$A$ particle of mass $m$ moves with a velocity $v$ along the rim of a disc of radius $R$ in the counter-clockwise direction. The disc has a moment of inertia $I$ and is rotating in the clockwise direction with an angular velocity $\omega$. If the particle stops moving,what will be the angular velocity of the disc?

$A$ thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega '$ equal to:

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre),and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident. What is the loss in kinetic energy of the system in the process?

Keeping the mass of the Earth constant, if its radius is halved, then what will be the duration of the day (in $\text{ hours}$)?

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \; kg \cdot m^{2}$ and $10 \; rad \cdot s^{-1}$ respectively,while those for the second one are $0.2 \; kg \cdot m^{2}$ and $5 \; rad \cdot s^{-1}$ respectively. At some instant,they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is ........... $J$.

$A$ horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally fall off. The angular velocity during this period