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This is a purely theoretical question, so I am unsure if it will even make sense:

  • Imagine that there are two planes which are inclined at say, 45° to gravity. One has infinite friction and the other has zero.

    There are also two perfectly round, and infinitely hard spheres (no deformation), and similarly, one has infinite friction and the other has zero.

    The sphere with zero friction is placed on the plane that also has zero friction; and vice versa.

So assuming gravity is the only force acting, and equal starting conditions (apart from friction), and that the one scenario would have perfect roll, and the other would have zero (and zero friction losses): I think that the sphere that rolls will always be behind the other since it has to convert some energy into rotational energy. Is this correct? I know this is probably trivial to prove but I don't think I am competent enough to do the math yet.

Further extending that idea, does that mean both spheres, at any point in time, would have the same total energy?

Dale
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Buretto
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1 Answers1

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I think that the sphere that rolls will always be behind the other since it has to convert some energy into rotational energy. Is this correct?

Yes, this is correct. The amount of gravitational energy is given by $mgh$ where $h$ is the change in height. So at a given height both objects will have the same kinetic energy. However, the rolling one will have some portion of the energy in rotation and correspondingly less in translation. Therefore it will be slower at every height.

Further extending that idea, does that mean both spheres, at any point in time, would have the same total energy?

Yes. The total energy is the sum of the linear KE, the rotational KE and the gravitational PE: $\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$ where the second term, the rotational KE, is zero for the frictionless object.

Dale
  • 117,350