What would happen if we roll a disc up a smooth surface?

Question

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?

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).

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:

$$\Delta U=\Delta K$$

Or:

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

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.