If $I = 50\,kg-m^2$,then how much torque (in $N-m$) must be applied to stop it in $10\,s$,given its initial angular speed is $20\,rad/s$?

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The angular deceleration required to bring it to rest in $20 \ s$ is:

$A$ wheel having a moment of inertia of $5 \times 10^{-3} \ kg \ m^2$ is rotating at a rate of $20 \ rev/s$. The torque required to stop the wheel in $10 \ s$ is $... \times 10^{-2} \ N \ m$. (in $\pi$)

$A$ solid disc of radius $20 \, \text{cm}$ and mass $10 \, \text{kg}$ is rotating with an angular velocity of $600 \, \text{rpm}$ about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc to rest in $10 \, \text{s}$ is $x \pi \times 10^{-1} \, \text{Nm}$. Find the value of $x$.

$A$ torque of $50\, Nm$ acting on a wheel at rest rotates it through $200\, rad$ in $5\, s$. Calculate the angular acceleration produced. ........... $rad\, s^{-2}$.

An automobile moves on a road with a speed of $54 \,km h^{-1}.$ The radius of its wheels is $0.45\, m$ and the moment of inertia of the wheel about its axis of rotation is $3\, kg m^2$. If the vehicle is brought to rest in $15\, s,$ the magnitude of average torque transmitted by its brakes to the wheel is .......... $kg \,m^2\, s^{-2}$.