Angular Momentum

Do you know what is common between a spinning top, a planet orbiting the sun, or an electron swirling around the nucleus of an atom? It is angular momentum, a fundamental concept of science explaining how objects behave and what powers them in a circular motion. It is an important concept of not only macroscopic physics but also microscopic physics for objects in circular motion. 

1.0Angular Momentum Definition 

Angular momentum is defined as the measure of the amount of rotation of an object in a certain circular path, considering its speed of rotation and distribution of its mass with respect to the axis of rotation. In other words, angular momentum is the tendency of an object to remain in a circular motion. It is a vector quantity, meaning it contains both the magnitude and direction.

Angular Momentum Symbol: L 

Unit: The SI unit of angular momentum is kgm²/s

2.0Angular Momentum Formula

Angular momentum is a key quantity in both classical and quantum mechanics and can be calculated by using various formulas, which differ based on certain conditions. Which includes:

General Formula for Angular Momentum (Classical Mechanics): 

Generally, the angular momentum (L) can be calculated by doing the cross-product of the position of the object (r) and its linear momentum (p): 

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

Angular Momentum for a Point Mass: 

This formula of angular momentum is used when the object is rotating around a central point or axis, such as a satellite orbiting a planet. The formula can be expressed as:

$L=m.r.v$

Here:

• m is the mass of the object,
• r is the distance from the centre of rotation (axis of rotation),
• v is the tangential velocity of the object.

Angular Momentum for a Rotating Rigid Body: 

When a rigid object, such as a disc, wheel, or planet, is rotating about an axis, the formula for angular momentum becomes:

$L=I\omega$

Here,

• I = moment of inertia of the object, which depends on the distribution of mass which is relative to the axis of rotation.
• ω = angular velocity, the rate at which the object is rotating (in radians per second).

3.0The Direction of Angular Momentum

Since angular momentum is a vector quantity, it has both magnitude and direction. Understanding the direction of angular momentum helps in describing the orientation of the rotational motion of an object. The direction of angular momentum can be determined with the help of the right-hand rule, a simple yet powerful rule for visualising rotational quantities. To use this rule: 

• Curl the fingers of the right hand in the direction of rotational motion 
• The direction of your thumb will symbolise the angular momentum. 

4.0Angular Momentum of Electron

In contrast to classical mechanics, the angular momentum of electrons is determined by the laws of quantum mechanics, which take into account both orbital and intrinsic spin angular momentum. This is an essential concept in understanding energy levels, atom structure, and the relationship between matter and light. 

Angular Momentum Quantum Number

Quantum numbers are used to describe the state of a particle or system in quantum mechanics. It is a key element in determining the angular momentum of an atom, defining the shape of an electron’s orbit, energy levels, and behaviour. 

• The angular momentum quantum number can be denoted with l and can range between 0 and n – 1, where n is the principal quantum number. Each principal quantum number defines the energy level. 
• Every value of l is connected with a specific type of orbital for example, if I = 0, then it corresponds to “s” orbital. 

Orbital Angular Momentum

The angular momentum of an electron, caused by its movement around the nucleus, is known as orbital angular momentum. According to quantum physics, an electron in an atom exhibits particle-like behaviour within a particular orbital, and this movement is linked to angular momentum. Based on the angular momentum quantum number (l), the orbital angular momentum formula is as follows:

$L=\sqrt{l(l+1)\bar{h}}$

• Here, l represents the orbital angular momentum quantum number,
• and ℏ is the reduced Planck's constant.

Spin and Angular Momentum

In addition to orbital angular momentum, electrons also consist of spin angular momentum, which is an intrinsic property of atoms that is independent of the electron's orbital motion. The formula to find the spin angular momentum is expressed as: 

$S=\sqrt{s(s+1)\bar{h}}$

Here, s is the spin quantum number, and for an electron, s = ½. 

The total angular momentum of any electron can be given by doing the vector sum of both the spin angular momentum and orbital angular momentum of an electron. Which can be expressed as: 

${\bar{L}}_{total}=\bar{L}+\bar{S}$

5.0Examples of Angular Momentum

• Satellites in Orbit: As satellites rotate around the gravitational centre of the Earth, satellites in orbit around the planet have angular momentum.
• Spinning Top: Since it rotates around its axis, a spinning top has angular momentum.
• Revolving Disk: The amount of angular momentum displayed by a revolving disk is determined by its mass distribution and rotational speed.
• Electron in Orbit: An atom's electron's orbital motion around the nucleus gives it angular momentum.
• Gyroscope: owing to the angular momentum created by its rapid spinning motion, a gyroscope resists orientation changes.
• Planetary Motion: The elliptical orbits of the solar system's planets provide angular momentum.

6.0Solved Examples on Angular Momentum

Problem 1: An electron in an atom has an angular momentum quantum number l=2. Calculate the magnitude of the electron's orbital angular momentum.

Solution: Given that l = 2, $h=1.055\times 10^{-34}J$

Using the formula: 

$L=\sqrt{l(l+1)\bar{h}}$

$L=\sqrt{2(2+1)}1.055\times 10^{-34}$

$L=\sqrt{6}\times 1.055\times 10^{-34}$

$L\approx 2.449\times 1.055\times 10^{-34}$

$L\approx 2.58\times 10^{-34}$

Problem 2: A figure skater initially spins with an angular velocity of 2 rad/s. She pulls her arms inward, and her moment of inertia decreases by half. What will be her new angular velocity?

Solution: It is given that 

$I_2=\frac{1}{2}{I}_1$

As per the law of conservation of angular momentum: 

$I_1{\omega }_1=I_2{\omega }_2$

$I_1{\omega }_1=\frac{1}{2}{I}_1{\omega }_2$

${\omega }_1=\frac{1}{2}{\omega }_2$

${\omega }_2=2{\omega }_1=2(2)=4rad/s$

Problem 3: A planet of mass 6.0 × 10²² kg is orbiting the sun at a distance of 1.5 × 10¹¹ m with an orbital velocity of 3.0 × 10³ m/s. Calculate the angular momentum of the planet.

Solution: Given that 

m = 6.0 × 10²² kg

v = 3.0 × 10³ m/s

r = 1.5 × 10¹¹ m

Using the formula, 

$L=m.r.v$

$L=(6.0\times 10^{22})(1.5\times 10^{11})(3.0\times 10^3)$

$L=2.7\times 10^{37}kg.m^2/s$