$A$ rope is wound around a hollow cylinder of mass $3\, kg$ and radius $40\, cm.$ What is the angular acceleration (in $rad/s^2$) of the cylinder if the rope is pulled with a force of $30\, N$?

$A$ torque of $50 \ N \cdot m$ causes a wheel to rotate from rest through $200 \ rad$ in $5 \ s$. The value of the angular acceleration produced is ...... $rad/s^2$.

$A$ rope of negligible mass is wound round a hollow cylinder of mass $3\; kg$ and radius $40\; cm$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30\; N$? What is the linear acceleration of the rope? Assume that there is no slipping.

$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

$A$ particle of mass $m$ moves in a circular path of radius $r$,under the action of a force which delivers constant power $P$ and increases its speed. The angular acceleration of the particle at time $t$ is proportional to:

$A$ string is wrapped around the rim of a wheel of moment of inertia $0.40 \ kg \cdot m^2$ and radius $10 \ cm$. The wheel is free to rotate about its axis. Initially,the wheel is at rest. The string is now pulled by a force of $40 \ N$. The angular velocity of the wheel after $10 \ s$ is $x \ rad/s$,where $x$ is $\qquad$