$A$ bullet of mass $10\, g$ and speed $500\, m/s$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0\, m$ wide and weighs $12\, kg$. It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it will be

$A$ man is sitting in a smooth groove on a horizontal circular table at the edge by holding a rope joined to the centre. The moment of inertia of the table is $I$. The mass of the man is $M$. The man now pulls the rope so that he comes to the centre. The angular velocity of the table:

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

$A$ hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

Find the minimum height $h$ of the obstacle so that the sphere of radius $R$ can stay in equilibrium on an inclined plane of angle $\theta$.

One of two identical cylinders,cylinder $A$,rotates at an angular speed of $50 \text{ revolutions per second}$. This rotating cylinder is brought into contact with a second stationary cylinder,$B$. Due to kinetic friction between the two cylinders,the stationary cylinder starts rotating with an angular acceleration,while cylinder $A$ undergoes angular deceleration. If the magnitude of the angular acceleration for both cylinders is $1 \text{ revolution per second}^2$,after how many seconds $(t)$ will the angular speeds of both cylinders become equal?