Recent content by demenius

Ok. So Fx = Fsinθ = ma_rad Fy = Fcosθ - mg = 0 (no vertical acceleration) a_rad = (4R∏^2)/T^2 R = Lsinθ F = mg/cosθ Sub into Fx (mg/cosθ)*sinθ = ma_rad a_rad = gtanθ gtanθ = (4R∏^2)/T^2 T = √((4R∏^2)/gtanθ) = 2∏√(R/(gtanθ)) Sub in Lsinθ for R T = 2∏√(Lcosθ/g)

The Speed at the bottom would be √(2gR) = 12.53. Assuming the frictional part was flat the velocity after the the frictional part would be 12.12. And the max height would be 7.499m. But if the frictional part was not flat and was part of the semi circle then I am lost on how to do it.

So would the height be equal to (v*mb/(mb+mw))^2 * 1/2g? So. (200*0.005/(0.005+1))^2 *1/2(9.81) = 0.05m?

http://imageshack.us/photo/my-images/412/freebodydiagram.png/" I found T to be equal to 2∏√(Lcos(θ)/g). But there is no m in that equation. Is it not needed? If that equation is right, then when θ increases, T would decrease.

I know how to calculate the speed at the bottom but the friction part is before the bottom and I cannot figure out how to calculate the speed just before the friction. I believe the arc length from the top to the friction section is 4∏ - 0.5m. But do not know where to go from there.

I cannot believe I did not see that. :S. Thank you.

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