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Imagine you are biking, and you come across a ledge. I want to figure out what minimum velocity do you need to bike at to overcome a ledge, of some height $H$.

Roughly, it looks something like this: enter image description here

The wheel is essentially a thin ring, with mass $M$, radius $R > H$, velocity (moving rightward) $v$, and angular velocity $\omega$. We know that the wheel is purely rolling so $v=\omega r$, as long as it is on the rough surface.

I know that energy at the bottom has to be more than the gravitational potential energy of the wheel at the top (the limiting case is when the wheel has no velocity when it mounts the edge). Therefore, $$\frac{1}{2}I\omega ^2 + \frac{1}{2} Mv^2 = MgH$$ where $I$ is the moment of inertia.

I do not think this all you need to know, though. As in you could be going really fast, but a bike won't be able to climb the Empire State Building. Can we conserve angular momentum about the point of contact between the ledge and the wheel? Does the normal force do any work on the wheel? How do you go about finding the minimum velocity required to climb the ledge?

bad_chemist
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  • I think that you're right in that more than just energy has to be considered. At the point in time that the wheel makes contact with the ledge, you basically have a lever problem with the point of contact between the wheel and ledge acting as the fulcrum. You should be able to see that if the height of the ledge is greater than the radius of the wheel, then there's no way to have a clockwise torque with respect to the fulcrum and so the wheel can never start to climb over the ledge. If the ledge is smaller than that, then it should be possible if the wheel has sufficient momentum. –  Oct 25 '18 at 18:58
  • One of the issues here is that when the wheel hits the ledge, its momentum will change in a manner that is difficult to predict, since the direction of the "normal" force depends on a lot of factors. – V.F. Oct 25 '18 at 19:30

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