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Loop-the-Loop and velocity problem

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  1. Jan 23, 2015 #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 10.1 m from the bottom of the loop-the-loop which has a radius 3 m. The acceleration of gravity is 9.8 m/s^2 .

    What is its speed at the top of the loop? Answer in units of m/s.

    2. Relevant equations
    w=Fd
    PE=mgh
    KE=1/2mv^2

    3. The attempt at a solution
    My first thought was to find the PE of the ball at the top, since the KE and the PE are always going to add up to that number, no matter where the ball is on the track. I did mgh = 1/2mv^2 = m(9.8)(10.1) = 98.98m

    So the PE = mass times 98.98

    then i did 98.98m=1/2mv^2 to get v=14.06982587, then i divided that by two since the ball is at the top of the loop, it's halfway through it i thought to get 7.034912935, which was wrong...

    So what am I doing wrong? Any help would be greatly appreciated
     
  2. jcsd
  3. Jan 23, 2015 #2

    mfb

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    I don't understand why you divided the speed by 2. How does it matter which fraction of the loop the bead made so far? Do you expect it to magically stop at the end, for example?
    Your general approach allows to calculate the speed for every height, if you adjust the height difference accordingly.

    Please do not drop units in the calculations, they are important tools to check what you are doing and leaving them out is simply wrong.
     
  4. Jan 23, 2015 #3

    haruspex

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    There's no basis for that step.
    Forget about the bottom of the loop. Just compare energies at start with energies at top of loop.
     
  5. Jan 23, 2015 #4
    Ah thank you haruspex i appreciate your help, i got the answer. Thank you for not being sarcastic and rude
     
  6. Jan 23, 2015 #5

    haruspex

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    Oh well, there's always next time.
     
  7. Dec 30, 2016 #6
    What do you mean by comparing the two energies?
     
  8. Dec 30, 2016 #7

    mfb

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    Something went wrong with the sign of the gR term. A larger radius will lead to a smaller velocity.

    Giving full solutions is against the forum rules, but this thread is two years old - just let it rest.

    Edit: Merged some posts.
     
    Last edited: Dec 30, 2016
  9. Dec 30, 2016 #8
    sorry

    for the mistake in sign
     
    Last edited by a moderator: Dec 30, 2016
  10. Dec 30, 2016 #9

    haruspex

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    PE+KE at release point versus PE+KE at top of loop.
     
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