Circular motion problem ON THANKSGIVING

[SchruteBucks]

Circular motion problem...ON THANKSGIVING!

A sled starts from rest at the top of a hemispherical frictionless hill of radius R.
http://desmond.imageshack.us/Himg37/scaled.php?server=37&filename=sled.png&res=medium

a)Find the velocity of the sled at angle theta in terms of theta, R, and g.

b) At what angle does the sled leave the hill?

Using conservation of mechanical energy, for part (a) my velocity was:
U_{i} + K_{i} = U_{f} + K_{f}
mgh_{i}i + 0 = mgh_{f} + (\frac{1}{2})mv^{2}
mgR = mgRcos(theta) + (\frac{1}{2})mv^{2}
...
v^{2} = 2gR(1-cos(theta)), v=\sqrt{2gR(1-cos(theta))}

As for part b, I'm guessing that I use F=ma and try to find the angle where N (normal force) approaches zero. The only way I could think of converting the acceleration to velocity was using the uniform circular motion equation (a=v^{2}/r) but that only applies to objects at a constant speed (and this one started from rest)...so I'm stuck.

Any help would be VERY much appreciated :D

 

[Delphi51]

Your value for V looks good.
I think you can use a=v²/r for the centripetal acceleration.
And your idea of finding where the normal force is zero is good.
Carry on!
If you get in trouble, enjoy reading Feynman:
http://www.feynmanlectures.info/solutions/particle_on_sphere_sol_1.pdf

 

[SchruteBucks]

Wow...so you CAN use that circular motion equation. That blows my mind! I mean, after drawing a free body/force diagram, it looks as if the net acceleration should be nearly tangential to the circle, not towards its center. Do those types of diagrams not apply to circular motion? I guess I just have to wrap my head around the concept, but thanks a lot for the help!

 

[Delphi51]

I guess the radial acceleration is independent of the tangential acceleration.
Anyway, the centripetal force is determined by the speed.

 

[JHamm]

The free body diagram works perfectly well in all mechanics situations, you just need to decompose the force into tangential/radial components and solve for when the radial component \leq m\frac{V^2}{R}
You can find V as a function of \theta from the work energy theorem (you can find the height that the sled has moved through and only gravity does work) and the radial component of the force can also be found as a function of \theta which leaves you with only one variable :)

 

[SchruteBucks]

@JHamm:
That's what I did in my free body diagram (with the object at angle theta).
The radial components are the normal force (N) and the radial component of the object's weight(Wr), which seem to cancel each other out, and there is a tangential component of the object's weight that is unopposed (Wt), and that unopposed force (from my limited knowledge of physics) should therefore be the direction of the acceleration, since it is the direction of the net force.

Maybe a crappy lil' paint diagram will help:
http://img846.imageshack.us/img846/7765/sled.png

So basically, the net force in the diagram is tangential, but the acceleration is actually towards the center of the circle...meaning the tangential acceleration is ignored...?

@Delphi51: Thanks for all your help, the Feynman lectures website is fantastic! Possibly more helpful than going to class :O

 

[Delphi51]

The Feynman Lectures on Physics (3 volumes) themselves are great, too. They will not match your course, of course. But they can be very inspiring, giving the feeling that physics is fun and that you are really understanding it. Check Amazon for used copies.