System of Particles and Rotational Motion - Result Question 20
20. A solid cylinder of mass $2 kg$ and radius $4 cm$ is rotating about its axis at the rate of $3 rpm$. The torque required to stop after $2 \pi$ revolutions is :
[2019]
(a) $2 \times 10^{-6} Nm$
(b) $2 \times 10^{-3} Nm$
(c) $12 \times 10^{-4} Nm$
(d) $2 \times 10^{6} Nm$
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Answer:
Correct Answer: 20. (a)
Solution:
- (a) According to the Work-energy theorem,
$W=\frac{1}{2} I(\omega_f^{2}-\omega_i^{2})$
Given that,
$\theta=2 \pi$ revolution/minute
$\theta=2 \pi \times 2 \pi=4 \pi^{2} rad$
$\omega_i=3 \times \frac{2 \pi}{60} rad / s$
$\omega_f=0 rad / s$
Putting the values of $\omega_f$ and $\omega_i$ we get
$\Rightarrow-\tau \theta=\frac{1}{2} \times \frac{1}{2} mr^{2}(0^{2}-\omega_i^{2})$ $\Rightarrow-\tau=\frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times(4 \times 10^{-2})(-3 \times \frac{2 \pi}{60})^{2}}{4 \pi^{2}}$
$\Rightarrow \tau=2 \times 10^{-6} N-m$
