• Consider a body, rotating with angular velocity \(\omega\) about an axis, as shown in figure.
• The velocity of each particle on body is different.

Angular Momentum of \(i^{th}\) Particle 

Angular momentum of \(i^{th}\) particle is given by:

\(L_i=m_iv_ir_i\)

where, \(r_i\) is the perpendicular distance of \(i^{th}\) particle from the axis and \(v_i\) is the velocity of \(i^{th}\) particle.

Angular momentum of the body 

If axis of rotation is same for all particles:

\(L_{body}=\Sigma L_i\)

\(=\Sigma m_iv_ir_i\)

\(=\Sigma m_i(r_i\omega)r_i\)

\(L_{body}=(\Sigma m_ir_i^2)\omega\)

\(L_{axis}=I_{axis}\,\omega\)

• If net torque acting on a body is zero, the angular momentum remains conserved.

\(\vec{\tau}_{net}=0\)

\(\dfrac{dL}{dt}=0\)

• If the moment of inertia of a body is changed without changing the axis of rotation i.e., mass distribution about the axis is changed,

then, \(I_1\omega_1=I_2\omega_2\)

Kinetic Energy of a Rotating Body

• Kinetic energy of a rotating body is given by

\(K=\dfrac{1}{2}I\omega^2\)

\(\Rightarrow K=\dfrac{1}{2}(I\omega)\omega\)  or  \(\dfrac{1}{2} \dfrac{(I^2\omega^2)}{I} \)

\(\Rightarrow K=\dfrac{1}{2}L\omega\)  or  \(\dfrac{1}{2} \dfrac{L^2}{I} \)

• When angular momentum (\(L\)) is conserved, then moment of inertia of a body increases as angular velocity decreases and hence its kinetic energy decreases.

• Angular momentum of individual disks change due to friction which is an internal force for the system of two disks.

\(\vec{\tau}_{sys}=0\)

Translational motion

• If the velocity of all the particles of a body in same, the body is said to be in translational motion.
• The path of the body does not need to be linear.

Rotational motion

• If all the particles of a body move in circular motion such that their centers lie on a straight line, it is called rotational motion.

Axis of rotation

• The straight line perpendicular to the circular path passing through all the centers is called the axis of rotation.

• If a body is in translational motion i.e., all the particles have same velocity \(\overrightarrow{v}\), then we can replace it with a point at its center of mass moving with \(\overrightarrow{v}\), as shown in figure.

• If a body is rotating about a fixed axis, its moment of inertia about the fixed axis \(I{axis}\) is always involved while writing the formulae.

• When torque along any axis is zero, then angular momentum about that axis remains constant.
• If  \(\tau_{axis}=0\)

\(\Rightarrow \dfrac{d\;L_{axis}}{dt}=0\)

\(\Rightarrow L_{axis}\) = constant

• Thus, \(L_{axis}\) is conserved.
• \(I_\omega\) is constant.
• If mass distribution of a rotating body with respect to axis of rotation changes during rotation, then its moment of inertia changes and therefore, its angular velocity changes.
• If moment of inertia increases, angular velocity decreases and if moment of inertia decreases, angular velocity increases.

• Consider two points on a body.

here,

\(\vec{F_{12}}\) is a force applied on particle 1 due to particle 2 

\(\vec{F_{21}}\) is a force applied on particle 2 due to particle 1

• The forces \(\vec{F_{12}}\) and \(\vec{F_{21}}\) will act along the line joining particles 1 and 2.
• By Newton's III^rd law -

if \(\vec{F_{12}}=\vec{F}\)

then \(\vec{F_{21}}=-\vec{F}\)

Torque on the Body due to these Two Internal Forces

Torque, \(\tau_0=\vec{r_1}×\vec{F_{12}}+\vec{r_2}×\vec{F_{21}}\)

\(=\vec{r_1}×\vec{F}+\vec{r_2}×(-\vec{F})\)

\(=\vec{r_1}×\vec{F}-\vec{r_2}×\vec{F}\)

\(=(\vec{r_1}-\vec{r_2})×\vec{F}\)

\(\because \) The vector \((\vec{r_1}-\vec{r_2})\) is along \(\vec{F}\)

\(\therefore\tau_0=0\)

Torque due to all Pairs of Internal Forces 

Torque \(\vec{\tau}\) due to all pairs of internal forces about any point is zero.

 \(\therefore\vec{\tau}_{int}=0\)

• If external torque is zero,

\(\vec{\tau}=\dfrac{d\vec{L}}{dt}\)

\(\Rightarrow\sum \vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}\)

If \(\sum \vec {\tau}_{ext}=0\)

then \(L_{sys}\) is constant.

Thus, if net external torque on a system is zero, its angular momentum remains conserved.

External torque \((\vec{\tau}_{ext})\)

\(\Sigma\vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}\)

• If  \(\vec{\tau}_{ext}=0\)

then \(\vec{L}_{sys}\) is a constant.

• If net external torque on a system is zero, then angular momentum of the system remains conserved.

Angular Momentum during Collision 

• During collision or impulsive collision events, the impulsive force and its torque become internal to the system and do not affect the momentum and angular momentum of the system.
• Thus, angular momentum remains conserved.

As \(\overrightarrow{\tau}_{ext}=0\)

\(L_{sys}\) remains conserved.

• When the event is impulsive

When the event is impulsive, then the forces become internal and do not produce any torque. Thus, angular momentum of a body does not change and remains conserved.

• Mechanical energy is conserved 

Mechanical energy of a system is conserved if there are no external forces and the work done by internal non-conservative forces, is zero.

• Mechanical energy is not conserved 

Mechanical energy of a system is not conserved during impulsive situations like collision, jumping and landing.

Conservation of Mechanical Energy can be Applied 

The conservation of mechanical energy can be applied after the collision/event has taken place and dissipative and non-conservative forces are absent.