System Of Particles And Rotational Motion Ques 3
3. A wheel having moment of inertia $2 \mathrm{kg}-\mathrm{m}^2$ about its vertical axis, rotates at the rate of $60 $ $\mathrm{rpm}$ about this axis. The torque which can stop the wheel’s rotation in one minute would be
[2004]
(a) $\frac{\pi}{18} $ $N-m$
(b) $\frac{2 \pi}{15}$ $ N-m$
(c) $\frac{\pi}{12} $ $N-m$
(d) $\frac{\pi}{15}$ $ N-m$
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Answer:
Correct Answer: 3.(d)
Solution: (d) $ \mathrm{I}=2 $ $\mathrm{kgm}^2, v_0=60 $ $\mathrm{rpm}=1 \mathrm{rps} $
$ \omega_0=2 \pi v_0=2 \pi $ $\mathrm{rad} / \mathrm{sec} $
$ \omega_{\mathrm{f}}=0 \text { and } \mathrm{t}=1 \mathrm{min}=60 $ $\mathrm{sec}$
So, $\propto=\frac{\omega_f-\omega_0}{t}=\frac{0-2 \pi}{60}$
$ \alpha=\frac{-\pi}{30}$ $ \mathrm{rad} / \mathrm{sec}^2 $
$ \text { and Torque, } \tau=\text { I. } \alpha $
$ \tau=2 \cdot\left(\frac{-\pi}{30}\right)=\frac{-\pi}{15} $
$ \tau=\frac{\pi}{15} \mathrm{N}-\mathrm{m}$
BETA
