Height of ramp given normal force in circular loop

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mastergaurang
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Homework Statement


The attachment shows the problem. The normal force at the top of the loop is 3 g's (as in g-force).


Homework Equations


PE = mgh
KE = .5mv2
Fc = [itex]\frac{mv<sup>2</sup>}{r}[/itex]
PE + KE = PE' + KE'
v = [itex]\sqrt{2gh}[/itex]


The Attempt at a Solution


The normal force would be: 30 * mass.

Since he's at the top of the loop, the formula would be:
Fnet = Fc = (mv2)/r = Fnormal + mg

I have tried to cancel the mass out and to formulate a system of equations but I think I'm missing something. One formula I got is by setting PE equal to KE (for h, but only at the bottom before the skater goes up the loop):
h = [itex]\frac{.5v<sup>2</sup>}{g}[/itex]

I am starting to think that I need the radius/velocity. Any ideas?
 

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mastergaurang said:

The Attempt at a Solution


The normal force would be: 30 * mass.

Since he's at the top of the loop, the formula would be:
Fnet = Fc = (mv2)/r = Fnormal + mg

Right so you can get v2 from this.

mastergaurang said:
I have tried to cancel the mass out and to formulate a system of equations but I think I'm missing something. One formula I got is by setting PE equal to KE (for h, but only at the bottom before the skater goes up the loop):
h = [itex]\frac{.5v<sup>2</sup>}{g}[/itex]

I am starting to think that I need the radius/velocity. Any ideas?

Yes you will need the radius since v2 will be in terms of the radius 'r'.

You are also correct in that h = ½v2/g, so you can use the v2 you found above here.