# Homework Help: Conservation of Mechanical Energy and Centripetal Acceleration Problem

1. Sep 28, 2008

### ctpengage

Show that 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 circular 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 your go through it.

My working for the first half of the problem, the 6g's part is as follows

Hieght from which it is released is h

The speed at bottom of the loop is determined by the conservation of mechanical energy
1/2 mvbottom2=2mgh

Apparent weight at the bottom of the loop is obtained by the below:

mvBot2= FNorm. Bot.-mg
Therefore apparent weight at bottom is
FNorm. Bot.=mvbot2/R+mg
FNorm. Bot.=2mgh/R+mg (using result obtained via conservation of energy)

To find speed at top of the loop we have from Conservation of Energy
1/2 mvtop2+mg(2R)=mgh
mvtop2=2mg(h-2R)
Therefore using the above the apparent weight at the top of the loop is

mvTop2/R = FNorm. Top.+mg
Therefore Apparent weight is :
FNorm. Top. = (2mg(h-2R))/R - mg

Hence
FNorm. Bot. - FNorm. Top. =
2mgh/R + mg - [((2mg(h-2R))/R - mg)]=
2mgh/R + mg - 2mgh/R + 4mg + mg=
6mg

That's how i proved the first section. Can anyone please tell me how to complete the problem, namely proving that as long as your speed is above the minimum needed, the answer doesn't depend on the size of the loop or how fast your go through it. This part of the problem is relaly bugging me and I've tried heaps of ways but can't come up with a definitive, good answer

Last edited: Sep 28, 2008
2. Sep 28, 2008

### ctpengage

the minimum speed needed is solved by mg = mv2 / R
Therefore the minimum speed in this instance is the square root of (gr)

But i can't find a way to tie this into the problem and prove the last part

3. Sep 28, 2008

### ctpengage

4. Sep 28, 2008