$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

$A$ horizontal platform is rotating with a uniform angular velocity about a vertical axis passing through its center. At some instant, a viscous liquid of mass $m$ is dropped at its center, which is free to spread and eventually falls off the edge. During this time interval, the angular velocity

$A$ wagon of $200\, kg$ is moving on a smooth track with a velocity of $2\, m/s$. $A$ man of $80\, kg$ runs in the wagon with a velocity such that the speed of the centre of mass of the system is zero. Find the relative velocity of the man with respect to the wagon in $m/s$.

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 the rings are rotating about the same axis,then the loss in kinetic energy is:

$A$ particle starts from the point $(0\,m, 8\,m)$ and moves with a uniform velocity of $3\, \hat{i} \,m/s$. After $5\,s$,the angular velocity of the particle about the origin will be:

$STATEMENT-1$ If there is no external torque on a body about its center of mass,then the velocity of the center of mass remains constant. because
$STATEMENT-2$ The linear momentum of an isolated system remains constant.