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Conservation of Energy beads

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A bead slides without friction around a loop-the-loop (Fig. P8.5). The bead is released from a height h = 3.30R.


    (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?

    3. The attempt at a solution
    Having some trouble with both parts. Thanks.
  2. jcsd
  3. Mar 3, 2008 #2
    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.
  4. Mar 4, 2008 #3
    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
  5. Mar 4, 2008 #4


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    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.
  6. Mar 4, 2008 #5
    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
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