$A$ solid sphere of mass $2 \,kg$ and radius $1 \,m$ is free to rotate about an axis passing through its centre. $A$ constant tangential force $F$ is required to rotate the sphere with $10 \,rad/s$ in $2 \,s$ starting from rest. The value of $F$ is . . . . . . (in $\,N$)

Find the ratio of linear acceleration $(a)$ to angular acceleration $(\alpha)$ of the center of mass $(COM)$ for the given diagram. Given: $m = 2 \, kg$,$r = 10 \, cm = 0.1 \, m$,and force $F = 20 \, N$ applied at the center.

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ stationary horizontal disc is free to rotate about its axis. When a torque is applied on it,its kinetic energy as a function of $\theta,$ where $\theta$ is the angle by which it has rotated,is given as $K\theta^2.$ If its moment of inertia is $I,$ then the angular acceleration of the disc is

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

The moment of inertia of a wheel about an axis passing through its center is $200 \, kg \cdot m^2$. $A$ constant torque of $1000 \, N \cdot m$ is applied to rotate the wheel. After $3 \, s$,the angular velocity of the wheel will be ........ $rad/s$.