Rotation Ques 4
4 A rectangular solid box of length $0.3 \mathrm{~m}$ is held horizontally, with one of its sides on the edge of a platform of height $5 \mathrm{~m}$. When released, it slips off the table in a very short time $\tau=0.01 \mathrm{~s}$, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to
(2019 Main, 8 April II)
(a) 0.02
(b) 0.3
(c) 0.5
(d) 0.28
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Answer:
Correct Answer: 4.( c )
Solution:
- Key Idea The rectangular box rotates due to torque of weight about its centre of mass. Now, angular impulse of weight $=$ Change in angular momentum.
$ \begin{aligned} & \therefore & m g \frac{l}{2} \times \tau & =\frac{m l^2}{3} \omega \\ & \Rightarrow & \omega & =\frac{3 g \times \tau}{2 \times l} \end{aligned} $
Substituting the given values, we get
$ =\frac{3 \times 10 \times 0.01}{2 \times 0.3}=0.5 \mathrm{rad} \mathrm{s}^{-1} $
Time of fall of box.
$ t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s} $
So, angle turned by box in reaching ground is
$ \theta=\omega t=0.5 \times 1=0.5 \mathrm{rad} $
BETA
