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Rotational Kinetic Energy and Conservation of Momentum

  1. Feb 18, 2016 #1
    I am having trouble wrapping my head around a physics concept.

    If we roll solid sphere down an inclined plane it will reach the bottom at a different time than if we were to say, roll a hoop down the same inclined plane. This is because they have different rotational inertias, and so more of the potential energy will go into rotational energy.

    But doesn't this conflict with the law that the acceleration of the center of mass of a system of particles depends only on the net force acting on the system of particles, and therefore the center of mass of a solid sphere and a hoop should reach the bottom of the inclined plane at the same time?

    Another thing that is confusing me. In a physics class we were shown an example where a rod experienced an impulse either at the center of mass of the rod, or at the top end of the rod. The question was which will have more kinetic energy.

    The answer was that because the acceleration of the center of mass was the same for both, and the rod pushed at the top also had some rotational energy, the kinetic energy of the rod pushed at the top (and not directly in the center) was greater. The extra energy came from the fact that although the impulse was the same, the work done was not the same because the rod which was pushed from the top had its force act over a greater distance.

    But doesn't this above scenario seem to conflict with the first scenario presented?

    Would appreciate any help,
    Thank You.
     
  2. jcsd
  3. Feb 18, 2016 #2

    Redbelly98

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    For your first scenario:

    There is also a (static) friction force acting on the rolling objects at the point of contact with the incline -- as there must be, in order to produce a toque about the object's center and cause it to rotate as it rolls.

    Anyway, that friction force is different for the two objects of different rotational inertia, so the net force and acceleration are different as well.
     
  4. Feb 18, 2016 #3
    If we held the surface area to be the same, why would the frictional force be different? Furthermore, if we assume that the frictional force is only the simplified version of the normal force and the coefficient of friction, then with similar materials the frictional force should be the same.
     
  5. Feb 19, 2016 #4

    A.T.

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    No, this is not how static friction works. Normal force times static friction coefficient gives you the maximal static friction force possible, not the actual force acting.
     
  6. Feb 19, 2016 #5
    I am not saying that the static frictional force is proportional to the coefficient. The coefficient merely gives a maximal static friction force possible, like you said. But this still does not explain why the static frictional force will be different for the two rolling objects.
     
  7. Feb 19, 2016 #6

    A.T.

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    Static frictional force will take whatever value is needed to prevent slippage (within the bounds given by the coefficient).
     
  8. Feb 19, 2016 #7

    rcgldr

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    The friction force times the radius of the hoop or sphere equals the torque applied to the hoop or sphere. The torque in turn is related to the acceleration in the direction of the plane. I did this recently, but now search doesn't show my last post on this. So redoing the math again:

    force in the direction of the plane related to gravity = fg = m g sin(θ)
    force in the direction of the plane related to friction = ff= α I / r, where α is angular acceleration, I is angular inertia, r is radius
    linear acceleration in the direction of the plane = a = α r = (fg - ff)/m = g sin(θ) - a I/(m r^2)
    a(1 + I/(m r^2)) = g sin(θ)
    a = g sin(θ) / (1 + I/(m r^2))
    ff = m g sin(θ) - m a

    for hoop or hollow cylinder I = m r^2, a = 1/2 g sin(θ), ff = 2/4 m g sin(θ)
    for hollow sphere I = 2/3 m r^2, a = 3/5 g sin(θ), ff = 2/5 m g sin(θ)
    for solid cylinder I = 1/2 m r^2, a = 2/3 g sin(θ), ff = 2/6 m g sin(θ)
    for solid sphere I = 2/5 m r^2, a = 5/7 g sin(θ), ff = 2/7 m g sin(θ)

    As for the rod, the linear acceleration equals the linear force / mass, regardless of where the force is applied, but if the force is applied at the end of the rod, then the rate of acceleration of the end of the rod is greater than the rate of acceleration of the center of mass of the rod, and the force is applied over a greater distance.
     
  9. Feb 19, 2016 #8

    Dale

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    So if the coefficient only determines the maximum force, then what determines the actual force? When you work one of these problems, what is the process you use to determine it?
     
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