Teardrop loop-de-loop radius function

[nagyn]

Homework Statement

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

Homework Equations

a_c = v²/r

v = r*ω

ΔPE = -ΔKErotational -ΔKEtranslational

The Attempt at a Solution

ΔPE = mgh
-ΔKEtranslational = (1/2)m(v₀² - v_f²)
-ΔKErotational = (1/2)m(r₀²ω₀² - r_f²ω_f²)

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

mgh = m(v₀² - v_f²)

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

v_f² = (v₀²*r_f)/r₀

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

r_f = r₀((v₀² - gh)/v₀²)

This is incorrect, and I can't figure out where I've gone wrong?

 

[Delta2]

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.