- #1

B3NR4Y

Gold Member

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## Homework Statement

A roller-coaster car initially at a position on the track a height h above the ground begins a downward run on a long, steeply sloping track and then goes into a circular loop-the-loop of radius R whose bottom is a distance, d, above the ground in the figure. Ignore friction. At what minimum value of the starting height h above the ground must the car begin its journey if it is to remain on the track at the top of the loop?

## Homework Equations

a=v

^{2}/r

Energy is conserved

## The Attempt at a Solution

The first parts of the problem had me calculate the speed at the bottom of the loop, a quarter of the way up the loop, the normal force at the bottom and the normal force at a quarter of the way up. I got the following answers:

[itex] v_{bottom} = \sqrt{2gh-2gd} [/itex] I got this by energy conservation. [itex] F^{N}_{bottom} = m\frac{2gh-2gd}{R}+mg [/itex] I got this by circular motion, where I know the acceleration is given by a=v

^{2}/r, which would be the sum of the forces in the y direction, I just solved for the normal force. [itex]v_{\frac{1}{4}} = \sqrt{2gh-2g(d+R)} [/itex] I got this by geometry and energy conservation. The normal force at that position is [itex] F^{N}_{\frac{1}{4}} = m\frac{2gh-2g(d+R)}{R} [/itex]. Using these I calculated the Normal force at the top of the loop and the velocity at the top of the loop. I got [itex] v_{top} = \sqrt{2gh-2g(2R+d)} [/itex]. The Normal force is [itex] F^{N}_{top} = mg - \frac{2gh-2g(2R+d)}{R} [/itex]. I'm not sure where to start. For it to stick to the top, it would have to have enough energy to make it to the top of the loop, so I tried an answer of 2R+d, but that's wrong. I just need a hint, really.