I was trying to figure out what would happen if we rolled a disc up a smooth inclined plane (considering pure rolling). I think the translation component of velocity of the sphere would die out due to gravity, but would the disc keep spinning? In the same direction? With what tangential velocity? The same velocity the center of mass had when we rolled it up? Then when it translates down, would it keep spinning it it's initial direction, while moving down? I can't seem to figure this out. What would the sequence of events be as the sphere rolled up and subsequently rolled down?

  • $\begingroup$ Are you assuming that the plane is frictionless? Or are you assuming a "rolling without slipping" situation? The results would be different in each case. $\endgroup$ Dec 8, 2017 at 15:02
  • $\begingroup$ Assuming no friction, there's no torque to change angular velocity, so the disc would keep spinning. It's velocity (translational) would of course decrease, assuming the object has mass. Rotational and tranlational motions are of course independent. $\endgroup$
    – Gert
    Dec 8, 2017 at 15:04

1 Answer 1


Let us consider two (extreme) scenarios. Lets say we got the cylinder rolling on a horizontal plane, with friction coefficient $\mu_1$ sufficiently high so that pure rolling (no slippage) occurs. The disc's velocity is $v$ and it's angular velocity is $\omega=\frac{v}{R}$.

Its total kinetic energy will be:

$$K=\frac12 mv^2+\frac12 I\omega^2$$

Then the disc starts rolling up the slope (we ignore the little bump it will experience).

Rolling disc

Scenario 1:

The incline has a friction coefficient so that $\mu_2\geq \mu_1$. In that case it will continue to roll without slipping.

Because the friction force does no work, energy is conserved and the disc will decelerate going up the incline to height $h$ where $v=0$, so that:

$$\Delta U=\Delta K$$


$$\frac12 mv^2+\frac12 I\omega^2=mgh$$

Both $v$ and $\omega$ vanish at $h$.

It will then start to roll back down, until at $h=0$ the original values of $v$ and $\omega$ are restored.

Scenario 2:

The incline provides no friction whatsoever ($\mu_2=0$).

Without friction, no torque is exerted on the disc and its angular velocity remain constant. Rotational kinetic energy $K_r=\frac12 I\omega^2$ remains constant. The disc keeps spinning.

The new energy conservation equation then becomes:

$$\frac12 mv^2=mgh$$

In practice this means that the disc will end up less high (smaller $h$) before its translational speed has completely vanished, compared to the first scenario. But even at that point it will still spin at the initial angular velocity $\omega$.

Then, still spinning clockwise, it will start sliding back down the slope until its translational kinetic energy is restored.

When it 'hits' the horizontal plane it is spinning in the 'wrong direction'. Friction then causes torque and this torque will gradually reverse the direction of rotation, causing also a decrease in translational speed. Here energy conservation must also take into account the work done by the friction force.

  • $\begingroup$ Thanks! Love the response! (It's my first time asking a question) Another doubt ... as the disc starts rolling down the frictionless incline, will the velocity of the top most point of the disc be in the upward direction (i.e., up the incline?), or not? Won't this disrupt the rolling? And what'll be the velocity of the point in contact with the incline? I can't seem to figure out the velocities of different points on the disc, and how it'll affect the rolling ... $\endgroup$ Dec 8, 2017 at 17:38
  • $\begingroup$ Hi! The translational and rotational motions are completely independent of each other. So the information you're requesting just isn't very useful. In reality a point on the perifery of the disc moves like a cycloid: en.wikipedia.org/wiki/Cycloid but that too isn't useful info here. $\endgroup$
    – Gert
    Dec 8, 2017 at 18:17
  • $\begingroup$ Think for example of rifling (en.wikipedia.org/wiki/Rifling) It confers upon the bullet also a rotational motion. But that doesn't cause the bullet to move upwards or sideways or whatever. These motions are independent. One could assign a complicate position or velocity vector to each point of the bullet but it simply wouldn't be useful, really... $\endgroup$
    – Gert
    Dec 8, 2017 at 19:00
  • $\begingroup$ But what if I just asked the direction of velocity of the top most or bottom most point? Would it be up the plane or down the plane, or in a different direction altogether? $\endgroup$ Dec 9, 2017 at 3:03
  • $\begingroup$ I think you're overthinking. The question doesn't make a lot of sense to me. Sorry. $\endgroup$
    – Gert
    Dec 9, 2017 at 14:01

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