8
$\begingroup$

I don't understand the concept of rotational potential energy.

During linear motion, when a force is applied, the work it does gets converted to kinetic energy and there is no change in the potential energy. Similarly, during rotational motion when a torque is applied to angularly accelerate a body, the work done by the torque leads to an increase in kinetic energy. Hence the total energy of the body also increases. This video says that the total energy of the body remains constant, hence there is a need for rotational potential energy in order to balance out the increase in the kinetic energy.

Shouldn't the total energy of the "system" including the person applying torque be conserved and the increase in kinetic energy be justified on the basis of the person's energy being decreased? Where does potential energy come into the picture?

$\endgroup$
2
  • $\begingroup$ "During linear motion, when a force is applied, the work it does gets converted to kinetic energy and there is no change in the potential energy." Is this really true, though? Where does the force come from? If the force is due to gravity or electrostatics, then it can be thought of as the negative gradient (directional derivative) of the potential. I.e. left alone, the system will feel forces that tend to push it to lower potential energy. $\endgroup$ Nov 22, 2020 at 1:06
  • $\begingroup$ If the force is due to some external agent, that can certainly also change the potential energy. For instance, lifting an object against gravity, or pushing an object against electrostatic repulsion both increase the potential energy of the whole system. $\endgroup$ Nov 22, 2020 at 1:08

2 Answers 2

2
$\begingroup$

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy.

The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \Delta PE$

$\endgroup$
5
  • $\begingroup$ But won't there always be some external work...? $\endgroup$
    – oshhh
    Jun 25, 2016 at 12:24
  • 1
    $\begingroup$ Imagine a spinning disk with a radial rail with a mass on it. If you release the mass it will move - and total angular momentum of system is unchanged but KE of system will change. $\endgroup$
    – Floris
    Jun 25, 2016 at 12:51
  • $\begingroup$ I know it's four years since you posted this, but, well, this is when I'm finally learning this stuff in school. The same way that the other variables and relations have analogues in rotational dynamics to their translational counterparts, would $\Delta PE=m\cdot a\cdot\Delta s$ correlate with $\Delta PE_{\theta}=I\cdot\alpha\cdot r\cdot\theta$? $\endgroup$
    – DonielF
    Jul 16, 2020 at 23:37
  • $\begingroup$ @DonielF not sure of your notation but if the first terms on the right hand side indicate torque, then I think the answer is yes. Because in the first equation $ma=F$, and change in work is of course $F\cdot ds$ (again, I have to assume things about your notation... so while what you wrote looks directionally correct I cannot be sure unless you define your variables). $\endgroup$
    – Floris
    Jul 16, 2020 at 23:43
  • $\begingroup$ @Floris I accidentally threw in an extra $r$ but yes, the idea'd be torque time displacement. $I=mr^2$ is moment of inertia, $\alpha$ is rotational acceleration, $r\theta$ is displacement. So it'd be $PE = Fs = mar\theta = mr^2\alpha\theta = I\alpha\theta$, where $F$ and $a$ are tangential force and acceleration. $\endgroup$
    – DonielF
    Jul 17, 2020 at 0:03
0
$\begingroup$

The formula for PE is not F /s it's Fy. That subtle difference is quite important because if you move something perpendicular to gravity, you get no change in PE. It then follows that there is no rotational PE unless the rotating body is connected to a spring (or other energy storage).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.