State the law of conservation of angular momentum - eSaral

Hey, do you want to know how to state the law of conservation of angular momentum? If so. Then you are at the right place.

Angular momentum

The moment of linear momentum of a moving particle with respect to the axis of rotation is known as angular momentum.

It is a vector quantity, which is often represented by $\overrightarrow{\mathrm{L}}$ or $\overrightarrow{\mathrm{J}}$.

It is an axial vector (i.e. always perpendicular to the plane of motion)

Angular momentum $\overrightarrow{\mathrm{J}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}$

$=\vec{r} \times(m \vec{v})$

$=m(\vec{r} \times \vec{v})$

Or $\overrightarrow{\mathrm{J}}=r p \sin \theta \hat{n}$

$\theta$ is angle between $\vec{r}$ and $\vec{v}$

$\hat{n}$ is unit vector perpendicular to plane of $\vec{r}$ and $\vec{v}$

The direction of angular momentum is perpendicular to the plane of $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{v}}$ and it is given by right-hand screw rule.

If $\theta=0^{\circ}$

J = 0 (Minimum)

If $\theta=90^{\circ}$

J = m v r (Maximum)

S.I. Unit of J is Jule × sec (same as that of Planck’s const.)

Dimension: $\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}$

If direction of rotation is anticlockwise, angular momentum is taken positive and if direction of rotation is clockwise, angular momentum is taken negative.

The angular momentum of a system of particles is equal to the vector sum of angular momentum of each particle

$\overrightarrow{\mathrm{J}}=\overrightarrow{\mathrm{J}}_{1}+\overrightarrow{\mathrm{J}}_{2}+\overrightarrow{\mathrm{J}}_{3}+\ldots \ldots$

$\vec{J}=\sum_{i=1}^{n} \vec{J}_{i}$

The relation between angular momentum and angular velocity

$J=I \omega$

I - Moment of inertia with respect to axis of rotation

$\omega$ - Angular velocity due to angular momentum

J - The moment of momentum which is causing rotational motion.

The rate of change of angular momentum is equal to the torque applied on the body.

$\vec{\tau}=\frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}$

In rotational motion angular momentum has equal importance as linear momentum in linear motion If torque acting of a particle is zero then

$\vec{\tau}=0 \Rightarrow \frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}=0$

Or

$\overrightarrow{\mathrm{J}}=$ constant

Which implies that the angular momentum remains conserved when no external torque acts on the body.

Law of Conservation of Angular Momentum

  If no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is,

$\overrightarrow{\mathrm{J}}=\overrightarrow{\mathrm{I} \omega}=$ constant

Proof : For a rigid body, rotating about a given axis.

$\vec{\tau}=\mathrm{I} \vec{\alpha}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}$

Now, for a rigid body, rotating about a given axis, I remains constant.

Because

$\vec{\tau}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}$

$=\frac{\mathrm{d}(\mathrm{I} \vec{\omega})}{\mathrm{dt}}=\frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}$

If the external torque acting on the body, $\vec{\tau}=0$,

$\frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}=0$

and hence $\vec{J}=$ constant.

For the equation $\mathrm{J}=\mathrm{I} \omega,$ it can be seen that if $\mathrm{J}$ remains constant, $\omega$ increases when I decreases and vice

versa. If I changed to $\mathrm{I}_{1}$, then, in the absence of an external torque, $\omega$ will change to $\omega_{1},$ such that

$\mathrm{I}_{\mathrm{i}} \omega_{\mathrm{i}}=\mathrm{I}_{\mathrm{f}} \omega_{\mathrm{f}}$

This is the law of conservation of angular momentum.

Example :

So, that's all from this article. If you get the idea about how to state the law of conservation of angular momentum then please share it with your friends.

For a better understanding of this chapter, please check the detailed notes of the Rotational Dynamics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.