The particle m in Fig. 40 is moving in a vertical circle of radius R inside a tr

Question

The particle m in Fig. 40 is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0. (a) What is the minimum value vm of v0 for which m will go completely around the circle without losing contact with the track? (b) Suppose v0 is 0.775vm. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path shown roughly by the dashed line. Find the angular position theta of point P.

Explanation / Answer

a ) mv^2 / R = mg at top point
energy conservation
mgh = 1/2 mVo^2 - 1/2 mv^2
g*2R = 1/2 Vo^2 - 1/2 v^2
g*2R = 1/2 Vm^2 - 1/2 (gR)
Vm = 5gR

b ) mg sin() = mv^2/R

mgh = 1/2 mVo^2 - 1/2 mv^2

g(R+Rsin() ) = 1/2 (0.775*5gR)^2 - 1/2 (Rgsin())

on solving we get = 19.53

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