
#1
Mar2508, 05:58 PM

P: 5

1. The problem statement, all variables and given/known data
A block mass M slides down the side of a frictionless circle Radius R. At an angle Theta the mass M flies off the circle, what is the angle? 2. Relevant equations PE(top) = KE(point it flies off) + PE(at that point) Arad = V^2 / R Sum Of Forces = Mass * Acceleration 3. The attempt at a solution Okay I actually did this one before and I was trying to do it again but somehow I don't seem to be able to get it. The answer was Inverse Cosine of 1/1.5 or 48 degrees. The problem was done with energy equations PE(top) = KE(point) + PE(point) I set 0 PE to be the middle of the circle so mgR = .5 mv^2 + mg(Rcos(theta)) mass cancels gR = .5v^2 + gRcos(theta) I think I'm going wrong here but I said the only force acting on the block is weight or mg, because at the point is leaves the normal force goes to 0. so sum forces = mass * acceleration mg = ma mass cancels g = a then v^2 / R = a , so gR = V^2 then plugging that in gR = .5v^2 + gRcos(theta) gR = .5gR + gR cos(theta) gR cancels 1 = .5 + cos(theta) and I end up with 60 degrees so I think I missed out a number somewhere but I don't know where. 



#2
Mar2508, 09:18 PM

HW Helper
P: 1,986

If [itex]\theta[/itex] is the angle made with the upward vertical with the radius at the position of the particle, then the centripetal force is given by, mv^2/r = mgcos[itex]\theta[/itex]  N, which will give you mv^2 when N=0. Also, KE = change in PE from top position, which gives, mv^2/2 = mg(...) [you find it, in terms of r and [itex]\theta[/itex], using geometry]. From this, you'll get [itex]\theta[/itex]. 


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