Why is the Height at Point A Considered to be 2R in a Loop-the-Loop Problem?

[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

Homework Equations

(a) Potential Energy( at height 'h')=Kinetic Energy( at A)+Potential Energy( at A)

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

The Attempt at a Solution

(a)​

mgh=(½ mv²) + mgh_A
gh= (½ v²) + gh_A
​

v²=2g( h- h_A )=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 don't at all understand the equation for part(b) of the problem. Please explain your solutions with inline comments.

Thanks in advance! :)https://www.dropbox.com/s/wsd8g5q9d87undj/Loop%20Quest.png?dl=0​

 

[Doc Al]

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

 

[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

Homework Equations

(a) Potential Energy( at height 'h')=Kinetic Energy( at A)+Potential Energy( at A)

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

The Attempt at a Solution

(a)​

mgh=(½ mv²) + mgh_A
gh= (½ v²) + gh_A
​

v²=2g( h- h_A )=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 don't at all understand the equation for part(b) of the problem. Please explain your solutions with inline comments.

Thanks in advance! :)https://www.dropbox.com/s/wsd8g5q9d87undj/Loop%20Quest.png?dl=0​

 

[Doc Al]

My Professor has suggested the above solution but I want to know how can you take the height at point A to be 2R?

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

Similarly, I don't at all understand the equation for part(b) of the problem.

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