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Rotational Dynamics Confusion

  1. Dec 6, 2012 #1
    I am right now studying Rigid Body dynamics..

    I have some doubts regarding Dynamics...

    1) When a body is rotating as well as translating....we say that every particle of the ..lets say ROD is moving with a resultant velocity = Vcm ( translational velocity of COM) + ω.R (where R is the distance from COM ) ( ω is angular velocity..)
    My doubt is that how do we knw that every particle of the rod is translating with a velocity with which the COM is Translating...

    2) My sir told that ω of a body is same whatever be the axis of rotation ..
    And thats why we very effectively use the Angular Momentum conservation theorem ..
  2. jcsd
  3. Dec 6, 2012 #2
    That's pretty much the definition of the rigid body model. The distance between all points of the particles that the body consists of remains constant eg. the body cannot deform.
  4. Dec 6, 2012 #3
    I knw that all the particles move with the same velocity...But can it be different than that of Vcm ...In case if COM is outside the body..?
  5. Dec 6, 2012 #4


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    You don't.
    If you inspect one particle of the body at some instant, without reference to any other parts of it, it will have a position. if you inspect it for a slightly longer instant, you discover it has a velocity. In isolation, there is no basis for breaking the velocity up into a translational and rotational velocity. That said, if you study for a yet longer instant, you'll find it has an acceleration, and you could choose to distinguish the component of the velocity orthogonal to the acceleration as 'rotational', but I'm not sure that makes sense in general.
    If the body as a whole is not being subjected to a net force then its CoM will have no acceleration. If it is rotating about a steady axis, those particles on the axis will have the same velocity as the CoM. The velocity of a particle of a rigid body in general will have a velocity v equal to vc+ω x r, where vc is the velocity of the mass centre, ω is the rotation vector and r is the position vector of the particle relative to the mass centre. That is the consequence of the fact that it is a rigid body. This does provide a basis for thinking of the particle's velocity as composed of those two components, but it's only a way of viewing things.
  6. Dec 7, 2012 #5

    So here we study the motion of centre of mass of the body and letz say it has a velocity Vcm ..then we know that the whole body has a translational velocity Vcm....

    But letz say that at some instant i obtained the net velocity of a particle say V...then the translational velocity of the body will be

    V(translational) = Net velocity - ω.R (R=Distance from COM)...

    I hav one mOre question.....Why dont we consider the rotation of COM..
    Does COM have an angular velocity ω ...
    Or is it that we study the motion w.r.t the COM and hence it does not have angular velocity and all other particles have their angular velocity measured w.r.t COM..??? ..
  7. Dec 7, 2012 #6


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    Hi D_DaYwAlKeR! :smile:

    We don't have to start with the c.o.m., we can start with any point P,

    the velocity of any other point X is then vP + ω x PX :wink:

    (after all, velocity of a rigid body is just geometry

    it depends only on shape, not on mass distribution …

    so it doesn't matter where the c.o.m is! :smile:)​
  8. Dec 7, 2012 #7


    Ok. So Since in most of the cases Bodies rotate along the COM axis ..that's why we study rotation with respect to COM...and that's why particles other than COM have angular velocity while COM only translates...More over its easy to calculate Moment of Inertia along COM axis...Am i correct here/????

    Now if choose to study the motion with respect to some other axis passing through a point P on the body ...then with respect to that axis the point P will have no angular velocity while other particles including COM have angular as well as translational velocity.,..

    Am i correct here??...
  9. Dec 7, 2012 #8


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    we study rotation wrt the c.o.m. because usually the c.o.m is the only point whose velocity we know (we get it from F = mac.o.m :wink:)

    (the exception is where one point is fixed, eg a pivot or a rolling point of contact … ie, it's the instantaneous centre of rotation … then obviously it's easier to use that point!)
    i'm not following that :confused:
    correct, but that doesn't make much difference …

    we can always apply the parallel axis theorem to get the moment of inertia about the centre of rotation instead
    yes, v = vP + ω x r
  10. Dec 7, 2012 #9
    ...and that's why particles other than COM have angular velocity while COM only translates

    By this I meant that when we study motion w.r.t COM, then COM will only have translational velocity whereas all other particles will have velocity = Vcm + w.R ...and COM will have no tangential velocity as V(tangential) = w.R (R here is 0)...
    Whatever axis we choose that point of the body will have w.R = 0 ..

    Am i correct???...

    Thnxxx a lot @tiny - tim ..
    DeIdeal ..
    You all helped a lOT...:)..:)
  11. Dec 7, 2012 #10


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    yes, it applies to any reference point, not just the c.o.m. :smile:

    (but it's all just maths, not physics …

    i'd forget the explanation, and just use the formula :wink:)
  12. Dec 7, 2012 #11
    @tiny -tim ...Thnxx A lot mate!...
    u cleared all my doubts ...
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