$A$ thin uniform circular disc of mass $M$ and radius $R$ is rotating with angular velocity $\omega$ in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of the same radius but of mass $\frac{M}{3}$ is placed gently on the first disc co-axially. The new angular velocity will be:

$A$ horizontal disc of moment of inertia $4.25 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning counter-clockwise at $15 \,rps$ about its axis,as viewed from above. $A$ second disc of moment of inertia $1.80 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning clockwise at $25 \,rps$ as viewed from above about the same axis and is dropped on top of the first disc. The two discs stick together and rotate as one about their axis of symmetry. The new angular velocity of the system as viewed from above is close to

Angular momentum of a particle rotating under a central force is constant due to

$A$ uniform rod $AB$ of length $L$ and mass $M$ is pivoted at its center $O$ such that it can rotate freely in a vertical plane. The rod is initially in a horizontal position. $A$ particle $S$ of the same mass $M$ falls vertically with velocity $v$ and strikes the rod at point $C$. $C$ is the midpoint between $O$ and $B$. Find the angular velocity of the rod immediately after the particle strikes it.

If the torque applied on a particle is zero,then its angular momentum will be:

Two discs of moment of inertia $I_1$ and $I_2$ and angular speeds $\omega_1$ and $\omega_2$ are rotating along collinear axes passing through their centers of mass and perpendicular to their planes. If the two discs are made to rotate together along the same axis,what will be the rotational kinetic energy of the system?