A disc of mass \(25 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating at 240 r.p.m. A retarding torque brings it to rest in \(20 \mathrm{~s}\). If the torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force in newton is

Using rotational kinematic equation,
\(\begin{aligned} & \omega=\omega_0-a t \\ & \Rightarrow 0=\omega_0-a t \\ & \Rightarrow \alpha=\left(\frac{\omega_0}{t}\right)\end{aligned}\)
We know torque is given by,
\(F \cdot R=\text { Torque }=I \alpha\)
and the moment of inertia is given by, \(I=\frac{M R^2}{2}\)
\(\therefore F=\frac{I \alpha}{R}=\left(\frac{M R^2}{2}\right) \cdot\left(\frac{1}{R}\right) \cdot\left(\frac{\omega_0}{t}\right)=\left(\frac{M R \omega_0}{2 t}\right)\)
Given, \(M=25 \mathrm{~kg}, R=0.2 \mathrm{~m}, \omega_0=\frac{240(2 \pi)}{(60) \mathrm{sec}}\) and \(t=20 \mathrm{sec}\)
\(\begin{aligned} & \therefore F=\frac{(25 \mathrm{~kg}) \times(0.2 \mathrm{~m}) \times\left(8 \pi \mathrm{sec}^{-1}\right)}{(2 \times 20 \mathrm{sec})} \\ & \Rightarrow F=\pi N\end{aligned}\)