- 16 May '09 23:58 / 5 edits

But it seems like you are only using translational energy when considering the kinetic energy of the wheel. If the wheel is rolling, I think there should also be a rotational energy term in the expression for kinetic energy, and this term will depend on the moment of inertia. Maybe as approximation we could treat the wheel as a disk. I think that would give something like Vmin = sqrt[(4/3)gh]. If instead we treated is as either a hoop or a thin cylindrical shell, I think it would give Vmin=sqrt(gh).*Originally posted by wolfgang59***Assuming no energy loss due to friction....**

Potential Energy gained = gMh

so initial Kinetic Energy must at least equal that

M/2 * v^2 = gMh

therefore v = sqrt(2gh)

EDIT: Also, I am pretty sure this type of approach is correct when considering, say, minimum speed for the wheel's getting to vertical height h going up some sort of ramp or something. But is this approach correct for considering getting over the rectangular bump mentioned here? My memory of what I learned in physics is for crap, but I am not yet convinced it is. If the bump height were R or greater, then the wheel shouldn't get over no matter how fast it is going (at least, I think I am correct in saying that -- or at least consider what would happen if h were much greater than R). Of course, he specifically stipulates in the problem that h<R, but my point is that your analysis approach doesn't suggest that there is anything interesting about when h approaches R, which really makes me question the approach. I am not sure, though. - 17 May '09 14:57

Not enough data. If the bump and wheel are very smooth (the friction between them is 0) then the wheel would never clear the bump. Not so?*Originally posted by PBE6***Imagine a wheel of mass M with radius R rolling along a horizontal surface at some speed V. There is a rectangular bump with height h (where h<R) built solidly into the surface. What is the minimum speed the wheel must be going in order to clear the bump?** - 18 May '09 11:37

My original post was from a tired, slightly drunk man taking a simplistic approach!*Originally posted by LemonJello***But it seems like you are only using translational energy when considering the kinetic energy of the wheel. If the wheel is rolling, I think there should also be a rotational energy term in the expression for kinetic energy, and this term will depend on the moment of inertia. Maybe as approximation we could treat the wheel as a disk. I think that would ...[text shortened]... out when h approaches R, which really makes me question the approach. I am not sure, though.**

My solution looks pretty lame when you consider h>r

But your reply made me think (which is unusual!) ... does a cylinder rolling at velocity V have more energy than a cylinder sliding at velocity V? - 18 May '09 16:23 / 1 edit

Yes.*Originally posted by wolfgang59***... does a cylinder rolling at velocity V have more energy than a cylinder sliding at velocity V?**

What is the rotational energy of a cylinder with a fixed centre?

What is the energy of a sliding cylinder with no rotation?

A rolling cylinder has the sum of energy of the above two answers. - 18 May '09 18:57

knowing PBE6, this is most likely not physics 1 material*Originally posted by wolfgang59***My original post was from a tired, slightly drunk man taking a simplistic approach!**

My solution looks pretty lame when you consider h>r

But your reply made me think (which is unusual!) ... does a cylinder rolling at velocity V have more energy than a cylinder sliding at velocity V? - 20 May '09 11:55The energy argument certainly gives you a lower bound.

*Assuming*the wheel has its mass evenly distributed, then the rotational energy is mv^2/4.

=> v = sqrt(4gh/3)

But I suspect it's more complicated than that. If h is very close to R then I suspect the wheel wouldn't necessarily make it over. Needs some more work... - 20 May '09 12:20OK, here are some further thoughts.

We need more information. There are two relevant factors:

- the friction between the wheel and the bump

- the coefficient of restitution between the wheel and the bump.

If it's frictionless, then the rotation isn't going to have an effect. We can resolve the velocities in directions parallel and perpendicular to OA (where O is the centre and A is the corner of the block).

Parallel = V cos a

Perdendicular = V sin a

[a = angle of OA with horizontal, so tan a = (R - h)/R]

Afterwards, perdendicular velocity is still V sin a.

Parallel velocity is - cV cos a (c = coefficient of restitution)

So forward velocity after collision is V [(sin a)^2 - c (cos a)^2]

Which is positive if tan a > (sqrt c)

=> h < R[1 - sqrt(c)]

Otherwise, the wheel will bounce the other way.

Note - this*still*isn't enough to show the wheel will get over (we still need to check if it will bounce high enough) but it gives us some further restrictions.

If there is some friction, then some of the rotational energy will be converted into translational energy, making it easier to get over, but we'd need to know how much. - 20 May '09 12:23

I get:*Originally posted by mtthw*

Note - this*still*isn't enough to show the wheel will get over (we still need to check if it will bounce high enough) but it gives us some further restrictions.

V >= sqrt [2gh/(1 + c^2)] . [(R - h)^2 + R^2]/[R(R - h)]

Fixing the errors in that is left as an exercise for the reader!

- 20 May '09 14:08 / 1 editIt might be easier to assume a certain height of the ramp and work out the calc for that, then apply what you learn to the generic height for the ramp.

Ramp height = half the radius of the wheel

This would mean the wheel edge would contact the ramp at the**45' angle**on the wheel (bottom right quandrant assuming left to right motion)

If the ramp height is R/2, what is the speed needed to get over the ramp?

Edit (my suspicion here is that the wheel can't get over a ramp height higher than 50% of the radius....based on nothing but gut instinct)