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. 

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

• Its direction is always perpendicular to the plane containing 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

• - $L_{\max }=r * P\left(\right.$ 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

$\mathrm{As   } \vec{ r} \perp \vec{v}$ 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 consisting of $n$ particles is equal to the vector sum of the angular momentum of each particle.

$$
\vec{L}_{n e t}=\vec{L}_1+\vec{L}_{2 \ldots \ldots}+\vec{L}_n
$$