What is the speed and normal force of a bead sliding around a loop-the-loop?

[Sheneron]

Homework Statement

A bead slides without friction around a loop-the-loop (Fig. P8.5). The bead is released from a height h = 3.30R.

http://img530.imageshack.us/my.php?image=p815ye0.gif

(a) What is its speed at point A? Answer in terms of R and g, the acceleration of gravity.
(b) How large is the normal force on it at this point if its mass is 4.40 g?

The Attempt at a Solution

Having some trouble with both parts. Thanks.

 

[username2]

Think of this problem as a falling loop, there is no friction, so just treat the loop as if it has rolled down and had its center of mass accelerated at 9.8m/s^2 down a wall height h. now simply use the equations for potential and kinetic energy and you're done.

 

[Sheneron]

Alright I think I got A.

mgy=1/2 mv^2 so g(3.3-2)=1/2v^2

so that leaves, v= (g2.6R)^.5

But still need help with B

 

[Dick]

Draw a force diagram at point A. The forces acting on the particle are mg (down) and normal force (down). The acceleration is v^2/r. F=ma.

 

[Sheneron]

Excellent, I found the answer.

Fn = ma - mg

Fn = m(2.6gR)/R - mg
Fn = .0044(2.6*9.8) - .0044(9.8) = .06899 N

Thanks for the help