Question

A rectangular solid box of length $0.3m$ is held horizontally, with one of its sides on the edge of a platform of height $5m$. When released, it slips off the table in a very short time $\tau =0.01s$, 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
$\tau \Delta t=\Delta L$
$mg\frac{\ell }{2}\times 0.1=\frac{m{\ell }^2}{3}\omega$
$\omega =\frac{3g\times 0.01}{2\ell }$
$=\frac{3\times 10\times .01}{2\times 0.3}$
$=\frac{1}{2}=0.5rad/s$
time taken by rod to hit the ground
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 5}{10}}=1$ sec
in this time angle rotate by rod
$\theta =\omega t=0.5\times 1=0.5$ radian