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Angular Momentum - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Angular Momentum is considered one of the most asked concept.
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39 Questions around this concept.
Solve by difficulty
A particle of mass $m$ moves along line PC with velocity $\nu$ as shown. What is the angular momentum of the particle about P?
A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s-1, the magnitude (in kg m2s-1) of its angular momentum about a point on the ground right under the centre of the circle is :
Angular momentum of a single particle moving with constant speed along circular path :
The position vector of 1 kg object is $\vec{r}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}) \mathrm{m}$ and its velocity $\overrightarrow{\mathrm{v}}=(3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \mathrm{ms}^{-1}$. The magnitude of its angular momentum is $\sqrt{\mathrm{x}} \mathrm{Nm}$ where X is:
A small particle of mass $m$ is projected at an angle $\Theta$ with the x-axis with an initial velocity $v_0$ in the $x$-y plane as shown in the figure. At a time
$
t<\frac{v_0 \sin \Theta}{g}
$
the angular momentum of the particle is
where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $\mathrm{x}, \mathrm{y}$ and z -axis respectively.
When a dancer folds her arms she spins faster on ice. This is due to
A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of this sphere:
The term moment of momentum is called
If $\overrightarrow{\mathrm{L}}$ and $\overrightarrow{\mathrm{P}}$ represent the angular momentum and linear momentum respectively of a particle of mass ' m ' having position vector $\overrightarrow{\mathrm{r}}=\mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t}+\hat{\mathrm{j}} \sin \omega \mathrm{t})$. The direction of force is
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Angular Momentum
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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
$
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