$A$ solid cylinder of mass $2 \ kg$ and radius $0.2 \ m$ is rotating with an angular velocity of $3 \ rad/s$. $A$ particle of mass $0.5 \ kg$ moving with a velocity of $5 \ m/s$ strikes its periphery and sticks to it. The loss in kinetic energy due to the collision is ....... $J$.

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is $I$. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both rings are rotating about the same axis, then the loss in kinetic energy is

$A$ solid metallic sphere of radius '$R$' having moment of inertia '$I$' about its diameter is melted and recast into a solid disc of radius '$r$' of uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is

An object has a moment of inertia of $3 \ kg \cdot m^2$. It is rotating with an angular velocity of $2 \ rad/s$. If a mass of $12 \ kg$ is moving with a velocity of $v \ m/s$,at what value of $v$ will their kinetic energies be equal?

$A$ wheel of radius $0.5 \ m$ and a moment of inertia of $10 \ kg \cdot m^2$ is rotating freely at an angular speed of $70 \ rev/min$. The wheel can be stopped in $5.0 \ s$ by pressing a wet cloth against the rim and exerting a radially inward force of $88 \ N$. The coefficient of kinetic friction between the wheel and wet cloth is:

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 speeds $\omega_{1}$ and $\omega_{2}$ are brought into contact face to face with their axes of rotation coincident. $(a)$ What is the angular speed of the two-disc system? $(b)$ Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take $\omega_{1} \neq \omega_{2}$.