$A$ rotating body has angular momentum $L$. If its frequency is doubled and its kinetic energy is halved,what will be its new angular momentum?

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

$A$ solid cylinder of mass $3 \ kg$ is rolling on a horizontal surface with velocity $4 \ m/s$. It collides with a horizontal spring whose one end is fixed to a rigid support. The force constant of the spring is $200 \ N/m$. The maximum compression produced in the spring will be (assume the collision between the cylinder and the spring is elastic). (in $m$)

$A$ uniform rod of length $L$ and mass $M$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicular to the rod. After the collision,the particle comes to rest. What is the angular velocity of the rod about its centre of mass after the collision?

$A$ uniform circular disc placed on a rough horizontal surface has initially a velocity $v_0$ and an angular velocity $\omega_0$ as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v_0}{r\omega_0}$ is

$A$ solid sphere of mass $M$,radius $R$ and having moment of inertia about an axis passing through the centre of mass as $I$,is recast into a disc of thickness $t$,whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. Then,the radius of the disc will be