1. The problem statement, all variables and given/known data A solid sphere rolls down a hemisphere from rest. Find the angle at which the sphere loses contact with the surface. R = radius of hemisphere a = radius of sphere 2. Relevant equations ΣFr = Macm,r N-mgcosθ = -mVcm2/(R+a) N = mgcosθ - mvcm2/(R+a) eq. (1) Conservation in mechanical energy mg(R+a) = 1/2mvcm2 + 1/2Icmω2 + mg(R+a)cosθ mg(R+a) = 1/2mvcm2 + 1/2(2/5ma2)(Vcm/a)2 + mg(R+a)cosθ mg(R+a) = 7/10mVcm2 + mg(R+a)cosθ Simplified 7/10mVcm2 + mg(R+a)(1-cosθ) Vcm2 = 10g/7 (R+a)(1-cosθ) eq. (2) eq. (2) into eq. (1) N = mgcosθ - m10g(R+a)/(R+a) (1-cosθ) N= 17mgcos/7 - 10mg/7 N goes to 0 since the sphere loses contact with the surface, thus: 0 = 17cosθ-10 θ = cos-1(10/17) θ = 54° 3. The attempt at a solution My attempt solution has been trying to prove this problem experimentally. By rolling a solid sphere on a hemisphere, the angle I found was much smaller than 54°. Any thoughts?