Physics Quiz 2

[Ether]

This is a continuation of the Physics Quiz posted here (http://www.chiefdelphi.com/forums/showthread.php?t=98333) yesterday.

The weight of the motor+wheel assembly is W.

The coefficient of static friction between the wheel and floor is mu_s.

How much motor torque "tau_s" is required to start the wheel rotating?

[SavtaKenneth]

we know that when the system is static the force that the wheel exerts on the surface is F = τ/(rsinθ). The minimum force for the wheel to start spinning is F = τ/(rsinθ) > W*mu_s.

and so we get

tau_s > W*mu_s*r*sinθ

[brndn]

F of friction = μmg which is equal to the F in τ=rFsinθ and mg = W

tau_s = r*mu_s*W*sinθ

But as SavtaKenneth said, it has to be greater than in order to overcome static friction.

[Ether]

we know that when the system is static the force that the wheel exerts on the surface is F = τ/(rsinθ). The minimum force for the wheel to start spinning is F = τ/(rsinθ) > W*mu_s.

and so we get

tau_s > W*mu_s*r*sinθ

Correct. Now look at Quiz 3 (http://www.chiefdelphi.com/forums/showthread.php?t=98338).

[Ether]

tau_s > W*mu_s*r*sinθ

Notice that when θ is small, it takes very little motor torque to break the static friction and start the wheel rotating. Perhaps you can see the analogy to a skid-steer vehicle with a wide trackwidth.

As θ increases toward 90 degrees, the required motor torque steadily increases. Think of a skid-steer vehicle with a long wheelbase.

[James Critchley]

I like where this is headed.

FYI: In order for the weight to interact with the ground under the wheel and not via a reaction at the "frictionless pivot" (assumed to be a revolute joint) it must instead be a cylindrical joint. Otherwise the normal force is indeterminate (as it can have any pre-load or none at all).