$A$ disc of mass $1 \, kg$ and radius $0.1 \, m$ is rotating with angular velocity $20 \, rad/s$. What is the angular velocity (in $rad/s$) if a mass of $0.5 \, kg$ is placed on the circumference of the disc?

$A$ horizontal disc of moment of inertia $4.25 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning counter-clockwise at $15 \,rps$ about its axis,as viewed from above. $A$ second disc of moment of inertia $1.80 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning clockwise at $25 \,rps$ as viewed from above about the same axis and is dropped on top of the first disc. The two discs stick together and rotate as one about their axis of symmetry. The new angular velocity of the system as viewed from above is close to

$A$ person standing on a rotating platform has his hands lowered. He suddenly outstretch his arms. The angular momentum

Consider a uniform rod of mass $M=4m$ and length $\ell$ pivoted about its centre. $A$ mass $m$ moving with velocity $v$ making an angle $\theta=\frac{\pi}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

$Assertion$ : For a system of particles under a central force field,the total angular momentum is conserved.
$Reason$ : The torque acting on such a system is zero.

$A$ person is rotating on a rotating table with an angular speed of $\omega$. He holds two identical masses in his outstretched hands. Without moving his hands,he simply drops the masses. How will his angular velocity change?