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benf.stokes
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Homework Statement
A point particle of mass m placed on top of a semi-sphere of radius R and mass M, which is based on a frictionless surface. At one instant, the particles begin to slide over the semi-sphere.
Obtain an equation for the angle at the point where the particle loses contact with the semi-sphere.
How do M and m get into the picture?
Thanks
Homework Equations
[tex]m\cdot g\cdot \cos(\theta) - N = m\cdot \frac{v^2}{R}[/tex]
[tex]m\cdot g\cdot R+0= \frac{1}{2}\cdot m\cdot v^2+m\cdot g\cdot R\cdot \cos(\varphi)[/tex]
The Attempt at a Solution
[tex]
\mbox{Using conservation of mechanical energy comes:}
[/tex]
[tex]
m\cdot g\cdot R+0= \frac{1}{2}\cdot m\cdot v^2+m\cdot g\cdot R\cdot \cos(\varphi)[/tex]
[tex]
\mbox{Solving for v^2 gives:}
[/tex]
[tex]
v^2 = 2\cdot g\cdot R\cdot (1-\cos(\varphi))
[/tex]
[tex]
\mbox{Using the centripetal force condition yields:}
[/tex]
[tex]
m\cdot g\cdot \cos(\theta) - N = m\cdot \frac{v^2}{R}
[/tex]
[tex]
\mbox{When contact is lost N = 0:}
[/tex]
[tex]
v^2 = g\cdot R\cdot \cos(\theta) \hspace{1 pt} \mbox{being \theta \hspace{0.5 pc} \mbox{the angle at which contact is lost and so:}}
[/tex]
[tex]
2\cdot g\cdot R\cdot (1-\cos(\theta)) = g\cdot R\cdot \cos(\theta) \hspace{0.5 pc} \mbox{which upon solving gives:}
[/tex]
[tex]
\theta=\cos^{-1}(\frac{2}{3})
[/tex]
[tex]
\mbox{But I don't know where do M and m figure in this solution.}
[/tex]
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