The problem statement, all variables and given/known data A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.40 m with negligible friction. When M is at the top, it has a speed of 5.27 m/s. Calculate size of the force with which the M pushes on the hoop when M is at an angle of 27.0 degrees. Picture attached at the bottom. The attempt at a solution E at the top of the circle should equal the energy at that particular part of the circle. So, mg(2r) + (1/2)mvtop2 = mg(r-rcosθ) + (1/2)mv^2 m's cancel, so g(2r) + (1/2)vtop2 = g(r-rcosθ) + (1/2)v^2 2(g(2r) + (1/2)vtop2 - g(r-rcosθ) = v^2 a = v^2/r F = ma When I plug everything in, I get 2(39.831) = v^2 v^2 = 79.66 a = 79.66/1.4 a = 56.90 F = (.52)(56.90) F = 29.59 N This answer is incorrect. Please, can someone let me know where I'm going wrong?