In orbital motion,the angular momentum vector is directed:

$A$ constant torque of $200 \, N-m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 \, kg-m^{2}$, then in $4 \, s$, what will be the change in its angular momentum?

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through its centre and perpendicular to its length is given by $\frac{ML^2}{12}$. Now,consider one such rod pivoted at its centre,free to rotate in a vertical plane. The rod is at rest in the vertical position. $A$ bullet of mass $M$ moving horizontally at a speed $v$ strikes and embeds itself in one end of the rod. The angular velocity of the rod just after the collision will be

Write the general formula of total angular momentum of rotational motion about a fixed axis.

Define angular momentum.

As shown in the figure,a particle of mass $m$ moves with a velocity $v$ along the line $PC$. What is the angular momentum of this particle with respect to point $P$?