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**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*= 3.50

*R.*(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

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

## The Attempt at a Solution

(a)

mgh=(½ mv

gh= (½ v

v^{2}) + mgh_{A}gh= (½ v

^{2}) + gh_{A}

^{2}=2g( h- h

_{A})=2(9.8)(3.5R-2R)=29.4R

v = √29.4R

N + mg = (mv

N = m {(v

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?^{2})/RN = m {(v

^{2}/R)-(g)}N = 0.005 {(29.4R/R)-(9.8)}

N = 0.098

Similarly, I dont 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**

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