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Conservation of energy - velocity problem

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A bead slides without friction around a loop-
    the-loop. The bead is released from a height
    of 26.7 m from the bottom of the loop-the-
    loop which has a radius 9 m.
    The acceleration of gravity is 9.8 m/s2 .

    http://img31.imageshack.us/img31/2377/19550530.png [Broken]

    What is its speed at point A?
    Answer in units of m/s.

    a of grav = 9.8 m/^2
    released from: 26.7 m
    loop radius 9 m

    2. Relevant equations

    PE = mgh
    Ke = .5mv^2

    3. The attempt at a solution

    Ok so this is what I did. I set the PE and KE equations equal to each other so mass can cancel out cuz its not given.

    MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)
    To get height i did 26.7-9= 17.7

    Then I plugged it into the equation:
    v = sqrt(2 * 9.81 * 17.7)
    = 18.64 m/s

    I submitted the answer and it was wrong still!!!

    omg, someone please explain what am i doing wrong! physics test soon :(
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 6, 2009 #2


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    What is the total energy of the system?

    At point A, what types of energy does the bead have?
  4. Dec 6, 2009 #3
    Wouldn't KE be 0? I'm not sure how to do this one..I'm really confused
  5. Dec 6, 2009 #4


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    nope. KE is not zero, so what is an expression for the energy at point A?
  6. Dec 6, 2009 #5

    Velocity is 0 at that point becuse its the maximum
  7. Dec 6, 2009 #6


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    No. The velocity is not zero. At point A, there is are two types of energy present. The initial energy is converted into these two types.
  8. Dec 6, 2009 #7
    what about the rotational kinetic energy?
  9. Dec 6, 2009 #8
    Potential Energy and Kinetic Energy yes????Thats why I set them equal to to cancel out mass. But its still incorrect.
  10. Dec 6, 2009 #9
    i think if the object is ROTATING ( because i havent understood if it is sliding means rotating or not) then
    PE= linear KE+ Rotational KE.
    maybe this can help us!
    i am not sure anyway..we had just learned about it.
  11. Dec 6, 2009 #10


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    Good, at A there is mgh+ 1/2 mv2 put this equal to the initial energy now.
  12. Dec 6, 2009 #11
    MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)

    wouldn't that be correct?
  13. Dec 6, 2009 #12


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    you'd have

  14. Dec 6, 2009 #13
    But we'd still have 1 mass left in the equation. Or do all 3 of them cancel out x_x
  15. Dec 6, 2009 #14


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    one bead...all the same mass :smile:
  16. Dec 6, 2009 #15
    I dont understand what you mean lol. It doesn't give us a mass. So why use mass in the equation when not even all of them cancel out. How am I supposed to solve it with that equation. I'm confused xD
  17. Dec 6, 2009 #16
    This discussion is getting too cluttered when you're already on the right path. The only mistake you made here is that you got the height wrong. It should be 26.7 - 18 = 8.7 meters. Remember, you want the height difference between your original point and your final point (so if radius is 9, the diameter of the hoop is 18). You should get 13.1 m/s.

    That's the easy way to do it. The harder way would be to calculate the velocity of the particle at the bottom with the equation v = sqrt(2 * 9.8 * 26.7). Then plug that into the kinematics equation: v(final)^2 = v(initial)^2 + 2ax, where v(initial) is what you calculated in the previous equation. Solve for v(final) and that would also give you the same answer.
    Last edited by a moderator: May 4, 2017
  18. Dec 6, 2009 #17
    :eek:, thanks a ton man. I think this was a trick question....Since I never thought of the diameter. It makes sense why its 26.7-18 instead of 9.

    Thanks for clearing it up, I was really starting to get a little confused there.
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