Rotational kinetic energy and angular momentum

In this lesson, we will learn:

• How to calculate rotational kinetic energy?
• Definition of angular momentum
• Newton’s 2^nd law for rotation
• The law of conservation of angular momentum

• Objects rotating about an axis possess “Rotational Kinetic Energy”
• Objects moving along a straight line possess “Translational Kinetic Energy”.
• Translational kinetic energy is calculated using, $\frac{1}{2}mv^{2}$ , we can convert this formula to rotational kinetic energy question using the rotational motion analogues:



If we consider a rotating ball, every point on the ball is rotating with some speed.
The ball is made up many tiny particles, each of mass “$m$”. let’s take “$r$” to be the distance of any one particle from the axis of rotation (O);



$\Rightarrow \,$ rotational $KE = \frac{1}{2}I\, \omega^{2}$

• The center of mass of a rotating object might undergo translational motion (a sphere rolling down an incline), in this case we have to consider both rotational and translational kinetic energy.


$KE = \frac{1}{2}M v^{2} \, + \, \frac{1}{2}I_{CM \;} \omega^{2}$

$M$: total mass of the object
$I_{CM}$: moment of inertia about the axis through center of mass 
$v$: translational speed 
$\omega$: angular speed 


Angular Momentum In the previous section, we used the rotational analogue to translate the translational kinetic energy to rotational kinetic energy.
In like manner, the linear momentum can be changed to angular momentum, using the rotational analogues;


$L$: is the angular momentum with a standard unit of kg.m²/s


Newton’s 2^nd Law for Rotation

The Newton’s 2^nd law also can be written in terms of rotational analogues;

$\sum F = \, ma \, =$ $\large \frac{m \Delta v}{\Delta t}$ $=$ $\large \frac{\Delta p}{\Delta t} \quad$, in translational motion “Force” causes linear acceleration

$\qquad \quad$ Similarly for rotational motion , $\sum =$ $\large \frac{\Delta L }{\Delta t}$

$\sum \tau = \,$ $\large \frac{\Delta L}{\Delta t}$ $=$ $\large \frac{I \Delta \omega}{\Delta t}$ $\, = I \propto \quad$, in rotational motion “Torque” causes rotational acceleration

$\sum F = ma \quad$ Newton’s 2^nd Law in Translational Motion
$\sum \tau = I \propto \quad$ Newton’s 2^nd Law in Rotational Motion


The law of Conservation of Angular Momentum
The law of conservation of angular moment states that;

“The total angular momentum of a rotating object remains constant if the net torque acting on it is zero.”

$\sum \tau = 0 \; \Rightarrow \; L_{i} = L_{f} \; \Rightarrow \; I_{i \omega i} = I_{f \omega f}$

Example: A skater doing a spin on ice, illustrating conservation of angular momentum.



Open arms: $\qquad L_{i} = I \omega_{i} = mr^{2}_{i} \omega_{i}$
Closed arms:   $\qquad L_{f} = I \omega_{f} = mr^{2}_{f} \omega_{f}$

When the arms of the skater are tucked in, the mass is not changing, but the radius of rotation decreases; $r_{f}$ < $r_{i}$, therefore;  $mr^{2}_{f}$ < $mr^{2}_{i}$

According to the law of conservation of angular momentum; $L_{i} = L_{f}$

$L_{i} = L_{f}$

$\qquad \Rightarrow \quad \omega_{f}$ > $\omega_{i} \;$ “the result would be higher angular velocity”
$mr^{2}_{f}$ < $mr^{2}_{i}$