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Combined translational and rotational motion of a rigid body

  1. Feb 10, 2014 #1
    The velocity of any point P of a rigid body in rotation plus translation is [itex]\vec{}v[/itex]p=[itex]\vec{}v[/itex]COM+[itex]\vec{}v[/itex]p,COM.
    Now |[itex]\vec{}v[/itex]COM|=v and [itex]\vec{}v[/itex]p,COM =rω .
    But v and rω are same thing as v=rω ,so velocity of the particle every time will be √2 v.Then what is the difference between [itex]\vec{}v[/itex]COM and [itex]\vec{}v[/itex]p,COM?
     

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    Last edited: Feb 10, 2014
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  3. Feb 10, 2014 #2

    Filip Larsen

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    You may want to reconsider this statement.
     
  4. Feb 10, 2014 #3
    The short answer to your question is that v=rω makes some assumptions about your frame of reference, specifically that it's not rotating and that is isn't moving with respect to the CoM. If the body is moving or the FoR is rotating, you have to adjust accordingly.
     
  5. Feb 10, 2014 #4

    jbriggs444

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    The velocity of the center of mass ([itex]\vec{}v[/itex]COM) could be given by v=rω in the case, for instance, of a bicycle wheel rolling along the road. However, in the general case it will be completely independent of the rotation rate of the rigid body.

    Suppose, for the sake of argument that we are considering the case of a bicycle wheel so that the velocity of the center of mass with respect to the road is given by rω. It will also be true that the velocity of a point on the treads of tire will have a velocity with respect to the center of mass that is given by rω. Will it be the case that these two velocities will sum to √2 v? Decidedly not. Unless those two velocities are at right angles, their sum could be anywhere between 0 and 2v. Only if the two equal velocities are perpendicular will their sum be √2 v.

    The cases where the velocity of a particular point on the treads of a bicycle tire has a velocity equal to √2 v are where that point is level with the hub, either at 90 degrees forward of the contact patch or at 90 degrees back. At the contact patch the two velocities will sum to zero; the contact patch is at rest with respect to the road. At the top, the two velocities will sum to 2v; the top of the tire is moving at twice the speed of the bicycle.
     
  6. Feb 10, 2014 #5
    That means as the bicycle wheel is circular so that two velocities are same,right? that is, the r is distance between centre of mass and the Point P and also the radius of the wheel.But if it is of irregular shape,then will →vCOM and →vp,COM be equal? In that case →vCOM will not be rw ?
     
    Last edited: Feb 10, 2014
  7. Feb 10, 2014 #6

    jbriggs444

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    Take, for example, a triangular bicycle wheel that is rolling on the road without slipping. If you define "r" as the distance from the center of mass to one of the points on the triangle then the velocity of the center of mass with respect to the road will be rω. If you define "r" as the distance from the center of mass to a point P in the middle of one of the sides then the velocity of point P with respect to the center of mass will also be given by rω.

    But if you try to argue that the two velocities are equal because they are both given by "rω" then you will have committed two errors:

    1. The fallacy of equivocation -- you will have used the expression "rω" in the same context with two different meanings for the letter r.

    2. The error of treating a vector as if it were a scalar and ignoring its direction.
     
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