Question

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

• A. 0.02
• B. 0.28
• C. 0.5
• D. 0.3

Answer 1

Answer: (C)

Angular impulse = change in angular momentum
$\tau \Delta t = \Delta L $
$ mg \frac{\ell}{2} \times0.1 = \frac{m\ell^{2}}{3} \omega$
$ \omega = \frac{3g\times0.01}{2\ell} $
$ = \frac{3\times10\times.01}{2 \times0.3} $
$= \frac{1}{2} = 0.5 \; rad /s$
time taken by rod to hit the ground
$ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times5}{10}} = 1 $ sec
in this time angle rotate by rod
$\theta =\omega t = 0.5 \times1 = 0.5 $ radian