Block sliding down frictionless semicircle

[platinumtucan]

Homework Statement

A block of mass, m, sits atop a semicircular bowl of radius, r, and the angle the radius makes with the horizontal is \theta. Find what angle, \theta, the block will slide off the bowl.

Homework Equations

\frac{1}{2}mv^2 + mgy = \frac{1}{2}mv^2 + mgy

\alpha=ar

a_{c}=\frac{v^2}{r}

The Attempt at a Solution

Drawing a free-body diagram, the forces I think at work are the normal force, gravity, and possibly a centripetal force. The resolution of gravity along a circular path is where my trouble begins. I don't think it's really equivalent to resolve gravity as the sum of two x and y vectors along an incline, since this is a circle. Maybe resolving into x=rcos\theta and y=rsin\theta and saying mgrcos\theta=X motion and Y motion is mgrsin\theta=y. Then I'm stuck how to incorporate the normal force into either of those two equations (assuming they're accurate representations of the motion of this object along this semicircle). Maybe \sum{F_{y}}=F_{n}-mgrsin\theta=mar and \sum{F_{x}}=mgrcos\theta=mar. I guess if I could by some mathematical bastardization equate mgrcos\theta=mgrsin\theta I'd get that \theta=\frac{\Pi}{4}

I hope you all can give me a little guidance so that I can solve this problem. I know it's been posted before, but I searched, and I couldn't any help toward a solution.

 

[PhanthomJay]

What's the normal force acting on the block as it slides off the bowl?

 

[platinumtucan]

so if the normal force equals zero then -mgrsin\theta=mgrcos\theta and \theta=\frac{-\pi}{4} which would have to become positive \frac{\pi}{4} since...?

 

[Doc Al]

so if the normal force equals zero then -mgrsin\theta=mgrcos\theta and \theta=\frac{-\pi}{4} which would have to become positive \frac{\pi}{4} since...?

The normal force is zero as it loses contact with the surface, but why in the world are you attempting to equate "vertical" and "horizontal" components of gravity? (At least I think that's what you're doing. To properly find the components of gravity, draw yourself a diagram and apply a bit of trig. Don't convert coordinates using x = r cosθ.)

Hint: Apply Newton's 2nd law in the radial direction.

 

[FedEx]

The most important equation is

N = 0

At theta what are the forces on the block in the radial direction? Equate them as N is zero.

You will get a relation in r g and theta and velocity. Find the velocity by energy conservation

 

[TaiwanCountry]

ac=gcos(theta)