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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?

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    $\begingroup$ @Sam: in “perfect” rolling without slipping, the rim of the wheel is never in motion with respect to the surface, and so friction does no work on the wheel. The reason a rolling object experiences rolling resistance is because of the inelastic compression & re-expansion of the ball as it rolls. $\endgroup$ Mar 1 '20 at 18:26
  • $\begingroup$ @MichaelSeifert, it is true that the static friction in pure rolling motion does not perform any translational work. However, let me make a note that it does perform rotational work. I agree with your comment, which is about the rolling friction or rolling resistance. $\endgroup$
    – kbakshi314
    Mar 1 '20 at 18:35
  • $\begingroup$ You are correct in that at any given height or time the speed of the ball that is rolling will lag behind due to conservation of mechanical energy. Also, if you make a forces analysis, the rolling ball will move slower because of the friction force that points up the plane. $\endgroup$
    – user65081
    Mar 1 '20 at 18:37
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    $\begingroup$ @kb314 the static friction performs no work at all. Neither linear nor rotational. All of the work is from gravity, both linear and rotational. $\endgroup$
    – Dale
    Mar 1 '20 at 18:45
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    $\begingroup$ @Dale, you're right. The definition of work makes no distinction between either. Sorry, late night! $\endgroup$
    – kbakshi314
    Mar 1 '20 at 20:55
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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.

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    $\begingroup$ Hmm.. thanks for confirming my suspicions. I was taking a look at this which is probably an explanation; I guess the subtraction of the friction terms is where the rotational energy comes in, but my head can't get why the cosine is there; I'll probably come back and get to it later. $\endgroup$
    – Buretto
    Mar 1 '20 at 18:58
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    $\begingroup$ OH never mind, I got out the pen and paper and I see know. Was a bit of a refresher for me. Perfect, thanks again. $\endgroup$
    – Buretto
    Mar 1 '20 at 19:12

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