# Gravitational potential and kinetic energy

1. Feb 23, 2009

### moy13

1. The problem statement, all variables and given/known data

A car in an amusement park ride rolls without friction around the track shown in the figure

http://session.masteringphysics.com/problemAsset/1041727/8/YF-07-32.jpg

It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?

2. Relevant equations

Ugrav1 + K1 = Ugrav2 + K2
circumference of a circle = 2*pi*R

3. The attempt at a solution

I thought that since there is no friction, energy is conserved so the energy used to roll up and down the loop is the energy transformed from gravitational potential energy to kinetic energy. So, I thought the height should be 2*pi*R to make it around the loop, but it is wrong. So, how can I decide what the minimum value of h should be?

2. Feb 23, 2009

### LowlyPion

You're right that PE becomes ½mv² but just getting the car back to h means that it has 0 speed, which isn't enough to hold it in contact with the tracks.

Hence you need mv²/r = mg at that point in order to maintain contact.

With that in mind work your PE to ½mv² magic.

3. Feb 23, 2009

In addition to using the conservation of energy you need to consider the forces on the car.There are two forces on the car at the top of the loop.Try sketching it out and marking in these forces.One of these forces remains constant,which one is it?What is the second force and what happens to this when the speed of the car reduces?Sorry LowlyPion you beat me to the click.I will leave this here anyway.

Last edited: Feb 23, 2009
4. Feb 25, 2009

### moy13

mv^2 / r = mg, then v^2 = rg where r = R

so I get mgh = mg(2R) + (1/2)mRg

mg cancels out and h = 2R + (1/2)R = (5/2)R

is this correct? I don't really understand.

Thanks for any help though.

5. Feb 25, 2009

### LowlyPion

mgh is your potential energy. You've got that.

Now at any point you will have PE and KE. (At the bottom PE is 0 and it's all KE.)

But now you have the additional constraint that the car stays on the track at the top of the loop.

At the top there is centripetal acceleration outward (up) and there is mg (down) and they need to add to 0. If centripetal acceleration is at least mg then it stays on the rail and completes the loop. Well you know then that mv²/r = mg. Or mv² = mg*R

Since then 1/2mv² is your kinetic energy and since at the top of the loop you also have the mg2R of potential energy then you have 2R + 1/2R as you found.

6. Feb 25, 2009

### moy13

oh wow, that does make sense, thank you LowlyPion.