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Angular momentum without rotational energy?

  1. Aug 24, 2006 #1
    Angular momentum without rotational kinetic energy?

    Consider a particle (a ball or electron or whatever you like) that moves in a perfect circle with constant angular and tangential velocity about a fixed point in space. This particle does not spin about itself. The movement is perfectly translational. (OR could we say its movement is rotational?)

    Obviously, this particle has angular momentum about this fixed point. Also, this particle has translational kinetic energy. However, could we say that this particle has rotational kinetic energy?
     
    Last edited: Aug 24, 2006
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  3. Aug 24, 2006 #2

    quasar987

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    I will not answer your questions officially as I don't recall even seeing a formal definition of the terms "rotation" and "rotationnal energy". Unoficially, however, I can casually say that yes, I would use the words "the particle is in rotation about that point" to describe the motion. And I would say that its kinetic energy is of rotational nature.

    I mainly wanted to reply to your post in response of its title, "Angular momentum without rotational energy?" to show you a trivial exemple in which "Angular momentum without rotational energy!"

    Consider a free particle of mass m moving in a straight line at unit velocity. Suppose we take a coordinate system in which its motion is given by (x(t)=t, y(t)=1, z(t)=0). The angular momentum of the particle about the origin is [itex]L=|\vec{r}\times \vec{p}|=mrsin\theta = mr(1/r)=m[/itex]

    There is no rotational motion whatesoever, yet, L is not zero. So, you see that angular momentum doesn't have that much to do with rotation. I had this prejudged about angular momentum at a time and this exemple "opened my eyes" so to speak, so now I pass it on to whoever may read this. Amen.
     
  4. Aug 24, 2006 #3

    Doc Al

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    As quasar987 explained, you don't need rotational or circular motion of any kind to have non-zero angular momentum about some point. (I would have given the same example.)

    The total KE of an (extended) object can be viewed as the sum of the translational KE of the center of mass plus the rotational KE about the center of mass. So, in that sense, the circularly moving (revolving, but not rotating) object in your example has pure translational KE.

    But don't get hung up on semantics.
     
  5. Aug 24, 2006 #4

    Gokul43201

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    Either way of calculating the KE ([tex]0.5I\omega ^2~or~ 0.5mv^2 [/tex]) gives you the same answer. So I guess it doesn't matter what you call it. Typically though, "translation" is used to describe linear displacements.
     
  6. Aug 24, 2006 #5
    I meant rotational kinetic energy. Pardon my omission of the word "kinetic"
    I edited the title of this thread to include "kinetic"
     
  7. Aug 24, 2006 #6
    Yes, I had understood that even an object moving in a straight line has an angular momentum with respect to a certain point.

    The red portion was what I thought it was. Yes, the object is "revolving" but not "rotating," hence no rotational kinetic energy. Revolving is just another case of translation, so the object has translational kinetic energy.

    So far, I had no confusion. But hereforth, my confusion starts.

    Consider a straight-moving particle. It has angular momentum about a certain reference point in space. Obviously, this particles angular monentum about the reference point is L= r x p

    But also, Angular momentum can be expressed as [itex]L = I \omega[/itex]

    Since angular momentum exists, angular velocity ([itex]\omega[/itex]) and moment of inertia (I) must exist. Since both exists, the rotational kinetic energy KE_r =0.5 Iw^2 must exist. This is contradictory to what we have first stated.

    Angular velocity in this case is easy to imagine. I just think of a line stretching from the reference point to the particle and imagine how much radian this line moves in a very short period of time.

    But moment of inertia.........hmm......??? This momen of inertia is NOT the moment of inertia of the particle that occurs when the particle spins about its center of mass. Then what are we thinking here?
     
    Last edited: Aug 24, 2006
  8. Aug 24, 2006 #7

    Andrew Mason

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    A discrete mass cannot rotate about a point. There has to be matter at the centre providing the centripetal force.

    I don't see a problem with saying that the mass has rotational kinetic energy with respect to the frame of reference of the centre of rotation. After all, the difference between that and a spinning hoop is just that the mass in the hoop is spread out.

    AM
     
  9. Aug 24, 2006 #8

    Doc Al

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    It's a matter of semantics. The object certainly has no rotational KE about its center of mass, which is what counts. Whether you treat the object as a whole as simply translating or as "rotating" about a point is up to you.


    No problem.

    I see your point, but I'd still call it semantics.

    It's the moment of inertia of a point mass about some reference point. In your example of a particle in circular motion you have two choices for calculating the KE. You can express it in terms of linear speed as: KE = 0.5 m v^2; or in terms of angular speed as: KE = 0.5 I w^2. Of course I for a point mass about the center of the circle is just: I = mR^2. Since v = wR, KE = 0.5 I w^2 = 0.5 m v^2. You get the same KE either way you slice it!
     
  10. Aug 24, 2006 #9
    Thanks. I got it.
     
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