Conservation of Angular Momentum

Rotational Dynamics · Maharashtra HSC

Theory

Statement

The angular momentum of an isolated system is conserved in the absence of an external unbalanced torque.

This is the rotational analogue of the conservation of linear momentum, which says that the linear momentum of an isolated system is conserved in the absence of an external unbalanced force.

Proof

The angular momentum, or moment of linear momentum, of a system is given by

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

where $\vec r$ is the position vector from the axis of rotation and $\vec p$ is the linear momentum.

Differentiating with respect to time,

$\dfrac{d\vec L}{dt} = \dfrac{d}{dt}\left(\vec r\times\vec p\right) = \vec r\times\dfrac{d\vec p}{dt} + \dfrac{d\vec r}{dt}\times\vec p$

Now $\dfrac{d\vec r}{dt} = \vec v$ and $\dfrac{d\vec p}{dt} = \vec F$ , and $\vec p = m\vec v$ , so

$\dfrac{d\vec L}{dt} = \vec r\times\vec F + m\left(\vec v\times\vec v\right)$

But $\vec v\times\vec v = 0$ , hence

$\dfrac{d\vec L}{dt} = \vec r\times\vec F$

And $\vec r\times\vec F$ is the moment of force, or torque, $\vec\tau$ .

$\therefore\ \vec\tau = \dfrac{d\vec L}{dt}$

Thus if $\vec\tau = 0$ , then $\dfrac{d\vec L}{dt} = 0$ , i.e. $\vec L =$ constant.

Hence the angular momentum $\vec L$ is conserved in the absence of an external unbalanced torque $\vec\tau$ . This is the principle of conservation of angular momentum.

Working Form

For a rigid body or a system rotating about a fixed axis, $L = I\omega = 2\pi I n$ is constant, so

$I_1\omega_1 = I_2\omega_2, \qquad I_1n_1 = I_2n_2$

Thus if the moment of inertia $I$ is increased, the angular speed and hence the frequency of revolution $n$ decreases; if $I$ is decreased, $n$ increases.

Where a particle of mass $m$ is added at a distance $r$ from the axis (a blob of wax landing on a rotating disc), the new moment of inertia is $I_2 = I_1 + mr^2$ .

Examples

Ballet dance, an acrobat in a circus, ice skating and diving in a swimming pool all realise this principle.

Ballet dancers

During ice ballet, the dancers have to undertake rounds of smaller and larger radii.
The dancers come together while taking the rounds of smaller radius, near the centre.
In this case the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling.
While taking the outer rounds, the dancers outstretch their legs and arms.
This increases their moment of inertia, which reduces the angular speed and hence the linear speed. This is essential to prevent slipping.

Diving in a swimming pool

While on the diving board, the divers stretch their body so as to increase the moment of inertia.
Immediately after leaving the board they fold their body, which reduces the moment of inertia considerably. As a result the frequency increases and they can complete more rounds in the air, making the show attractive.
While entering the water they stretch their body into a streamlined shape, which allows a smooth entry.

Other standard examples

The angular velocity of revolution of a planet around the sun in an elliptical orbit increases when the planet comes closer to the sun, and decreases when it moves away — gravity is a central force and exerts no torque about the sun. The principle applies to both elliptical and circular orbits.
A person carrying heavy weights in the hands and standing on a rotating platform can change the speed of the platform by drawing the weights in or stretching them out.

Standard Results

$\vec\tau = \dfrac{d\vec L}{dt}$ ; if $\vec\tau = 0$ then $\vec L$ = constant
$I_1\omega_1 = I_2\omega_2$ , $\quad I_1n_1 = I_2n_2$
Adding a point mass: $I_2 = I_1 + mr^2$
For a sphere whose mass changes at constant density, $\dfrac{R_1}{R_2} = \left(\dfrac{m_1}{m_2}\right)^{1/3}$ and $I \propto mR^2$

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
State the law of conservation of angular momentum and explain with a suitable example.
Maharashtra HSC board question — asked in October 2014Answer 1 ↓
2
State and prove: law of conservation of angular momentum.
Maharashtra HSC board question — asked in October 2015Answer 2 ↓
3
State and prove principle of conservation of angular momentum.
Maharashtra HSC board question — asked in February 2023; March 2018Answer 3 ↓
4
Explain the principle of conservation of angular momentum with the help of two appropriate examples.
Maharashtra HSC board question — asked in July 2019Answer 4 ↓
5
State the law of conservation of angular momentum.
Maharashtra HSC board question — asked in July 2022Answer 5 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

