- #1
Adriano25
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Homework Statement
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
Homework 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°
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?