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
Answer: (A) $0.5$
While its end rests on the edge, the box turns about that edge under the torque of its weight acting at the centre, $l/2$ away. The angular impulse gives it an angular velocity:
$$mg\frac l2\,\tau = \frac{ml^2}{3}\,\omega \Rightarrow \omega = \frac{3g\tau}{2l} = \frac{3\times10\times0.01}{2\times0.3} = 0.5\ \text{rad/s}$$
Time to fall $5\ \text{m}$: $t = \sqrt{2h/g} = 1\ \text{s}$. Angle turned before hitting the ground:
$$\theta = \omega t \approx 0.5\ \text{rad}$$
Solution by Sreeraj P, M.Sc Physics