A small particle is placed at rest on the very top of a fixed smooth…

Question

A small particle is placed at rest on the very top of a fixed smooth (frictionless) sphere of radius 5 m and given a negligible nudge so that it slides down the outside surface. It leaves the sphere at a certain point. The vertical height of the leaving point below the top of the sphere equals m/n metres in lowest terms, where m and n are positive integers. Find m + n.

Accepted Answer

Let theta be the angle from the top. The particle leaves when the normal force vanishes, i.e. when mg cos(theta) = m v^2 / R. Energy conservation from the top gives v^2 = 2 g R (1 - cos theta). Substituting: g cos theta = 2 g (1 - cos theta) => cos theta = 2 - 2 cos theta => 3 cos theta = 2 => cos theta = 2/3. The drop in height is R(1 - cos theta) = R(1 - 2/3) = R/3 = 5/3 m. So m/n = 5/3, m + n = 8.
Final answer: 8

Final answer: 8

A small particle is placed at rest on the very top of a fixed smooth…

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