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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.
Consider a rigid body of mass M rotating about an axis YY’ as shown in the figure. Let particles of the body of masses m_{1}, m_{2}, m_{3} …..m_{n} are situated at distances r_{1}, r_{2}, r_{3}, …..r_{n} 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 v_{1}, v_{2}, v_{3}, ….. v_{n} be the linear velocities of the particles of masses m_{1}, m_{2}, m_{3} …..m_{n }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 m_{2}, m_{3}, …. are m_{2}r_{2}^{2} ω, m_{3}r_{3}^{2} 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.
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.
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*}
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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ASK ANY QUESTION ON Angular Momentum and Principle of Conservation of Angular Momentum
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