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## Homework Statement

http://imgur.com/b7y3Khb

A small cart slides without friction on the loop-the-loop track, as shown, starting from rest at initial height h. The weight of the cart is W. When the cart is at the point A --- the top of the loop --- the force on the track is 0.6 W. Then what was the initial height h?

Data: W = 111.0 N; R = 16.2 m.

## Homework Equations

E

_{k}=[itex]\frac{1}{2}[/itex]mv

^{2}

E

_{u}=mgh

Centripetal force = [itex]\frac{mv^2}{r}[/itex]

## The Attempt at a Solution

If I can find the kinetic energy at the given point I can then solve for v. With v, which contains h, I can use the equation for centripetal force to solve for h.

m=w

I know the original potential energy = wgh

The potential at the point of interest is = 2Rwg

Thus the kinetic energy = (original)-(current) = wgh-2Rwg = mg(h-2R) = [itex]\frac{1}{2}[/itex]wv

^{2}

Thus v=[itex]\sqrt{2g(h-2R)}[/itex]

Plugging v into the equation for centripetal force I get

Force Centripetal = [itex]\frac{2gw(h-2R)}{r}[/itex]

I think this is where I am messing up:

The force on the object is equal to 0.6w and the only two forces acting on the object are the loop and gravity.

0.6w= [Force Centripetal]-[Force Gravity] ?

F

_{c}=0.6w+F

_{g}=> [itex]\frac{2gw(h-2R)}{r}[/itex]=.6w+wg

Solving that for h:

h=[itex]\frac{3(5gR+R)}{10g}[/itex]

Plugging in the known values I get h=24.8m which is wrong. Any suggestions?

Thank you!