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    Angular Momentum - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • Angular Momentum is considered one of the most asked concept.

    • 37 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:

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    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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    Concepts Covered - 1

    Angular Momentum
    • 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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    Angular Momentum

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