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Re: Physics Quiz 3
I wanted to think about this kind of extension after I figured out part one, but I had to learn how to drive instead.
I'm going to go very slowly... and try to not be too confused...
1. Centripetal acceleration is constant so:
Centripetal force on the wheel = (WL/g)(omega_p)^2 = Tension in the rod - Friction parallel to the rod
2. Tangential acceleration is zero, so net force perpendicular to the rod is zero.
3. And I think brndn was right in saying F=(mu_k)(W), yes?
4. And then friction parallel plus friction perpendicular would still equal tau_k/(rsintheta) if there was no wheel slip. However, there is wheel slip, which is why point 2 is true.
So, the only forces are tension and friction parallel, so friction is anti-parallel to the rod. At least I think so...?
Is there another way to get the magnitude of friction using the difference in wheel speed and rod speed?
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Last edited by Ninja_Bait : 19-11-2011 at 14:45.
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