$A$ cylinder of mass $5 \ kg$ and radius $30 \ cm$ is free to rotate about its axis. It receives an initial angular impulse of $3 \ kg \ m^2 s^{-1}$ and subsequently receives the same impulse every $4 \ s$. What will be the angular speed of the cylinder $30 \ s$ after the initial impulse? The cylinder is initially at rest.

The angular momentum of a single particle moving with constant speed along a circular path:

$A$ uniform rod of mass $m$ and length $l$ is moving with velocity $u$ in a direction perpendicular to its length. $A$ blow of impulse $J$ is applied in the direction of $u$,perpendicular to its length at a distance $l/4$ from its center at point $P$,such that the instantaneous velocity of point $P$ is $2u$. Then the velocity of its center of mass will be:

$A$ torque of $50 \ Nm$ acts on a body for $8 \ s$,which is initially at rest. The change in its angular momentum is:

$A$ particle of mass $2 \ kg$ located at the position $(\hat{i} + \hat{j}) \ m$ has a velocity $2(\hat{i} - \hat{j} + \hat{k}) \ m/s$. Its angular momentum about the $z$-axis in $kg \cdot m^2/s$ is:

$A$ particle of mass $0.01 \ kg$ has a position vector $\vec{r} = (10\hat{i} + 6\hat{j}) \ m$ and moves with a velocity $\vec{v} = 5\hat{i} \ m/s$. Calculate its angular momentum about the origin in $\hat{k} \ J \cdot s$.