$A$ homogeneous disc with a radius $0.2\, m$ and mass $5\, kg$ rotates around an axis passing through its centre. The angular velocity of the rotation of the disc as a function of time is given by the formula $\omega = 2 + 6t$. The tangential force applied to the rim of the disc is ........ $N.$

$A$ pulley of radius $1.5\,m$ is rotated about its axis by a force $F = (12t - 3t^2)\,N$ applied tangentially (where $t$ is measured in seconds). If the moment of inertia of the pulley about its axis of rotation is $4.5\,kg\cdot m^2$,the number of rotations made by the pulley before its direction of motion is reversed will be $\frac{K}{\pi}$. The value of $K$ is $.....$

$A$ 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 ...... $rad/s^2$.

$A$ disc has mass $M$ and radius $R$. How much tangential force should be applied to the rim of the disc so as to rotate the disc with angular velocity $\omega$ in time $t$?

$A$ disc is rotating with an angular velocity $\omega_0$. $A$ constant retarding torque is applied on it to stop the disc. The angular velocity becomes $\frac{\omega_0}{2}$ after $n$ rotations. How many more rotations will it make before coming to rest?

$A$ wheel of radius $0.2 \ m$ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $10 \ N$ as shown in the figure. The established torque produces an angular acceleration of $2 \ rad/s^2$. The moment of inertia of the wheel is . . . . . . $kg \ m^2$. (Acceleration due to gravity $= 10 \ m/s^2$)