• nagyn
In summary: When we have the wheel rolling without sliding the energy equation is ##mgh=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2## (the same equation if the wheel was sliding on the surface with velocity ##v=v_0## without rotational motion), and the condition ##v=\omega R##. So the kinetic energy of the wheel in this case is not the sum of rotational and translational kinetic energy, but something in between.In summary, when designing vertical loops, the radius of curvature must change as the car's height above the ground increases in order to maintain a constant centripetal acceleration. The energy equation in this scenario involves the mass, height, and initial and final
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 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 - 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?

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.

## What is a Teardrop loop-de-loop radius function?

A Teardrop loop-de-loop radius function is a mathematical formula used to calculate the radius of a loop-de-loop shape, commonly used in roller coasters and other amusement park rides.

## How is the Teardrop loop-de-loop radius function calculated?

The Teardrop loop-de-loop radius function takes into account the velocity of the object, the force of gravity, and the centripetal force required to keep the object in a circular motion. It also considers the angle and height of the loop-de-loop.

## What factors affect the Teardrop loop-de-loop radius function?

The Teardrop loop-de-loop radius function is affected by the velocity of the object, the force of gravity, the angle and height of the loop-de-loop, and the mass of the object. Changes in any of these factors can alter the required radius.

## What is the purpose of the Teardrop loop-de-loop radius function?

The Teardrop loop-de-loop radius function is used to ensure the safety and stability of roller coasters and other amusement park rides. It helps engineers and designers determine the appropriate size and shape of the loop-de-loop to prevent any accidents or discomfort for riders.

## Can the Teardrop loop-de-loop radius function be applied to other objects besides roller coasters?

Yes, the Teardrop loop-de-loop radius function can be applied to any object that follows a circular path, such as a car going around a curve or a satellite orbiting the Earth. The same principles of centripetal force and velocity apply in these situations.

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