Wheels and Friction Part 3

I want to finish this topic and move to one of the gazillions of other topics I want to discuss. I'm supposed to talk qualitatively about some examples of how a wheel under friction can reach steady state. Let's start with a wheel with some arbitrary linear and rotational velocities and consider what happens when one is greater than the other.

When the forward linear velocity is greater than the backward rotational velocity of the wheel's bottom, it has a net forward velocity relative to ground, so friction acts backward on the wheel. Backward friction decreases the forward linear velocity while generating a torque that increases the backward rotational velocity until the two velocities are equal. Vice versa for when the forward linear velocity is less than the backward rotational velocity.

And that's basically it - simple right? No need for nauseating math. But let me go a teeny bit further and highlight some cases of interest to maybe bowlers and billards players.

Case 1: 0 rotational velocity (no spin). You get this state by tossing the wheel forward without spin. Then friction rotates the wheel while slowing it down until steady state is reached.

Case 2: backward rotational velocity (top spin) - toss the wheel forward while brushing forward over its top. If the spin is less than the forward linear velocity, friction increases the spin while slowing the wheel. The wheel is slowed less than if it had no spin to start with. On the other hand, if the spin is larger than the forward linear velocity, friction decreases the spin while speeding up the wheel. So use top spin if you want more speed!

Case 3: forward rotational velocity (back/under spin) - toss the wheel forward while brushing backward over its top. Here, friction decreases both the spin and the speed! As will be shown in the appendix (for the more technical), if the rotational velocity is larger in magnitude than twice the linear velocity, friction will decrease the linear velocity so much that the wheel eventually moves backward in steady state. If you're good enough to pick a forward rotational velocity exactly twice the forward linear velocity, friction decreases the linear velocity just the right amount to stop the wheel in steady state!

Finally I want to clarify a point about the wheel's energy. Although energy is conserved in steady state, it isn't when the wheel is getting to steady state. In that case, its bottom moves relative to ground so friction does work against it, and energy is lost at a rate given by the product of its net velocity and friction. Unfortuately, you can't win against friction here, and striking even is only possible if you start exactly in steady state.

I'm going to modify the wheel demo, at some point, to let you choose non-zero initial velocities. With that feature, you can look at the 3 types of spin without having to figure out what force you need to apply to get the wheel to that state. Other than that, I hope this trilogy has helped clear any confusion you might have had about wheels and friction, and I look forward to starting my next "adventure"! Questions please!

Appendix:
To prove the claim in case 3, look back to part 2 to find

\theta''[T] = -2X''[T]

Integrating from T=0 to steady state, where ɵ'[T] = X'[T],

\theta'[T] - \theta'[0] = 2(X'[0]-X[T])

or

X'[T] = \frac{2X'[0] + \theta'[0]}{3}

which is negative if ɵ'[0] < 0 and |ɵ'[0]| > 2X'[0].