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## Homework Statement

The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 12 m, as shown in the figures. Assume that friction can be ignored. Also assume that, in order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.5 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

## Homework Equations

f

_{cp}= ma

_{cp}= (mv

^{2}) / r

K = (1/2)mv

^{2}

(1/2)mv

^{2}= mgh

## The Attempt at a Solution

If the cart is to go around safely, F

_{n}is greater than or equal to .5 mg. F

_{cp}is equal to the total forces acting on the cart at the top of the loop (I think this is where I'm going wrong...). The problem says that the normal force is (.5)500 = 250.

So I set f

_{cp}= 250 (again...I have a feeling this is wrong), so

250 = (mv

^{2})/r = 500 kg * v

^{2}/ 12

v = 2.44 m/s

I then tried plugging this into the equation mv

^{2}= 2mgh (which I used because the energy will be the same on both sides) and got:

500 kg * 6 m/s = 2 (500 kg) * 9.81 m/s

^{2}* h

h = .305 m

And this is obviously the wrong answer, which is why I'm here. =]