The Question is: A mass M of 5.00×10-1 kg slides inside a hoop of radius R=1.50 m with negligible friction. When M is at the top, it has a speed of 5.65 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. My workings thus far: 1st i know i have to use the radius and theta to get the distance of the mass below the hoop then i have to use the speed at the top and the conservation of mechanical energy to get the speed of the mass at theta then i make a fbd and get the normal force so using trig 1.5 cos 27= x x= 1.3365m (distance under center of hoop) distance from bottem then equals 0.1635 m as the hoop is 1.5 m in radius so using conservation of energy i get total energy top= total energy bottem potentical energy top + kE top = pot Energy bottem + Ke bottem mgh + 1/2 mv^2 = mgh+1/2mv^2 (all masses cancel out (9.81)(3) +1/2(5.65)^2 = (9.81)(0.1635) + 1/2 v^2 solve for v = 5.88223 m/s so the speed at theata is 5.88223 m/s, then using centripical force =mv^2/r we can get 23.067 N for the force. Then we draw a free body diagram and we get that Fc=Fn (what were looking for)+ Fg This is were i get stuck.. Help from here??