A wheel having moment of inertia 2   kg   m 2 about its axis, rotates at 50 rpm about the same…

Given,
Moment of inertia of wheel $\left(I\right)=2 \mathrm{kg} {\mathrm{m}}^2$

Initial angular velocity

$\left({\omega }_0\right)=\frac{2\mathrm{\pi N}}{60}=\frac{2\mathrm{\pi }\times 50}{60}=\frac{10}{6}\mathrm{\pi } \mathrm{rad} {\mathrm{s}}^{-1}$

Final angular velocity $\left(\omega \right)=0$

Time required $\left(t\right)=1 \min =60 \mathrm{s}$

Using first equation of motion-

$\omega ={\omega }_0+\alpha t$

$\Rightarrow 0=\frac{10}{6}\pi +\mathrm{\alpha } \left(60\right)$
$\Rightarrow \alpha =-\frac{\pi }{36} \mathrm{rad} {\mathrm{s}}^{-2}$

Torque required $\left(\tau \right)=I\alpha =2\times \frac{\pi }{36}=\frac{\pi }{18} \mathrm{Nm}$