A disc has radius r<<R and mass m = 7.8 g. When released, it rolls on its edge without skidding on the track in the sketch. The circular section has radius R = 6.5 cm. The initial height of m above the bottom of the track is h = 39.0 cm. When the ball is rolling at the top of the circular region, what is the magnitude of the force the disc exerts on the track? (Hints: how fast is it going? and don't forget to draw a free body diagram)
There is a sketch with this that is a disc positioned on a vertical inclination at the bottom of which is a circular loop of radius R.
Answer = 0.33N
The Attempt at a Solution
So my understanding is that the disc will initially have a potential energy of mgh. This will then be converted into kinetic energy once this disc is at the bottom. Once the disc reaches the top of the circular region some of this kinetic energy will be converted into potential energy. This being mgh where in this case h=0.13m(diameter).
The remaining energy must be in the form of kinetic energy. I believe some energy would be in the rotational energy of the disc however i assume the statement r<<R is indicating its negligible.
So (0.0078*9.8*0.39)-(0.0078*9.8*0.13) = 0.5mv^2
So (2*KE)/r is the centripetal force. Where r=0.065m
This results in the answer 0.61152N
I am not too sure what i am missing. Any help is greatly appreciated.