Question

A rectangular solid box of length $0.3\text{m}$ is held horizontally, with one of its sides on the edge of a platform of height $5\text{m}$. When released, it slips off the table in a very short time $\tau =0.01\text{s}$, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

• A. $0.28$
• B. $0.02$
• C. $0.3$
• D. $0.5$

Answer 1

The correct option is D $0.5$

Here,
Angular impulse = change in angular momentum

$\tau \Delta t=\Delta L$

$mg\frac{l}{2}\times 0.01=\frac{m{l}^2}{3}\omega$

$\omega =\frac{3g\times 0.01}{2l}=\frac{3\times 10\times 0.01}{2\times 0.3}=0.5\text{rad/s}$

As the change in angular momentum occures in a very small interval of time, $\omega$ can be assumed to remain constant.

Now, $\theta =\omega T=\omega \sqrt{\frac{2h}{g}}$

$=0.5\times \sqrt{\frac{2\times 5}{10}}=0.5\text{rad.}$

Hence, $\left(A\right)$ is the correct answer.