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Homework Help: Teardrop loop-de-loop radius function

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data
    To avoid this stress, vertical loops are teardrop-shaped rather than circular, designed so that the centripetal acceleration is constant all around the loop. How must the radius of curvature R change as the car's height h above the ground increases in order to have this constant centripetal acceleration? Express your answer as a function: R=R(h).

    Given variables:
    Initial radius R0
    Initial velocity v0

    2. Relevant equations
    a_c = v2/r

    v = r*ω

    ΔPE = -ΔKErotational -ΔKEtranslational

    3. The attempt at a solution
    ΔPE = mgh
    -ΔKEtranslational = (1/2)m(v02 - vf2)
    -ΔKErotational = (1/2)m(r02ω02 - rf2ωf2)

    if v=r*ω, then unless I'm mistaken, ΔKErotational = ΔKEtranslational (substitution), and therefore:

    mgh = m(v02 - vf2)

    Mass cancels out. Since centripetal acceleration is constant, I know that a_c initial = a_c final, therefore:

    vf2 = (v02*rf)/r0

    Substituting this back into the energy equation and then doing algebra,

    rf = r0((v02 - gh)/v02)

    This is incorrect, and I can't figure out where I've gone wrong?
  2. jcsd
  3. Nov 30, 2016 #2


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    Homework Helper
    Gold Member

    You are doing a really "weird" mistake, you consider the same energy two times. The car has rotational kinetic energy which in this case is the "same" as its translational kinetic energy. When I say "same" I mean its the same thing (and not two different things but equal). So in your equation of energy you should keep the factor of 1/2, that is ##mgh=\frac{1}{2}m(v_0^2-v_f^2)##.

    The rotational kinetic energy is "different thing" and not equal to the translational kinetic energy in the case we have a wheel that is rolling with translational velocity ##v## and has rotational velocity ##\omega## around its center and the additional equation ##v=\omega R## holds (that is the condition of rolling without sliding, R the radius of wheel). In this case the total kinetic energy of the wheel is ##\frac{1}{2}I\omega^2+\frac{1}{2}mv^2## where I the moment of inertia of the wheel.
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