# Roller coaster Conservation of Energy problem

## Homework Statement

Show that on a roller coaster with a circular vertical loop, the difference in your apparent weight at the top of the circular loop and the bottom of the loop is 6 g's--that is six times your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast you go through it.

## Homework Equations

PE(1)+KE(1)=PE(2)+KE(2)
F=ma

## The Attempt at a Solution

I'm not really sure what it is that they are asking me to solve for. I set up my problem from the the bottom of the loop. The bottom of the loop is 1 and the top of the loop is 2.

KE(1)=PE(2)+KE(2)

.5mv2=2mgr+.5mv2.

The mass cancels out but I don't know how to prove that at the top of the loop a persons weight at the top of the loop is 6x greater.

Things I tried

.5mv^2=12mgr+3mv^2

2.5mv^2=12mgr

2.5v^2=12gr

I gave up on that because I really didn't see how it was useful for what I needed to find. Can someone help me please?

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Astronuc
Staff Emeritus
One must realize that the speed at the bottom v1 is greater than the speed at the top v2.

One must find v2, for which one uses the constraint the v2 is the minimum needed to maintain contact with the track, i.e. not fall. So what is v2?

Then apply the conservation of energy .5mv12=2mgr+.5mv22.

Think of centripetal force.

Ok so v22=gr.
I substitute that into my energy equation and I get:
.5mv12=2mgr+.5mgr
After all the algebra I get
v=$$\sqrt{5gr}$$
I don't see how this is a step in proving that a person's weight is 6 times their normal weight at the top of the loop. Could you explain what is happening please?

Astronuc
Staff Emeritus
At the bottom of the loop, what are the forces on the person?

Normal force and the force of gravity.
N-mg=ma
N-mg=mv2/r
N-mg=m5gr/r
N=5mg+mg
N=6mg

I see now! Thank you very much! =)