# Rotational Kinetic Energy

So I'm having a bit of an issue in solving equations that involve rotational energy and how to distinguish it from mechanical. For instance, how to set up the energy equation of an atwoods machine, like pictured, in the event of m2 just before it hits the ground and m1 is on a frictionless surface. How do I know if KErot is initial or final is it simply if it is moving? I have this eqn, not sure if it's right:

• A note to the wording: Instead of saying "rotational and mechanical kinetic energy" then say "rotational and translational kinetic energy". Both of them are mechanical energies. – Steeven Apr 26 '17 at 10:01

You know if it is the initial or final situation from which values you plug in. The energy conservation equation you have made holds for both situations, only the values change. The only rule is to pick a point in time and then to keep it consistently:

• Are you picking the initial point in time, then you input the initial speeds $v$ (maybe it is zero?) and initial height $h$. The rotational (angular speed) $\omega$ will then also be for this moment.

• Are you picking the final point in time, then you input the final speeds $v$ and final height $h$ (maybe zero). You might have to calculate them first in other ways. And the $\omega$ is as well the angular speed in that final moment.

When you ask if there's only a $K_{rot}$ during rotation, then the answer is the same as for the usual $K_{translation}$: Yes, it is only there during rotation, but you don't have to think about that; the equation solves it for you if you just plug in the right values. Just plug in a zero as $\omega$ if there is no rotation, and then $K_{rot}$ becomes zero as well.

In principle if you were ever in doubt, you could include both a rotational and a translational kinetic energy term as well as a potential energy term for every object into the equation. All three for the rotating disc and all three for each of the boxes. But because you can easily see that the disc doesn't translate and that the boxes don't rotate and that the first box and the disc don't change their heights, you skip those terms already in the beginning. But when something moves, there is a kinetic energy term present and maybe also a potential, and are you in doubt which or if, then include all and plug in values.

• Thanks!! So does this mean if this were an inclined atwoods machine with the same masses and pulley, would initial energy include mgh for m2 and m1 (if they were at rest) and then Ef add mgh for m1? – Mad Apr 26 '17 at 10:04
• @Mad If an object changes it's height, then there will be a potential energy term for that one as well, yes. And even if there isn't, you can still include that term when in doubt, because it just becomes zero. See the last paragraph I just added. – Steeven Apr 26 '17 at 10:09

$M_{\rm p}$ is a term when evaluating the moment of inertia of the pulley.

Here is the edit which was done by @Mad

• That's not what I'm asking @farcher I know that, I'm asking how to set it up, is my energy eqn correct and how does one know if KErot is initial or final – Mad Apr 26 '17 at 8:52
• If $M_{\rm p}$ becomes larger the moment of inertia of the pulley is increased and so the rotational kinetic energy component is large hence the linear speed of the masses will be less. – Farcher Apr 26 '17 at 8:57
• Yeah, I know, that isn't what I'm asking. Is the energy equation that I have up there set up correctly @farcher – Mad Apr 26 '17 at 9:01
• Yes, provided the system started at rest. – Farcher Apr 26 '17 at 9:16
• @Steeven The original question was this" . . . . . . it hits the ground and m1 is on a frictionless surface. Where does mp come into play? I have this: . . . . . . . " As you can see it is significantly different from the current question. You can see what was edited by clicking in the "edited . . . . . hour ago" link at the bottom of the question. – Farcher Apr 26 '17 at 10:36