Angular Momentum
The angular momentum of a particle about a given axis of rotation is defined as the product of the linear momentum and the perpendicular distance of its line of action from the axis of rotation. It is represented by $L$.
Angular Momentum $=$ Linear Momentum $×$ Perpendicular distance from the axis of rotation
Let a mass $m$ be rotating with velocity $v$ and is distance $r$ apart from the axis of rotation. \[L=mvr\] \[∴L=mωr²\;\;\;[\because v = \omega r]\]
Consider a rigid body consisting of $n$ particles of masses $m_1, m_2, m_3, …, m_n$ situated at distances $r_1, r_2, r_3, …, r_n$ respectively from the axis of rotation. Let the body be moving with uniform angular velocity $ω$ about the axis. Then, the total sum of the angular momentum of the particles give the angular momentum of the rigid body. \[L=L_1+L_2+L_3+…+L_n\] \[L=m_1r_1^2ω+m_2r_2^2ω+m_3r_3^2ω+…+m_nr_n^2ω\] \[L=(m_1r_1^2+m_2r_2^2+m_3r_3^2+…+m_nr_n^2)ω\]
\[L=\left(\sum_{i=1}^n m_ir_i^2\right)ω\] \[∴L=Iω\] \[\text{Where, }\sum_{i=1}^n m_ir_i^2=I \text{ (I = Moment of Inertia)}\]
Its SI unit is $\text{kg m}^2\text{s}^{-1}$. \[\text{When }ω=1, I=L\]
Therefore, moment of inertia of a rigid body is numerically equal to the angular momentum of the rigid body when rotating with unit angular velocity about that axis.
Relation Between Torque And Angular Momentum
The angular momentum of a body about an axis is given by, \[L=Iω\]
Differentiating with respect to time $t$, \[\frac{dL}{dt}=\frac{d}{dt}(Iω)\]
If moment of inertia does not change, then, \[\frac{dL}{dt}=I\frac{dω}{dt}\] \[\text{Here, }\frac{dω}{dt}=α\text{ (Angular Acceleration of the body)}\] \[∴\frac{dL}{dt}=Iα\] \[\text{But }Iα=τ\text{ (Torque acting on the body)}\] \[∴\frac{dL}{dt}=τ\]
Hence, torque may be defined as the time rate of change of angular momentum.
Law of Conservation of Angular Momentum
We know, torque is equal to the time rate of change of angular momentum. \[τ=\frac{dL}{dt}\]
If no external torque acts, \[\frac{dL}{dt}=0\] \[∴L=\text{constant}\] \[∴Iω=\text{constant}\] It is the law of conservation of linear momentum.
It states that if no external torque acts on a system, the total angular momentum of the system remains conserved.
In a body if moment of inertia changes from $I_1$ to $I_2$. Then, the angular velocity of the body must change from $ω_1$ to $ω_2$ so that \[I_1ω_1=I_2ω_2\]
If no external torque acts, then if moment of inertia decreases, then angular velocity increases and vice versa.
Some examples;
- The angular velocity of a planet revolving in an elliptical orbit around the sun increases when it comes near the sun and vice versa. It is because when the planet comes near the sun, moment of inertia decreases due to decreases in distance between them, which results in increase in angular velocity.
- A ballet dancer increases her angular velocity by bringing her hands and legs close to her body. When she wants to decrease her angular velocity, she stretches her hand and her leg outwards.
Rotational Kinetic Energy
Consider a rigid body consisting of $n$ particles of masses $m_1, m_2, m_3, …, m_n$ situated at distances $r_1, r_2, r_3, …, r_n$ respectively from the axis of rotation. Let the body be rotating with uniform angular velocity $ω$ about the axis. The angular velocity $ω$ of all the particles will be same but their linear velocities will be different. Then, \[K.E._{\text{rot}}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2+…+\frac{1}{2}m_nv_n^2\] \[K.E._{\text{rot}}=\frac{1}{2}m_1r_1^2ω_1^2+\frac{1}{2}m_2r_2^2ω_2^2+\frac{1}{2}m_3r_3^2ω_3^2+…+\frac{1}{2}m_nr_n^2ω_n^2\] \[K.E._{\text{rot}}=\frac{1}{2}\left(\sum_{i=1}^nm_ir_i^2\right)ω^2\] \[Here,\sum_{i=1}^nm_ir_i^2=I\text{ (Moment of Inertia of the body)}\] \[∴K.E._{\text{rot}}=\frac{1}{2}Iω^2\]
Kinetic Energy of a Rolling Body
When a body rolls, it has two types of kinetic energy; translation kinetic energy and rotational kinetic energy.
Consider a body of mass $m$ and radius $r$ rolling over a horizontal surface. Let $I$ be the moment of inertia, $v$ be the linear velocity and $ω$ be the angular velocity of the body.
Then, Kinetic energy of the rolling body $=$ Rotational K.E. + Translation K.E. \[∴K.E._{\text{rolling}}=\frac{1}{2}Iω^2+\frac{1}{2}mv^2\]