$A$ ring of radius $R$ rolls without sliding with a constant velocity $u$. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be

$A$ uniform rod of mass $8m$ and length $6a$ is placed on a horizontal table. Two point masses $m$ and $2m$ are moving with speeds $2v$ and $v$ respectively,as shown in the figure. They strike the rod and stick to it after the collision. Calculate the speed of the centre of mass of the system after the collision.

$A$ disc of radius $R$ is rolling purely on a flat horizontal surface with a constant angular velocity $\omega$. The angle between the velocity and acceleration vectors of point $P$ (which is at the same horizontal level as the center $C$) is

$A$ pendulum consists of a bob of mass $m=0.1 \ kg$ and a massless inextensible string of length $L=1.0 \ m$. It is suspended from a fixed point at height $H=0.9 \ m$ above a frictionless horizontal floor. Initially,the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. $A$ horizontal impulse $P=0.2 \ kg \cdot m/s$ is imparted to the bob at some instant. After the bob slides for some distance,the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \ kg \cdot m^2/s$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. $(1)$ The value of $J$ is. . . . . . $(2)$ The value of $K$ is. . . . . Give the answers of the questions $(1)$ and $(2)$.

$A$ wheel starting from rest is uniformly accelerated at $2 \, rad/s^2$ for $20 \, s$. It is allowed to rotate uniformly for the next $10 \, s$ and is finally brought to rest in the next $20 \, s$. The total angle rotated by the wheel (in radians) is ............

Statement-$1$: $A$ body is rotating about an axis with angular velocity $\omega$ and moment of inertia $I$. Its angular momentum $L$ remains constant,but its rotational kinetic energy $K$ decreases,provided no external torque is applied.
Statement-$2$: $L = I\omega$ and $K = \frac{L^2}{2I} = \frac{1}{2} I\omega^2$.