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Conservation of linear or angular momentum + center of mass.

  1. Dec 12, 2009 #1
    So I have been reading about conservation of linear momentum and conservation of angular momentum of a rotating object. For example a thin rod resting horizontally on the a flat surface. If this rod is being hit by a small object(collision), the rod will rotate about it's center of mass and will have both translational and rotational motion and therefore both linear and angular momentum are conserved.

    But how about if this rod is now pinned at it's center(still resting on the flat horizontal surface), if the small mass comes to collide again, I think angular momentum is still conserved but linear momentum is NOT conserved because the rod cannot have translational motion.

    Am I right???
  2. jcsd
  3. Dec 12, 2009 #2
    You are right

    The explanation (if you were looking for one) is that the pin applies force on the rod, so linear momentum isn't conserved.
  4. Dec 12, 2009 #3
    That's true.
    The pinning will transfer some momentum to the surface. So total linear momentum and total angular momentum are always conserved, but that doesn't have to be true for individual components. You always have to include all components that forces are acted on.
  5. Dec 12, 2009 #4
    What do you mean by total linear and angular momentum? Would that be like momentum "before" and "after" say in a collision? So if we consider this as a collision between a small object and the resting rod(pinned at center). I will say total linear momentum of the collision is NOT conserved because the rod cannot have any translational motion. But the total angular momentum of the collision IS conserved because the the rod is free to rotate freely about it's center of mass.
  6. Dec 12, 2009 #5
    For conservation laws (which are never violated) you have to include all components such as the surface where the rod is attached. The rod considered alone of course does not conserve linear momentum (this linear momentun is not zero however because the rod does in fact have a time varying translational motion). The linear momentum will instead transfer to the the surface on which the rod is attached.
  7. Dec 12, 2009 #6
    So I did more reading and this is what I am understanding:

    In a closed system(system upon which no external forces act), both linear and angular momentum are conserved.

    So if we say this is a collision between a "resting thin rod(pinned at it's center)" and a "small object" on a frictionless horizontal surface. Both angular and linear momentum are NOT conserved because we have an external force from the pin.

    In the case where the thin rod is resting on the same horizontal frictionless surface and NOT pinned down, then since there is no external force on this close system, both angular and linear momentum will be conserved.
  8. Dec 12, 2009 #7
    I am not sure what your statement is. Maybe you should reread the answers here and think what they more. Or maybe formulate which issue you want to understand.
  9. Dec 12, 2009 #8


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    The issue here is what is considered as part of the closed system. If whatever the pin is attached to, for example, the earth, is considered as part of the closed system, then earth, pin, rod, and small object are all part of a closed system, and both linear and angular momentum are conserved.
  10. Dec 12, 2009 #9
    Well maybe I should design a question to explain what my statement means.

    Let's say a projectile(mass Mp, initial velocity Vpi, final velocity Vpf) collides with a thin rod(mass Mr, length L, initial velocity is zero, Inertial mass I, angular speed W). The rod is pinned at it's center on a horizontal frictionless surface. Let "X" be the distance between where the projectile hit the rod and the center of the rod. Let's say after the collision, the 2 objects do NOT stick. and the

    In my opinion, linear momentum is not conserved in this system because of the external force from the pin. But we can say angular momentum is conserved because there is no external torque; therefore conservation of angular momentum equation stands:

    Li = Lf : Mp*Vpi*X + 0 = Mp*Vpf*X + I*w

    How does that look?
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