# Block sliding down frictionless semicircle

## 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
Homework Helper
Gold Member
What's the normal force acting on the block as it slides off the bowl?

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
Mentor
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.

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

ac=gcos(theta)