Question

A uniform disc of mass $20kg$ and radius $0.5m$ can tum about a smooth axis through its centre and perpendicular to the disc. A constant torque is applied to the disc for $3s$ from rest and the angular velocity at the end of that time is $\frac{240}{\pi }$ rev/min. Find the magnitude of the torque. If the torque is then removed and the disc is brought to rest in t seconds by a constant force of $10N$ applied tangentially at a point on the rim of the disc, find $t$ in seconds.

Answer 1

Angular impulse $=$ change in angular momentum
$\therefore \tau t=I\omega =\frac{1}{2}m{R}^2\omega .....\left(i\right)$
$\therefore \tau =\frac{m{R}^2\omega }{2t}$
$=\frac{\left(20\right)(0.5{)}^2(2\pi \times \frac{240}{\pi \times 60})}{2\times 3}$
$=\frac{20}{3}N.m$
During retardation using Eq.(1)
$t=\frac{m{R}^2\omega }{2\tau }=\frac{m{R}^2\omega }{2\left(FR\right)}$
$=\frac{mR\omega }{2F}$
$=\frac{\left(20\right)\left(0.5\right)(2\pi \times \frac{240}{\pi \times 60})}{2\times 10}$
$=4s$