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Conservation of Energy question

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A roller coaster car on the frictionless track shown in the figure below starts from rest at height h. The track's valley and hill consist of circular-shaped segments of radius R. Find the formula for the maximum height h_max for the car to start so as to not fly off the track when going over the hill. Give you're answer as a multiple of R. Show that when R = 10m, h_max =15m.

    http://img152.imageshack.us/img152/181/physquestion.jpg [Broken]

    2. Relevant equations

    U=mgh
    K=1/2 mv2
    K=1/2 I [tex]\omega[/tex]2
    I = mr2
    [tex]\omega[/tex] = v/r
    where [tex]\omega[/tex] is angular velocity
    3. The attempt at a solution

    I attempted to use simple substitutions for I and [tex]\omega[/tex] and then combine the three equations so as to have a total energy of 0, but I can't seem to find a proper equation that works. Any suggestions/tips?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 11, 2009 #2

    Doc Al

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    Staff: Mentor

    You need more than conservation of energy. First figure out the maximum KE it can have at the top of the hill. Hint: Newton's 2nd law applied to circular motion.

    (FYI: You don't need rotational inertia for this problem.)
     
  4. Apr 11, 2009 #3
    So, at the top of the hill, KE = mgh-mg(10) if I have the slightest clue as to what I'm doing in this question. I'm really unsure as to what to do after this however. I'm not seeing the relationship between the 2 separate circles and how they affect the starting height. I can understand the second "circle" affecting it as it creates a loss in kinetic and a gain in gravitational, but why does the first "circle" affect the height, if it even does.
     
  5. Apr 11, 2009 #4

    Doc Al

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    Staff: Mentor

    Just worry about what the maximum speed (and thus maximum KE) can be at the top of the hill (the second circle). See my previous hint.

    Once you find that max speed, then you can use conservation of energy.
     
  6. Apr 11, 2009 #5
    Ok, so using the Newton's second law, F=ma where a in this case is radial acceleration.

    a=v2/r

    From here, I get v2 = (F*r)/m so that the maximum KE = 1/2F*r?

    Am I supposed to be able to get an actual numerical value for v? or am I completely missing something about this problem.
     
  7. Apr 11, 2009 #6

    Doc Al

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    Staff: Mentor

    Excellent.

    Now analyze the forces acting on the car at the top of the hill. What criteria must be met for the car to just barely maintain contact with the road? That will tell you what F (the net force) is.
     
  8. Apr 11, 2009 #7
    So, the forces at the top of the hill acting on the car are:

    Fg - Force of gravity
    and the force caused by the radial acceleration of the car, F=mv2/r. These two forces need to equal zero for the car to maintain contact with the road at the very top of the hill so, Fnet = mg+mv = 0.

    mg = -mv2/r or, **I am slightly confused as to why I have a negative sign. I dropped it in the next equation as you can't take the sqrt of a negative number to get a real speed. My final equation seems to be correct, but the negative sign has to go somewhere. Any thoughts?

    v=sqrt(gr).

    subbing this into a conservation of energy equation.

    mgh = 1/2 mv2 + mgr gives us

    mgh = 1/2 m(gr)+mgr. m and g cancel leaving us with h = 1.5r.
    thanks Doc Al.
     
  9. Apr 11, 2009 #8

    Doc Al

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    Staff: Mentor

    The forces acting on the car are: (1) gravity, which acts down, and (2) the normal force, which acts up. So the net force F = N - mg.

    According to Newton, this net force must equal ma = mv^2/r, thus:
    N - mg = m(-v^2/r) = -mv^2/r (note that the acceleration points down, thus is negative).
    The point where the car just starts to lose contact with the road will be where N = 0, thus:
    N - mg = -mv^2/r →
    mg = mv^2/r

    (Note: mv^2/r is not a separate force acting, it's just a statement of Newton's 2nd law for circular motion.)

    You just made a sign error. See above for how the signs come out right if you're careful.

    Good!
     
  10. Apr 11, 2009 #9
    Ah, gotcha. Thanks for the help, studying for first year final and this question just had me stumped.
     
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