Question

A cubical block of side $10\text{cm}$ is moving with velocity $20\text{cm/s}$ on a horizontal smooth plane as shown. It hits a ridge at point $\text{O}$ as shown in the figure. What is the angular speed of the block after it hits $\text{O}$?

• A. $3.5\text{rad/sec}$
• B. $3\text{rad/sec}$
• C. $2\text{rad/sec}$
• D. $1.5\text{rad/sec}$

Answer 1

The correct option is D $1.5\text{rad/sec}$

Given that
Side of cubical block $\left(a\right)=10\text{cm}$
Speed of block $\left(v\right)=20\text{cm/s}$
Distance OM $=r=\frac{10}{\sqrt{2}}\text{cm}$


Net torque about O is zero, so angular momentum about O will be conserved.
Initial angular momentum $\left(L_i\right)=mv\times \frac{a}{2}$
Final angular momentum $L_f=I_0\omega$
$\Rightarrow mv\frac{a}{2}=(I_{cm}+m{r}^2)\omega (\because r^2=\frac{a^2}{2})$
$\Rightarrow mv\frac{a}{2}=(\frac{m{a}^2}{6}+m\left(\frac{a^2}{2}\right))\omega$
$\Rightarrow mv\frac{a}{2}=\frac{2}{3}m{a}^2\omega$
$\Rightarrow \omega =\frac{3v}{4a}$
$\Rightarrow \omega =\frac{3\times 20}{4\times 10}$
$\Rightarrow \omega =1.5\text{rad/s}$