Angular Momentum - Definition, Formula, Examples, FAQs

Momentum is the product of mass and velocity. It is a vector quantity. A closed system's momentum doesn't change unless an outside force is applied. Angular momentum is the characteristic that describes the rotatory inertia of an object in motion about an axis that may or may not pass through that particular object.

In this article, we will cover the concept of angular momentum. This topic falls under the broader category of rotational motion, a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), almost thirteen questions have been asked on this concept. And for NEET three questions were asked from this concept.

Let's read this entire article to gain an in-depth understanding of angular momentum.

What Is Angular Momentum

Angular momentum definition: The moment of linear momentum of a body concerning 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.

What Is The Formula Of Angular Momentum

The angular momentum formula is,

$\begin{aligned}
& \vec{L}=\vec{r} \times \vec{P}\\
& \vec{L}=\vec{r} \times \vec{P}=\vec{r} \times(m \vec{V})=m(\vec{r} \times \vec{V})
\end{aligned}$

$|\vec{L}|=r p \sin \theta$, where $\theta$ is the angle between $\mathrm{r}$ and $\mathrm{p}$.
$
|\vec{L}|=m v r \sin \theta
$

Its direction is always perpendicular to the plane containing vector 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.

$
\begin{aligned}
& L_{\max }=r * P\left(\text { when } \theta=90^{\circ}\right) \\
& L_{\min }=0\left(\text { when } \theta=0^0\right)
\end{aligned}
$

• SI Unit Of Angular Momentum

• SI Unit is Joule-sec or $\mathrm{kg}-\mathrm{m}^2 / \mathrm{s}$

• Dimension Of Angular Momentum

• Dimension $M L^2 T^{-1}$

Relation Between Angular Momentum And Moment Of Inertia

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 $\vec{L}=I \vec{\omega}$

This is the relation between angular momentum and moment of inertia.

The net angular momentum of a system consisting of n particles is equal to the vector sum of the angular momentum of each particle.

$\vec{L}_{n e t}=\vec{L}_1+\vec{L}_2 \ldots \ldots .+\vec{L}_n$

Conservation Of Angular Momentum

Conservation of angular momentum states that the system remains constant if no external torque acts on it.

$L_{\text {initial }}=L_{\text {final }}$

Relationship Between Torque And Angular Momentum

The equation connecting torque and angular momentum:

$\tau=\frac{d L}{d t}$

where,

• $\tau$ is the torque applied
• $\frac{d L}{d t}$ is the rate of change of angular momentum over time

When a torque is applied externally, there is a change in angular momentum. The direction of torque determines the direction in which the angular momentum changes.

Applications of Angular Momentum

1. Figure skating spins

Performers bring their arms close to their body to increase the angular momentum as there is a reduction in moment of inertia.

2. Bicycle Wheels

The spinning of wheels generates angular momentum which helps in balancing.

3. Earth's rotation

Rotating on its axis produces angular momentum which helps to spin steadily.

Other applications are:

• Orbits of planets and satellites
• Formation of neutron stars
• Divers and gymnasts
• Spacecraft rotation

Relation Between Linear Momentum And Angular Momentum

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Solved Example Based on Angular Momentum

Example 1: A particle of mass $m$ moves along line PC with velocity $\nu$ as shown. What is the angular momentum of the particle about $\mathrm{P}$?

1) $m \nu L$
2) $m \nu l$
3) $m \nu r$
4) zero

Solution:

Angular momentum

$
\vec{L}=\vec{r} \times \vec{p} \mid
$

The particle moves along the line PC.
$
\begin{aligned}
& \theta=0^{\circ} \\
& L=m v r \sin \theta^{\circ} \\
& \sin \theta^{\circ}=0 \\
& \mathrm{~L}=0
\end{aligned}
$

Hence, the answer is option (4).

Example 2: 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 center of the circle is :

1) 8.64

2) 11.52

3) 14.4

4) 20.16

Solution:

We know,

$\vec{L}=\vec{r} \times \vec{p}$

Angular momentum

$
\begin{aligned}
& L_0=m v r \sin \theta \\
& \theta=90^{\circ} \\
& L_0=m v r \sin 90^{\circ}
\end{aligned}
$

After calculation we get
$
L_0=14.4 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{sec}
$

Hence, the answer is option (3).

Example 3: A ball of mass 160 g is thrown up at an angle of 60⁰ to the horizontal at a speed of 10 ms^-1. The angular momentum (in kg m²/s) of the ball at the highest point of the trajectory concerning the point from which the .ball is thrown is nearly (g=10 ms^-2)

1) 1.73

2) 3

3) 3.46

4) 6

Solution:

$\begin{aligned} & (m v) L_{\perp} \\ & \text { Angular momentum about point } 0 \text { is }=\left(m u_x\right) H \\ & u_x=u / 2=5 \\ & H=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{100 \times\left(\frac{\sqrt{3}}{2}\right)}{2 \times 10}=30 / 8=3.75 \mathrm{~m} \\ & \text { Angular momentum }=0.16 \times 5 \times 3.75 \mathrm{Kgm}^2 / \mathrm{s}=3.000 \mathrm{Kgm}^2 / \mathrm{s} \\ & \end{aligned}$

Example 4: A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed of rad/s about the vertical. About the point of suspension :

1) angular momentum is conserved.

2) angular momentum changes in magnitude but not in direction.

3) angular momentum changes in direction but not in magnitude.

4) angular momentum changes both in direction and magnitude.

Solution:

Angular momentum
$
\vec{L}=\vec{r} \times \vec{p}
$
wherein
$\vec{L}$ represents the angular momentum of a moving particle about a point.
it can be calculated as $L=r_1 P=r P_1$
$r_1=$ Length of perpendicular on the line of motion
$P_1=$ component of momentum along perpendicular to $r$

Angular momentum = mvr is constant as the distance of the bob from the point of suspensions and its speed both are constant.

The direction of angular momentum changes continually as shown in the figure

Example 5: A particle is moving in a circular path of radius a, with a constant velocity as shown in the figure. The center of the circle is marked by ‘C’. The angular momentum from the origin O can be written as :

1) $v a(1+\cos 2 \theta)$
2) $v a(1+\cos \theta)$
3) va $\cos 2 \theta$
4) $v a$

Solution:

Angular momentum

$
\vec{L}=\vec{r} \times \vec{p}
$
wherein
$\vec{L}$ represents the angular momentum of a moving particle about a point.
it can be calculated as $L=r_1 P=r P_1$
$r_1=$ Length of perpendicular on the line of motion
$P_1=$ component of momentum along perpendicular to $r$

$\begin{aligned} & \cos \theta=\frac{a^2+L^2-a^2}{2 a L} \text { or } L=2 a \cos \theta \\ & \text { component of length } \perp^r \text { to velocity } \\ & =L \cos \theta \\ & =2 a \cos ^2 \theta \\ & \cos 2 \theta=2 \cos ^2 \theta-1 \\ & =2 \cos ^2 \theta=1+\cos 2 \theta \\ & L_{\perp}=a(1+\cos 2 \theta) \\ & \text { Angular momentum } L=m v L_{\perp}=m v a(1+\cos 2 \theta)\end{aligned}$