- #1

f todd baker

- 61

- 22

*f=μN*) in a hemispherical bowl. Start at the top at rest. So far I have -

*μN+mg*cos

*θ=ma*and

*N=mg*sin

*θ+mv*

^{2}/

*R*where

*θ*is the angle below horizontal from the center of the sphere and

*R*is the radius of the sphere. I know it likely does not have a simple solution but would be happy with an approximate solution for small

*μ*.