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Conservation of Energy problem

  1. Apr 30, 2015 #1
    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.

    Thanks in advance! :)https://www.dropbox.com/s/wsd8g5q9d87undj/Loop%20Quest.png?dl=0
     
    Last edited by a moderator: Apr 30, 2015
  2. jcsd
  3. Apr 30, 2015 #2

    Doc Al

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    Staff: Mentor

    Can you attach the diagram? (For those of us who cannot access dropbox.)
     
  4. Apr 30, 2015 #3
     

    Attached Files:

  5. May 1, 2015 #4

    Doc Al

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    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.
     
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