$A$ 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.

$A$ mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is $L$. If the string is now halved,keeping the angular velocity the same,then the new angular momentum will be:

$A$ cylinder of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. $A$ particle of mass $m$ moving with velocity $v$ hits the cylinder and sticks to its rim. Calculate the angular velocity of the cylinder after the impact.

The time dependence of the position of a particle of mass $m = 2 \ kg$ is given by $\vec r(t) = 2t \hat i - 3t^2 \hat j$. Its angular momentum with respect to the origin at time $t = 2 \ s$ is:

$A$ particle of mass $m$ is moving along a straight line parallel to the $x$-axis with a constant velocity $v$ at a distance $b$ from the origin. Find the angular momentum about the origin in vector form.

The position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$, where $b$ is a constant. At an instant, $\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be: