Angular Momentum

Rotational Dynamics, Mechanics, Physics / August 16, 2022

Table Of Contents
  1. Angular Momentum Definition
  2. Conservation of Angular Momentum

Angular Momentum Definition

A body possessing linear velocity has linear momentum. In the same way, a rotational body has rotational momentum called angular momentum.

Angular momentum is the moment of linear momentum. It defines the quantity of motion of rigid rotational bodies. Angular momentum is defined as the vector product( or cross product) of distance from the axis of rotation with the linear momentum. It is denoted by \overrightarrow{L}

In Vector form, It is written as \overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{P}

    \[\overirghtarrow{L} = rP \sin\theta \hat{n}\]

where \hat{n} is a unit vector along the direction of \overrightarrow{L} and \theta is the angle between \overrightarrow{r} and \overrightarrow{P}

In magnitude: \boxed{L = rP\sin\theta}

Consider a rigid body rotating with uniform angular velocity ‘\omega‘ and one particle of mass m.

Angular momentum of that particle = rP

But P = mv and v = \omega r So, angular momentum becomes rP = rmv = rm\omega r = \omega mr^2

If there are ‘n’ particles in a rigid body, the total angular momentum can be calculated by adding the angular momentum of all particles. So, Total angular momentum can be written as:

    \[ L = \omega m_1r_1^2 + \omega m_2r_2^2 + \omega m_3r_3^2 + \ldots + m_nr_n^2 \]

    \[ L = \omega \left( m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + \ldots + m_nr_n^2\right) \]

    \[ L = \omega \sum_{i=1}^n m_ir_i^2 \]

    \[\boxed{L = I\omega} \]

Where I = \sum_{i=1}^n m_ir_i^2 is defined as moment of inertia of rigid body.

Differentiating L with respect to time, we get:

    \[ \frac{dL}{dt} = \frac{d}{dt}(I\omega) \]

    \[ \frac{dL}{dt} = I \frac{d}{dt}\omega \]

    \[\frac{dL}{dt} = I \alpha \]

    \[ \boxed{\frac{dL}{dt} = \tau}\]

Hence, the rate of change of angular momentum is called torque.

Conservation of Angular Momentum

Conservation of Angular Momentum states that “If no external torque is acted on the system, the total angular momentum always remains constant.”

i.e. If \tau = 0 then, L = constant

We have L = I \omega

Differentiating with respect to t we get:

    \[ \frac{dL}{dt} = \frac{d}{dt}(I\omega) \]

    \[ \frac{dL}{dt} = I \frac{d}{dt}\omega \]

    \[\frac{dL}{dt} = I \alpha \]

    \[ \frac{dL}{dt} = \tau\]

If \tau = 0 then,

    \[ \frac{dL}{dt} = 0\]

L = constant \Rightarrow I\omega = constant

i.e. \boxed{I_1\omega_1 = I_2 \omega_2 = \ldots}

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