Simple Energy conservation, PE = KE + PE

[Scrum]

Homework Statement

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?

Homework Equations

PE(top) = KE(point it flies off) + PE(at that point)

Arad = V^2 / R

Sum Of Forces = Mass * Acceleration

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.

 

[Shooting Star]

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.

It'll leave the surface of the sphere when the normal reaction N becomes zero.

If \theta 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\theta - 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 \theta, using geometry].

From this, you'll get \theta.