1. The problem statement, all variables and given/known data A) A roller coaster has a vertical drop but followed by a loop. The total vertical height of the drop is 66m. It starts as a steep slope at an angle of 75 degrees relative to the horizontal and then enters into a circular path of radius R. In order to stay below 4 times the force of gravity at the bottom of the bend what is the minimum radius of this bottom section of the loop. B) After the bottom of the dip of radius R (from above) the track gradually transitions to the inverted (top) part of the loop. What is the force on the rider at the top of the loop if the loops height is 33 meters? Would the change fall out of the riders pockets? 2. Relevant equations Fc = mV^2/r PEg (gravitational potential energy) = mgh KE = 1/2 mv^2 Conservation of Energy : PEf-PEi = KEf-KEi 3. The attempt at a solution I apologize if what I did here seems completely wrong! -- I tried! A) Fc < 4Fg mv^2/r = 4mg v^2/r = 4g r = v^2/4g v=? EK (bottom) = PEg (top) 1/2mv^2 = mgh v = √(2x9.8m/s^2x66m) v = Approx 36m/s 36m/sxcos(75°) = 9.31 m/s (9.31m/s)^2/9.8m/s^2 = r 2.21m = r B) For this question I'm not sure whether or not to assume Fc = Fg. I found that by using the consrevation of Energy theory, I can equate the changes in Potential and kinetic energies (by subbing the v found in the previous question) to get Vf, the velocity at the top of the loop. (Right?) I'm stuck at this question, because there is no mass. Fc = mV^2/r F = mg F = ma ...I'm not too sure how to go about it. If I use the conservation of energy by equating the changes in both Kinetic/potential energy again, I was a bit stumped on how to solve for 'm'. I hope this is in the right section.