about axis passing through $\text{O}$.
From the diagrams shown below:




When the block collides with the ridge, the impulsive force which operates will pass through point $\text{O}$.
Therefore, net torque will be zero about $\text{O}$.
On applying law of conservation of angular momentum.
$L_i=L_f$
$\Rightarrow mv\frac{l}{2}=I_O\omega$
where $l$ is the side of the cube.

MOI of cubical block about point $\text{O}$ just as it hits the ridge and starts to rotate is given by parallel axis theorem:
$I_O=m\frac{l^2}{6}+m{\left(\frac{l}{\sqrt{2}}\right)}^2$

$\Rightarrow 5\times 20\times \frac{1}{2}=[\frac{5\times 1^2}{6}+5\times {\left(\frac{1}{\sqrt{2}}\right)}^2]\omega$
$\Rightarrow 50=\frac{10}{3}\omega$
$\Rightarrow \omega =15\text{rad/s}$
The angular speed of block after it hits $\text{O}$ is $15\text{rad/s}$.
Hence, option (d) is the correct answer.