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 : 

2 × 10^–6 N m

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