1. The problem statement, all variables and given/known data A small disc, radius r and mass m = 7.9 g, rolls on its edge. The friction with the track is enough to prevent slipping. When released, it rolls down the track (sketch) and reaches a circular section with radius R = 5.1 cm, which is very much greater than r. The initial height of m above the lowest point of the track is h = 30.4 cm. The disc reaches the top of the circular part of the track. When it does so, 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) 2. Relevant equations ½mvi2+mghi = ½mvf2+mghf Fc=mv2/r 3. The attempt at a solution The initial height is 30.4cm from the bottom while the final height is 10.2cm from the bottom, therefore the change in height is 20.2cm = 0.202m At the top, the force of the disk on the track would be the centrifugal force pushing it against the track which is equal to the centripetal force Subbing into conservation of energy to get the velocity: Initial v is 0, therefore: mghi = ½mvf2 + mghf divide by m and rearrange v = √2*g*(hi-hf) sub in values v = √2*9.8*0.202 = 1.98977 subbing it into mv2/r: (0.0079*1.989772)/0.051 = 0.61N What did I do incorrectly?