6
A horizontal disc is freely rotating about a transverse axis passing through its centre at the rate of 100 revolutions per minute. A 20 gram blob of wax falls on the disc and sticks to the disc at a distance of 5 cm from its axis. Moment of inertia of the disc about its axis passing through its centre of mass is $2\times10^{-4}$ kg m $^2$ . Calculate the new frequency of rotation of the disc.
Maharashtra HSC board question — asked in March 2015Answer 6 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.
2.↑ back to the question
Statement: [From 1.10, Q1 (Statement only)]
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping. Proof: i. Angular momentum or the moment of linear momentum of a system is given by, $\vec L=\vec r\times\vec p$ where $\vec r$ is the position vector from the axis of rotation and $\vec p$ is the linear momentum. ii. Differentiating with respect to time, we get, $\dfrac{d\vec L}{dt}=\dfrac{d}{dt}(\vec r\times\vec p)=\vec r\times\dfrac{d\vec p}{dt}+\dfrac{d\vec r}{dt}\times\vec p$ iii. Now, $\dfrac{d\vec r}{dt}=\vec v$ and $\dfrac{d\vec p}{dt}=\vec F$ $\therefore \dfrac{d\vec L}{dt}=\vec r\times\vec F+m(\vec v\times\vec v)$ Now $(\vec v\times\vec v)=0$ $\therefore \dfrac{d\vec L}{dt}=\vec r\times\vec F$ iv. But $\vec r\times\vec F$ is the moment of force or torque $\vec\tau$ . $\therefore \vec\tau=\dfrac{d\vec L}{dt}$ Thus, if $\vec\tau=0$ , $\dfrac{d\vec L}{dt}=0$ or $\vec L=$ constant. Hence, angular momentum $\vec L$ is conserved in the absence of external unbalanced torque $\vec\tau$ .
3.↑ back to the question
Statement: [From 1.10, Q1]
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.Proof: [From 1.10, Q2] Statement: [From 1.10, Q1 (Statement only)]
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping. Proof: i. Angular momentum or the moment of linear momentum of a system is given by, $\vec L=\vec r\times\vec p$ where $\vec r$ is the position vector from the axis of rotation and $\vec p$ is the linear momentum. ii. Differentiating with respect to time, we get, $\dfrac{d\vec L}{dt}=\dfrac{d}{dt}(\vec r\times\vec p)=\vec r\times\dfrac{d\vec p}{dt}+\dfrac{d\vec r}{dt}\times\vec p$ iii. Now, $\dfrac{d\vec r}{dt}=\vec v$ and $\dfrac{d\vec p}{dt}=\vec F$ $\therefore \dfrac{d\vec L}{dt}=\vec r\times\vec F+m(\vec v\times\vec v)$ Now $(\vec v\times\vec v)=0$ $\therefore \dfrac{d\vec L}{dt}=\vec r\times\vec F$ iv. But $\vec r\times\vec F$ is the moment of force or torque $\vec\tau$ . $\therefore \vec\tau=\dfrac{d\vec L}{dt}$ Thus, if $\vec\tau=0$ , $\dfrac{d\vec L}{dt}=0$ or $\vec L=$ constant. Hence, angular momentum $\vec L$ is conserved in the absence of external unbalanced torque $\vec\tau$ .
4.↑ back to the question
Statement and explanation: [From 1.10, Q1]
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.and Q. No. 2 Examples: i. The angular velocity of revolution of a planet around the sun in an elliptical orbit increases, when the planet comes closer to the sun and vice-versa. ii. A person carrying heavy weights in his hands and standing on a rotating platform can change the speed of the platform. iii. A diver performs somersaults by jumping from a high diving board keeping his legs and arms out stretched first, and then curling his body. (Any two examples)
5.↑ back to the question
Statement: [From 1.10, Q1]
i. Statement: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque. ii. Illustration — Ballet dancers: a. During ice ballet, the dancers have to undertake rounds of smaller and larger radii. b. The dancers come together by taking rounds of smaller radius (near the centre). c. In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. d. While during outer rounds, the dancers outstretch their legs and arms. e. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.
6.↑ back to the question
Given: $n_1$ = 100 r.p.m., $I_1=2\times10^{-4}$ kg m $^2$ , $\omega_1=2\pi n_1=(2\pi\times100)$ rad/min, m = 20 g $=20\times10^{-3}$ kg, r = 5 cm $=5\times10^{-2}$ m To find: New frequency ( $n_2$ ) Formula: $I_1\omega_1=I_2\omega_2$ Calculation: Using parallel axes theorem, $I_2=I_1+mr^2=2\times10^{-4}+[20\times10^{-3}\times(5\times10^{-2})^2]$ $\therefore I_2=2.5\times10^{-4}$ kg/m $^2$ From formula, $2\times10^{-4}\times2\pi\times100=2.5\times10^{-4}\times2\pi\times n_2$ $\therefore n_2=80$ r.p.m. $=\dfrac{4}{3}$ r.p.s. Ans: The new frequency of rotation of the disc is $\dfrac{4}{3}$ r.p.s.