# Simple Energy conservation, PE = KE + PE

1. Mar 25, 2008

### Scrum

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)

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. Mar 25, 2008

### Shooting Star

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 $\theta$.