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 :
Answer & explanation
Correct answer: option 2
W = \(\frac{1}{2} I (\omega_f^2 - \omega_i^2)\)
\(\theta\) = 2 \(\pi\) revolutions
1 revolution = 2 \(\pi\) rad ; 2 \(\pi\) revolutions = 2 \(\pi\) x 2 \(\pi\) rad = 4 \(\pi^2\) rad
\(\theta\) = 4 \(\pi^2\) rad
\(\omega_i\) = 3 x \(\frac{2 \pi}{60}\) rad/s
-\(\tau\theta\) = \(\frac{1}{2}\) \(\frac{1}{2} m r^2 (0^2 - \omega_i^2)\)
-\(\tau\theta\) = 2 x 10-6 N m