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The role of center of mass in motions

  1. Jun 4, 2014 #1
    Consider the motion of a system composed of a rigid bar and a rigid ring attached to it as is shown in the attached figure.The rigid system is rotating around the point O.
    Why we can not treat this motion as rotational motion of "the center of mass"? Namely why we can not get the correct result if we treat the motion as pure rotation of the point mass M located at CM with rotational inertia I=M (rCM)2 ?
     

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    Last edited: Jun 4, 2014
  2. jcsd
  3. Jun 4, 2014 #2
  4. Jun 4, 2014 #3

    I think I don't understand your question very well.

    The net (total) exterior force acting on your system (rigid bar + ring) will be equal (if we consider internal forces in the bar + ring to be "newtonian") to the total mass M times the vector acceleration of the center of mass. That is:

    [tex]\vec{F_{ext}} = M \vec{a_{cm}}[/tex]

    Given that the rigid system is attached to point O, the center of mass will describe a circle with center in point O. (The center of mass could be moving with uniform circular motion, uniform accelerated circular motion, or non-uniform accelerated circular motion, depending on the total exterior force upon our system).

    For example, it the center of mass is moving with uniform circular motion (with respect to inertial frame with origin in point O ), then the total exterior force is "radial" (pointing to O at all times), and its magnitude will be:

    [tex]F_{ext}=M \frac{v^2_{cm}}{r_{cm}}[/tex]

    This total exterior force will be the force arising because of the attachment of our system to the point O, plus maybe another exterior forces whose total vector sum is also radial.


    If the center of mass is moving with non-uniform circular motion, then the total exterior force is the force arising because of the attachment of the system to point O (which is always radial), plus another exterior force (which is causing the tangential acceleration of the center of mass along its trajectory).


    In any case, the total angular momentum vector of our system with respect to point O will be:

    [tex]\vec{L}(t) = I \vec{\omega}(t)[/tex]


    Where I is the moment of inertia of our system (bar + ring) with respect to the axis passing through point O and perpendicular to the plane of motion, and vector omega is the angular vector velocity of each particle that make up the bar + ring (which can vary with time if the circular motion is not uniform) with respect to our inertial frame with origin in O.


    Also the moment of the total exterior force with respect to point O (which is just the moment of the "other" exterior force, because the exterior force arising because of the attachment is radial and its moment with respect to point O will be zero) will be (if interior forces are "newtonian" and "central") :

    [tex]\vec{M}_{ext}(t) = I \vec{\alpha}(t)[/tex]


    The moment of inertia of your system with respect to the axis passing through O and perpendicular to the plane of motion, will be:


    [tex]I = \frac{M_{b} L^2}{3} + M_{r} R^2_{r} + M_{r} (L + R_r)^2 = \frac{M_{b} L^2}{3} + M_{r}(R^2_{r} + (L + R_r)^2)[/tex]


    where [tex]M_{b}[/tex] is the mass of the bar (supposed homogeneous), [tex]M_{r}[/tex] is the mass of the ring (supposed homogeneous) and [tex]R_r[/tex] is the radius of the ring.


    EDIT: sorry, are you just asking if I = (M_b + M_r) (r_CM)^2 ?

    Then I already gave you I, so you now can check if it is what you think or it is not. :-)
     
  5. Jun 4, 2014 #4
    I would think you can if you calculate r cm correctly.
     
  6. Jun 4, 2014 #5
    Thanks all.
    Let me ask my question clearly:
    We have a system composed of "rigid bar + rigid ring" as in the left of the attached picture. In right of the picture, sum of two masses are put as point mass in the center of mass.
    Suppose we were to calculate the angular velocity of the left system:
    why we can not treat the left system as a point mass rotating around O, as the right system, in which the torque is vector product of "center of mass vector" and "mg"?



    View attachment untitled.bmp
     
  7. Jun 4, 2014 #6

    D H

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    You can if you make a lot of extra computational work for yourself.

    The center of mass is moving from the perspective of an inertial frame, and it is doing so non-uniformly. Your center of mass frame is a non-inertial frame. There are problems where looking at things from the perspective of a non-inertial frame makes things easier.

    This isn't one of them.
     
  8. Jun 4, 2014 #7
    Because the correct equation is:


    [tex]\vec{r}_{cm}\times (M\vec{g}) = I \vec{\alpha}[/tex]

    Where [tex]\vec{r}_{cm}[/tex] is the vector position (with respect to intertial frame with origin in O ) of the center of mass of the system,

    [tex]M[/tex] is the total mass of the system,

    [tex]I[/tex] is the moment of inertia of the whole system with respect to the axis that passes through O and is perpendicular to the plane of motion,

    [tex]\vec{\alpha}[/tex] is the angular acceleration vector.


    And you want to use this other equation:

    [tex]\vec{r}_{cm}\times (M\vec{g}) = (M r^2_{cm}) \vec{\alpha}[/tex]


    This other equation is not correct because

    [tex]M r^2_{cm}[/tex]

    is not the moment of inertia of the system.

    The correct moment of inertia [tex]I[/tex]

    is what I wrote in my previous post.
     
  9. Jun 4, 2014 #8
    proposed problem

    This might help clear up the question. Suppose a force F is applied perpendicular to the rod at a distance of 1/2 the location of the center of mass from the center. What is the angular acceleration?
     
  10. Jun 4, 2014 #9
    Yes, as you say, we could solve it from the CM-frame. And thinking about this I have just realized that in this example ( only weight force and contact force between system and point attachment O ) the contact force IS NOT radial (contrary to what I previously stated) because if it were radial, the moment of this force with respect to the CM-frame would be zero, and the moment of the weight force with respect to the CM-frame is also zero, and so the system would not spin with respect to the CM-frame, which is false.

    So the contact force (at attachment point O ) is not always radial (depends on the situation).
     
  11. Jun 4, 2014 #10
    I think the initial question is already answered. The correct equation to use is in this case is "Moment of total exterior force (with respect to O ) is equal to moment of inertia of the whole system (with respect to O ) times angular acceleration vector"

    You can also use an analogous correct equation with respect to the center of mass of the whole system.


    EDIT: sorry, I thought you were the OP :-)
     
    Last edited: Jun 4, 2014
  12. Jun 4, 2014 #11
    Exactly Right.
    So I think we can not trust CM frame and we have to take care when solving problems in CM frame.
     
  13. Jun 4, 2014 #12
    Suppose you had 2 equal point masses, one at r and one at 2 r. The moment of inertia is m rsq + 4 m rsq or 5 m rsq. If I place both masses at the center of mass at 3/2 r, then the moment of inertia is 2 m x 9/4 rsq or 4.5 m rsq. I think this is where the confusion lies?
     
  14. Jun 4, 2014 #13
    Thanks, I know it but I like to know why the CM method fails here in solving the problem.
     
  15. Jun 4, 2014 #14
    Yes. Perfectly explained.

    The "CM method" does not fail. You can solve the problem from the CM-frame and it will produce the same result.

    It is just that you must use the correct moment of inertia with respect to the concrete frame you are using to solve the problem.
     
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