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Homework Help: Ball going around loop

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    a ball with:

    mass- .06722kg
    diameter- 2.53cm

    rolling down a track of length 1.26m.

    Find a) the minimum height the track must be for the ball to go around the loop. Loop has a diameter/height of .3m

    Like this:

    3. The attempt at a solution
    I did conservation of mechanical energy with just the loop by doing KE total(bottom) = PE(top) to find the initial velocity of the ball before going up loop, giving me a tangential velocity of 1.879 m/s. I used this number to put into the kinematics equation vf=sqrrt(vi^2 + 2ad), to give me an acceleration of 1.401 m/s^2. I then used dynamics F=ma to get a force of the ball going down the slope of .09418 N, used the weight force to get a Fwt = .6594 N, and set up a triangle to find theta (call it @ here). did sin@=.09418/.6594, and solved to get a @ of 8.21 deg. I then put that angle back into the first ramp, and used trig to get a height of .18 m, or 18 cm.

    If you can follow that, I have 2 questions:

    1) Did i do this problem right, or am i all wrong?

    2) When I try to solve for friction, I use the conservation of ME formula, to get PE - Ff = KE...however when I plug my numbers back in, I get a bigger number on the KE side then the PE side...which leads me to think I did this problem wrong in the first place? I've been working on this for about 1.5 hours now and can't seem to figure it out..past what I have done already. :( ...and yes I did add rotational KE to my totatl KE when I did the problem...
  2. jcsd
  3. Dec 10, 2008 #2
    OK, so at the top of the loop Fg <= Fc for it to not fall.

    So if we say...

    Fg= Fc

    Once we know the velocity it needs to reach at the top of the loop....

    Ek(at top of loop)=Eg(at top of ramp)
    1/2mv^2 = mgh
    [tex] \sqrt{gr}^2=2gh[/tex]

    .....I'm probably wrong though. It would make sense that the value needs to be greater than 0.3, simply because of conservation of energy. Good problem, interested in solution.

    Thats how I would have attacked it though
    Last edited: Dec 10, 2008
  4. Dec 10, 2008 #3
    EDIT: yea, thats why i thought i was wrong in the beginning...oh well. thanks for your attempt tho. :) I appreciate it.

    ...anyone else?
  5. Dec 10, 2008 #4
    Yeah, either way it doesn't make sense. The height has to be at least 0.3m. Hmmmm I like this.

    I think I have it 1 sec.
    Last edited: Dec 10, 2008
  6. Dec 10, 2008 #5
    Fg= Fc

    Eg(initial)= Eg(at top of ramp) + Ek
    mgh1 = 2mgr + 1/2 mv^2
    gh1=2gr + 1/2 v^2
    gh1= 2gr +1/2 rg (subbed in earlier value of v)
    h1=2r +1/2r
    h1= 2.5r
    h1= 2.5(0.15)


    The ball will have gravitational potential energy AND kinetic energy, in the first attempt, I only said it had kinetic energy. Let me know if my work is to messy and I'll latex it.

    Can anyone confirm this solution???
    Last edited: Dec 10, 2008
  7. Dec 10, 2008 #6
    OHHH...it would make sense that the ball would have BOTH KE and PE at the top of the loop! I never took that into consideration! I think you are right! :D
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