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Homework Help: Loop the loop problem

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    So, this problem is about energy. The question is: what must the minimum height (in R) of a height be, for a ball to go down it, through a loop the loop (with radius R), and then through a rough surface 6R's long with u= 0.5.

    2. Relevant equations
    gravitational potential energy= mgh
    kinetic energy= 1/2mv^2
    work of friction= umgs

    3. The attempt at a solution
    Hopefully this is right: I did initial potential energy= mgh= 1/2mv^2 (kinetic needed to get past the top of the loop) + umgs (work of friction of the surface) + mgh (potential energy at top of loop). Also, I got v^2=Rg, since using the centripetal force equation mv^2/R, I made it equal to the normal force, mg, and v^2=Rg.
    The mg's cancel out, so you're left with h=1/2R + (0.5)6R + 2R, which equals 5.5R. Is this right??
  2. jcsd
  3. Nov 12, 2008 #2
  4. Nov 13, 2008 #3
    [tex]m\vec{g}NR-\vec{W}=0[/tex]. Solve. So long as [tex]N>\frac{5}{2}[/tex], you're fine (because that's the min. height for an object to go through any loop-the-loop).
  5. Nov 13, 2008 #4


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    Homework Helper

    There are 2 constraints that need to be met aren't there?

    One is the height must be sufficient for the golf ball to stay in the loop the loop. The other is that you must start with enough potential energy to make it through the rough patch. Consider them separately.

    Figure first then the work to get through the rough patch.
    W = f * d = u*mg*d = .5*6R*mg = 3R*mg
    So long as PE > W it will make it right? So Constraint 2 is mg*h > mg*3R or h > 3R

    What about the loop the loop?
    3R*mg = mV2/2 means V2 will be at least 6*R*g.

    At the top of the loop the outward force of the mv2/R must be greater than the weight mg.
    mV2/R > mg ? Well substitute the necessary V2 for the rough patch of constraint 2.

    m*6*R*g/R = 6*mg > mg ?

    If constraint 2 is met then is that sufficient?
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