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Mass slides on loop-the-loop track, find original height.

  1. Sep 25, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/b7y3Khb

    A small cart slides without friction on the loop-the-loop track, as shown, starting from rest at initial height h. The weight of the cart is W. When the cart is at the point A --- the top of the loop --- the force on the track is 0.6 W. Then what was the initial height h?
    Data: W = 111.0 N; R = 16.2 m.


    2. Relevant equations

    Ek=[itex]\frac{1}{2}[/itex]mv2
    Eu=mgh
    Centripetal force = [itex]\frac{mv^2}{r}[/itex]

    3. The attempt at a solution
    If I can find the kinetic energy at the given point I can then solve for v. With v, which contains h, I can use the equation for centripetal force to solve for h.

    m=w
    I know the original potential energy = wgh

    The potential at the point of interest is = 2Rwg

    Thus the kinetic energy = (original)-(current) = wgh-2Rwg = mg(h-2R) = [itex]\frac{1}{2}[/itex]wv2

    Thus v=[itex]\sqrt{2g(h-2R)}[/itex]

    Plugging v into the equation for centripetal force I get

    Force Centripetal = [itex]\frac{2gw(h-2R)}{r}[/itex]

    I think this is where I am messing up:

    The force on the object is equal to 0.6w and the only two forces acting on the object are the loop and gravity.

    0.6w= [Force Centripetal]-[Force Gravity] ?

    Fc=0.6w+Fg => [itex]\frac{2gw(h-2R)}{r}[/itex]=.6w+wg

    Solving that for h:

    h=[itex]\frac{3(5gR+R)}{10g}[/itex]

    Plugging in the known values I get h=24.8m which is wrong. Any suggestions?

    Thank you!
     
  2. jcsd
  3. Sep 25, 2013 #2

    TSny

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    Gold Member

    Mass and weight are different concepts.

    Note how in your final answer: h=[itex]\frac{3(5gR+R)}{10g}[/itex] you have 5gR added to R. But these quantities have different dimensions (or units). This shows something is wrong and it's due to taking m = w.

    Otherwise, your work looks good.
     
  4. Sep 25, 2013 #3

    collinsmark

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    I'm not following you there. If w is the weight, then w = mg.

    You threw an extra g in there, but g is included in the weight. It's not necessary to put it in there again.

    original potential energy = wh

    2Rw

    I would say,

    w(h - 2R) = ½(w/g)v2.

    Your expression for v is correct. :smile:

    You have an extra g in there.

    And that's where things make a difference, when treating weight as mass.

    So far you've been equating mass and weight, and it comes to bite back here. If you are treating w as mass, then 0.6w is also a mass, and wg is a force. You're mixing mass and force.

    Instead, treat w, the weight, as a force; don't treat it as a mass.

    That gives you, on the right hand side of the equation, 0.6w + w

    I'll let you take it from there.

    [Edit: TSny beat me to the response.]
     
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