Imagine two rolls with the same diameter and mass. The mass of one roll is concentrated to the center of the roll while the mass of the other roll is concentrated to the edge of the roll. If the two are released from the same slope and the same height continuing to a flat platform, the roll whose mass is concentrated to the edge of the roll rolls considerably farther than the other. The two rolls have the same potential energy at the beginning. The roll with a smaller moment of inertia has greater speed at the end while the other roll has greater rotational energy. Does the other roll roll farther due to the air resistance being greater for the faster roll or because the moment of inertia resists changes in rotational speed more? Or maybe because the roll with a greater moment of inertia has more time to "load" energy on the slope? If you made a formula to count the distance rolled, would time play a part in the formula, or only the starting height and the velocity at the end of the slope?

  • $\begingroup$ Both rolls will have the same kinetic energy. Roll A will have a larger fraction of its kinetic energy in the translation of the center of mass. Roll B will have a larger fraction of its kinetic energy in the rotation of its mass around the symmetry axis. Both translation and rotation contribute to the kinetic energy. $\endgroup$
    – Bill N
    Apr 5 '16 at 20:52
  • $\begingroup$ Thank you for noting, english is not my first laguage so the translation may be a bit off. Changed. $\endgroup$ Apr 6 '16 at 10:27
  • $\begingroup$ for formulas see my answer to a similar question here physics.stackexchange.com/q/248357 $\endgroup$
    – anna v
    Apr 10 '16 at 7:35

An easy way to see this, assuming that they both roll without slipping, and that no friction is present is with conservation of energy as hinted in the comment above. The no-slip condition is $v=\omega r$. The total energy of one cylinder is

$$E = \frac{1}{2}(I\omega^2+mv^2)$$

Plugging the condition for no slip and solving for v, with $E=mgh$ :

$$v^2 = \sqrt{\frac{2gh}{\frac{I}{mR^2}+1}}$$

For a hollow cylinder, $I = mR^2$, while for a filled one, it is $I=\frac{mR^2}{2}$. As we can see, the ratio of the speeds when all the potential energy is converted to kinetic will be : $$\frac{v_{full}}{v_{hollow}} = \sqrt{4/3} \approx 1.15$$ So the filled cylinder is indeed faster just by energetic considerations.

To answer your question, neglecting friction, the "time" of the interaction does not matter, and the cylinders will roll infinitely far.

I think however that even if we introduce friction, the result should be roughly the same, assuming friction is not speed-dependent (hence not air resistance), since they both have the same mass and same geometry. That is just a hunch, though, but I think in a lab the primary effect is not due to friction, rather to the different moments of intertia.

If we introduce the effect of speed-dependent friction, then the problem becomes non-trivial. I think that a given roll won't always win given different initial conditions. In other words, with air friction, either roll could be the winner in terms of distance, given the right conditions.


Okay let's call the roll with the centralised mass Roll A, and Roll B is the roll with the more radial, outwardly dispersed mass.

When released, the rolls will begin to experience an acceleration (a torque), which is however many radians per second squared.

Net: each roll will receive the same amount of radians, which will be the circumference of the roll divided by the length of the slope. For future reference, we'll call this phi. If the diameter remains constant, then the roll will always be displaced phi radians, regardless of mass.

Roll A will reach the bottom of the slope first. It has a lower moment of inertia and accelerates quicker. In other words, it takes less torque to displace roll A phi radians than it does for roll B. Just like how it take less force to push a bowl 1 meter than it does to push a car 1 meter.

Each roll is experiencing the same torque (brought on by gravity) for every second it's on the slope. Roll A is being accelerated more than roll B because it has a lower moment of inertia (it's easier to spin because it's mass is all tucked in).

So it all comes down to how long it stays on the slope. The longer it's on the slope, the more rotational energy (angular momentum) it acquires. Since roll B stays on longer, it gets more energy than roll A, which translates into a farther travel.

  • 1
    $\begingroup$ No. While the rotational part of the KE of roll B is larger, that doesn't fully explain why roll B will travel farther. After all, it is moving slower than roll A. In order to answer the question "which one rolls farther" you must discuss why they don't keep rolling forever. $\endgroup$
    – Bill N
    Apr 5 '16 at 3:56
  • $\begingroup$ Granted, roll B is moving slower, but it has more energy stored, which means it will take more resistance (which translates into distance) to stop it. $\endgroup$ Apr 5 '16 at 19:43
  • $\begingroup$ Yes. You didn't say anything about resistance or stopping in your first version of your answer. If two objects are moving and don't lose their energy they both will travel the same distance $\to \infty$. Moment of inertia tells us how difficult it is to change the rotational motion, so the object with the larger moment of inertia will roll farther if both rolls are subjected to the same resistive torque. $\endgroup$
    – Bill N
    Apr 5 '16 at 20:47
  • $\begingroup$ OP points out, however, that the air resistance will be less for roll B than for roll A, so the rate of power loss is not the same for each roll. It's not obvious that B will go farther. If the main source of power loss is stickiness of the roll to the surface, then the resistance will probably be the same, and B will go farther. $\endgroup$
    – Bill N
    Apr 5 '16 at 20:50
  • 2
    $\begingroup$ It's important to realise how wrong this is: if both rolls have mass $m$ then initially they have PE $mgh$ and finally they have PE $0$. Conservation of energy (and assuming no frictional losses) says immediately that at the bottom of the slope they both have KE $mgh$. Their translational velocities will be different because one of them is storing more of this KE in rotation, but one does not have more energy than the other. $\endgroup$
    – user107153
    Apr 6 '16 at 11:11

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