#### Angular momentum

The rotational equivalent of linear momentum is *angular momentum*. It is defined as the product of moment of inertia and angular velocity.

\(L=I\omega\)

- \(L\) is angular momentum (kgm
^{2}s^{-1}) - \(I\) is moment of inertia (kgm
^{2}) - \(\omega\) is angular velocity (rads
^{-1})

Angular momentum is a *conserved* quantity: total angular momentum remains constant.

Ice skaters make good use of this effect when spinning.

With limbs spread out, the moment of inertia is high as more mass is at a distance from the pivot. This results in a low angular velocity.

With the limbs closer to the body, the moment of inertia decreases. Since angular momentum is conserved, this increases the angular velocity:

\(I_1\omega_1=I_2\omega_2\)

Conservation of angular momentum also explains why the addition of a child to a spinning roundabout results in a reduction in angular velocity:

\(L_\textrm{initial}=L_\textrm{final}\)

\(I_\textrm{roundabout}\omega_\textrm{initial}=(I_\textrm{child} +I_\textrm{roundabout})\omega_\textrm{final}\)

#### Rotational kinetic energy

Rotational kinetic energy is calculated as follows:

\({E_\textrm{K}}_\textrm{rot}={1\over 2}I\omega ^2\)

- \({E_\textrm{K}}_\textrm{rot}\) is rotational kinetic energy (J)
- \(I\) is moment of inertia (kgm
^{2}) - \(\omega\) is angular velocity (rads
^{-1})

Energy is another conserved quantity, with energy changed equal to work done:

\(\Delta {E_\textrm{K}}_\textrm{rot}=\Delta W\)

Reconsidering our roundabout example, a child would need to do work to move from position A to position B:

- The moment of inertia decreases as the child moves from A to B due to the reduction in radius.
- As a result of conservation of angular momentum, the angular velocity increases by the same proportion.
- Since rotational kinetic energy is proportional to the square of the angular velocity, this effect dominates, resulting in an increase in the rotational kinetic energy.
- The child must do work equal to this change in energy.

#### Rolling

Although not articulated, in the Mechanics topic it was assumed that all moving bodies were sliding. In the Engineering topic, we assume that bodies roll instead.

Bodies are made to roll by the action of a torque:

\(\Gamma ={\Delta L \over t}\)

- \(\Gamma\) is the torque applied (Nm)
- \(\Delta L\) is change in angular momentum (kgm
^{2}s^{-1}) - \(t\) is the time taken (s)

This equation is another statement of Newton's second law: \(\Gamma=I\alpha = I{(\omega_f-\omega_i)\over t}={I\omega_f-I\omega_i\over t}\)

If given the same push, a cylinder of equal mass to a hoop would roll more quickly. This is because:

- the push gives each body an equal torque over an equal time, and so each receives the same angular momentum
- the hoop has the larger moment of inertia (more mass is at a larger distance from the centre)
- since angular momentum is conserved, this means that the hoop will have the lower angular velocity