# Block sliding down frictionless semicircle

1. Dec 26, 2008

### platinumtucan

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\frac{1}{2}mv^2 + mgy = \frac{1}{2}mv^2 + mgy$$

$$\alpha=ar$$

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

3. 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.

2. Dec 26, 2008

### PhanthomJay

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

3. Dec 27, 2008

### 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...?

4. Dec 27, 2008

### Staff: Mentor

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.

5. Dec 27, 2008

### 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

6. Dec 27, 2008

### TaiwanCountry

ac=gcos(theta)