Notes on Angular Momentum and Principle of Conservation of Angular Momentum | Grade 11 > Physics > Rotational Dynamics | KULLABS.COM

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#### Angular Momentum (L)

It is defined as, “ The cross product of perpendicular distance and linear momentum.” It is denoted by ‘L’ and is given as

Suppose an object of mass m revolving round a circle of radius r with speed v amount an axis passing through the centre O as shown in the figure.

\begin{align*} \vec L &= \vec r \times \vec P \\ \text {where} \: \vec p = \text {linear momentum} \\ \vec L &= rp\sin \theta \widehat n \\ \end{align*}

where , $$\widehat n$$ is unit vector along the direction of normal to the plane containing r and p.

Since, $$\: \vec r \: \text {and} \: \vec p \:$$ are perpendicular to each other. So, the angular momentum in terms of magnitude is

\begin{align*} \text {or,} \: L &= rp \\ \therefore L &= mvr \dots (i) \\ \text {Since} v = \omega r, where \omega \text {is the angular velocity of the object, the angular momentum can be written as}\\ L &= m(\omega r) r \\ \text {or,} L &= mr^2\omega \dots (ii) \\ \end{align*}

The equation (i) and (ii) are the expressions for the angular momentum of the body. It is a vector quantity.

##### Relation between Angular Momentum and Moment of Inertia

Consider a rigid body of mass M rotating about an axis YY’ as shown in the figure. Let particles of the body of masses m1, m2, m3 …..mn are situated at distances r1, r2, r3, …..rn respectively from an axis of rotation YY’. Suppose the body be rotated with uniform angular velocity $$\omega$$ about the axis. Although, each particles within the body has the same angular velocity $$\omega$$ but their linear velocity will be different. Let v1, v2, v3, ….. vn be the linear velocities of the particles of masses m1, m2, m3 …..mn respectively. Then,

\begin{align*} v_1 = \omega r_1, \: v_2 =\omega r_2, \dots , v_n =\omega r_n \\ \text {The magnitude of the angular momentum of particle of mass} m_1 \text {about the axis of rotation} \\ &= \text {linear momentum} \times r_1 \\ &= m_1v_1r_1 \\ &= m_1r_1^2 \omega \\ \end{align*}

Similarly, the magnitude of angular momentum of particles of masses m2, m3, …. are m2r22 ω, m3r32 ω, …. respectively. Now, the total angular moment, l of the body about the axis YY’ is given by the sum of the angular momentum of the constituting particles about that axis,

\begin{align*} L &= m_1r_1^2 \omega + m_2r_2^2 \omega + m_3r_3^2 \omega + \dots \\ &= (m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + \dots ) \omega \\ &= (\sum m_ir_i^2) \omega \\ \therefore L &= I\omega \end{align*}

where, $$I = \sum m_ir_i^2$$, moment of inertia of the body.

This equation gives the relation between angular momentum and moment of inertia of the body. Therefore, the magnitude of the angular momentum of a body about a given axis is equal to the product of the moment of inertia, I of the body and its angular velocity ω about that axis. The angular momentum in rotational motion is similar to the linear momentum of linear motion.

##### Relation between Angular Momentum and Torque

The angular momentum of a rigid body rotating about an axis with an angular velocity ω is

\begin{align*} L &= I\omega \dots (i) \\ \text {where I is the moment of inertia of the body.} \\ \text {Differentiating both sides of equation} (i) \text {with respect to time, we get} \\ \frac {dL}{dt} &= \frac {d}{dt} (I\omega ) \\ \text {Moment of inertia I remain constant only when the axis of rotation is stationary. In this case, we have} \\ \frac {dL}{dt} &= I\frac {d\omega }{dt} \\ \text {or,} \: \frac {dL}{dt} &= I\alpha \dots (ii) \\ \text {where} \alpha = \frac {d\omega } {dt} , \text {the angular acceleration of the body.} \\ \text {But, torque acting on the body is} \\ \tau&= I\alpha \dots (iii) \\ \text {From equations} (ii) \text {and} (iii), \text {we get} \\ \tau &= \frac {dL}{dt} \\ \end{align*}

Hence, the torque acting on a body is equal to the time rate of change of angular momentum of the body.

#### Principle of Conservation of Angular Momentum

It states that if no external torque acts on a system, the total angular momentum of the system remains constant.

If I be the moment of inertia of a body about a given axis of rotation and ω, its angular velocity, then

$$I\omega = constant$$

Proof:

We know that the torque acting on a system is equal to the time rate of change of angular momentum of the system about the axis, i.e.

\begin{align*} \tau = \frac {dL}{dt} \\ \text {If no external torque acts on the system,} \tau = 0 \\ \text {or,} \: \frac {dL}{dt} &= 0 \\ \text {Integrating this equation, we have} \\ L &= \text {constant} \\ \text {or,} I\omega &= \text {constant} \\ \text {which is the principle of conservation of angular momentum. In general,} \\ I_1\omega _1 &= I_2\omega _2 \\ \end{align*}

##### Examples of conservation of Angular Momentum
1. When planet is near the sun, its moment of inertia about the axis of rotation decreases and its angular ω increases, whereas at places where it is far from the sun, its moment of inertia increases and hence angular velocity decreases. In both cases the angular momentum of the planet remains constant in the motion of the planet revolving in elliptical orbit round the sun.
2. A diver uses the principle of conservation of angular momentum while diving into a swimming pool. The diver spins quicker by curling himself as moment of inertia decreases and angular velocity increases keeping total momentum constant. But before entering water surface, he stretches his arms and legs so that his moment of inertia increases and the angular velocity decreases appreciably and he is able to touch the water surface with a reduced speed without hurting him.
3. A ballet dancer uses this principle to increase or decrease her spinning rate during the performance. When she stretches her hands and leg outwards as shown in the figure. The moment of inertia increases and hence the angular velocity decreases such that angular momentum, I ω remain constant. When she wants to increases the spinning rate or angular velocity she brings her arms and legs closer.

1,   \begin{align*} \vec L &= \vec r \times \vec P \\ \text {where} \: \vec p = \text {linear momentum} \\ \vec L &= rp\sin \theta \widehat n \\ \end{align*}

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