Conservation of Energy problem

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1. Apr 30, 2015

Dr. Who

Question:-
A bead slides without friction around a loop-the-loop (Please click on the below link for diagram). The bead is released from a height h = 3.50R. (a) What is its speed at point A? (b) How large is the normal force on it if its mass is 5.00 g?
https://www.dropbox.com/s/wsd8g5q9d87undj/Loop Quest.png?dl=0

2. Relevant equations
(a) Potential Energy( at height 'h')=Kinetic Energy( at A)+Potential Energy( at A)

(b) Normal force + Weight of the bead = Centripetal force​

3. The attempt at a solution
(a) ​
mgh=(½ mv2) + mghA
gh= (½ v2) + ghA
v2=2g( h- hA )=2(9.8)(3.5R-2R)=29.4R

v = √29.4R
(b)

N + mg = (mv2)/R
N = m {(v2/R)-(g)}
N = 0.005 {(29.4R/R)-(9.8)}
N = 0.098
My Professor has suggested the above solution but I want to know how can you take the height at point A to be 2R?
Similarly, I dont at all understand the equation for part(b) of the problem. Please explain your solutions with inline comments.

Last edited by a moderator: Apr 30, 2015
2. Apr 30, 2015

Staff: Mentor

Can you attach the diagram? (For those of us who cannot access dropbox.)

3. Apr 30, 2015

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4. May 1, 2015

Staff: Mentor

Point A is at the top of the circle of radius R. Height is measured from the bottom of the circle.

It's just an application of Newton's 2nd law.