Unit 6: Rotational Kinetic Energy and Momentum

Rotational Kinetic Energy:

The linear kinematic energy equation is as follows:

$$K{E}_t=\frac{1}{2}m{v}^2$$

If we replace mass and linear velocity with their rotational/angular counterparts, we get:

$$K{E}_r=\frac{1}{2}I{\omega }^2$$

Work Done By Torque:

The linear equation for work is:

$$W=F\Delta d_{parallel}$$

If we replace force and distance with their rotational/angular counterparts:

$$W=\tau \Delta \theta$$

Note that we no longer need the parallel distance only as torque is always perpendicular

Total Kinetic Energy:

To find the total Kinetic energy, we need to find the sum of translational kinetic energy and rotational kinetic energy

$$\begin{gathered} K{E}_{total}=K{E}_{translational}+K{E}_{rotational} \\ K{E}_T=K{E}_t+K{E}_r \end{gathered}$$

Finding total energy in the system, (GPE and KE) is still done in the same way, just with KE being split into two different parts:

$$\begin{gathered} (K{E}_T+GPE{)}_0=(K{E}_T+GPE{)}_f \\ (K{E}_t+K{E}_r+GPE{)}_0=(K{E}_t+K{E}_r+GPE{)}_f \end{gathered}$$

Remember that $GPE=mgh$

It is important to pay attention and see if the object is slipping or not. If it is NOT, then $v=\omega r$. This is not the case when the object is slipping as $\omega =0$, while $v$ isn’t

An object that has a greater moment of inertia will have a smaller linear velocity. This is because more kinetic energy is needed to make the object rotate, so less kinetic energy is available for translational motion

Angular Momentum:

As we have done above, we will take the linear equation for momentum:

$$p=mv$$

And replace the variables with their linear counterparts:

$$L=I\omega$$

Conservation of Angular Momentum

The conservation of angular momentum states that without any external forces acting on a system, momentum will be conserved. This is from Newton’s second law:

$$\sum F=\frac{\Delta p}{\Delta t}$$

If there is no force acting on the system $(\sum F=0)$, then $\Delta p=0$, meaning no change in momentum

Angular Version:

$$\sum \tau =\frac{\Delta L}{\Delta t}$$

If there is no torque acting on the system $(\sum \tau =0)$, then $\Delta L=0$, meaing no change in momentum

The Inverse Relationship between Inertia and Angular Velocity:

In a system with no external torque/force being applied to it, when you decrease the moment of inertia, you increase the Angular Velocity. This is because the momentum will be constant, forcing both variables to increase/decrease when the other increases/decreases.

$$\begin{gathered} \Delta L=0 \\ {L}_0=L_f \\ {I}_0{\omega }_0=I_f{\omega }_f \end{gathered}$$

Angular Momentum of an Object Moving in a Straight Line:

We get the equation of angular momentum in a straight line from the angular momentum equation:

$$\begin{gathered} L=I\omega \\ \omega =\frac{v}{r} \\ I=m{r}^2 \\ {L}_{straight}=m{r}^2\frac{v}{r}sin(\theta ) \\ {L}_{straight}=mrvsin(\theta ) \end{gathered}$$

The $sin(\theta )$ comes into the equation as a straight moving object makes an angle of $\theta$