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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 = M

_{acm},r

N-mgcosθ = -mV

_{cm}

^{2}/(R+a)

N = mgcosθ - m

_{vcm}

^{2}/(R+a) eq. (1)

Conservation in mechanical energy

mg(R+a) = 1/2m

_{vcm}

^{2}+ 1/2I

_{cm}ω

^{2}+ mg(R+a)cosθ

mg(R+a) = 1/2m

_{vcm}

^{2}+ 1/2(2/5ma

^{2})(V

_{cm}/a)

^{2}+ mg(R+a)cosθ

mg(R+a) = 7/10mV

_{cm}

^{2}+ mg(R+a)cosθ

Simplified

7/10mV

_{cm}

^{2}+ mg(R+a)(1-cosθ)

V

_{cm}

^{2}= 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?