- #1

nagyn

- 26

- 0

## 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 R

_{0}

Initial velocity v

_{0}

## Homework Equations

a_c = v

^{2}/r

v = r*ω

ΔPE = -ΔKErotational -ΔKEtranslational

## The Attempt at a Solution

ΔPE = mgh

-ΔKEtranslational = (1/2)m(v

_{0}

^{2}- v

_{f}

^{2})

-ΔKErotational = (1/2)m(r

_{0}

^{2}ω

_{0}

^{2}- r

_{f}

^{2}ω

_{f}

^{2})

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

mgh = m(v

_{0}

^{2}- v

_{f}

^{2})

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

v

_{f}

^{2}= (v

_{0}

^{2}*r

_{f})/r

_{0}

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

r

_{f}= r

_{0}((v

_{0}

^{2}- gh)/v

_{0}

^{2})

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