Angular Momentum - Practice Questions & MCQ

• The moment of linear momentum of a body with respect to any axis of rotation is known as angular momentum. If  P is the linear momentum of a particle and its position vector from the point of rotation is r then angular momentum is given by the vector product of  linear momentum and position vector. 

$
\begin{aligned}
\vec{L} & =\vec{r} \times \vec{P} \\
\vec{L} & =\vec{r} \times \vec{P}=\vec{r} \times(m \vec{V})=m(\vec{r} \times \vec{V}) \\
|\vec{L}| & =r p \sin \theta, \text { where } \theta \text { is the angle between } \mathrm{r} \text { and } \mathrm{p} . \\
|\vec{L}| & =m v r \sin \theta
\end{aligned}
$

- Its direction is always perpendicular to the plane containing vector r and P and with the help of right hand screw rule we can find it.

Its direction will be perpendicular to the plane of rotation and along the axis of rotation
- $L_{\text {max }}=r * P\left(\right.$ when $\left.\theta=90^{\circ}\right)$
- $L_{\text {min }}=0\left(\right.$ when $\left.\theta=0^0\right)$
- SI Unit- Joule-sec or $\mathrm{kg}-\mathrm{m}^2 / \mathrm{s}$
- Dimension- $M L^2 T^{-1}$
- In case of circular motion

As $\vec{r} \perp \vec{v}$ and $v=\omega r$ and $I=m r^2$

$
L=m v r=m r^2 \omega=I \omega
$

So in vector form $\vec{L}=I \vec{\omega}$
- The net angular momentum of a system consists of $n$ particles is equal to the vector sum of angular momentum of each particle.

$
\vec{L}_{n e t}=\vec{L}_1+\vec{L}_2 \ldots \ldots+\vec{L}_n
$