Angular Momentum MCQ - Practice Questions with Answers

• 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 then angular momentum is given by the vector product of linear momentum and position vector. 

$\begin{array}{rl}\overrightarrow{L}& =\overrightarrow{r}\times \overrightarrow{P}\\ \overrightarrow{L}& =\overrightarrow{r}\times \overrightarrow{P}=\overrightarrow{r}\times (m\overrightarrow{V})=m(\overrightarrow{r}\times \overrightarrow{V})\\ |\overrightarrow{L}|& =rp\sin \theta ,\text{ where }\theta \text{ is the angle between }\mathrm{r}\text{ and }\mathrm{p}.\\ |\overrightarrow{L}|& =mvr\sin \theta \end{array}$

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

 Its direction will be  perpendicular to the plane of rotation and along the axis of rotation

• - $L_{max}=r\ast P($ when $\theta ={90}^{\circ })$
• - $L_{\text{min }}=0($ when $\theta =0^0)$
• - SI Unit- Joule-sec or $\mathrm{kg}-{\mathrm{m}}^2/\mathrm{s}$
• - Dimension- $M{L}^2{T}^{-1}$
• - In case of circular motion

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

$$L=mvr=m{r}^2\omega =I\omega$$

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

$${\overrightarrow{L}}_{net}={\overrightarrow{L}}_1+{\overrightarrow{L}}_{2\dots \dots }+{\overrightarrow{L}}_n$$