A hemispherical bowl of radius R is sitting upside down on a table. A small bead

Question

A hemispherical bowl of radius R is sitting upside down on a table. A small bead of mass M is placed at the top of the upside down bowl and given a (negligible) push to get it moving. The bead then slides along the surface of the bowl without friction. At some point, h, above the table, the bead will lose contact with the surface.

a. What is the radial force?
b. What math statement describes the situation of the bead losing contact with the bowl?
c. Find the angle, measured from the vertical, at which point the bead loses contact with the bowl.

I have been told it is possible to get a numerical answer even though no numbers are given.

Explanation / Answer

mv^2 / R = m g sin theta - N where N is the normal force When N = 0 the bead leaves the surface - theta measured above horizontal m g h = 1/2 m v^2 where h is the distance the bead falls h = R ( 1 - sin theta) where R is the radius of the surface v ^2 = 2 g R (1 - sin theta) v^2 = R g sin theta from the first equation above Then 2 (1 - sin theta) = sin theta equating the equations for v^2 sin theta = 2/3 theta = 41.8 deg above the horizontal or 48.2 deg from vertical The radial force is N (when theta = 90 N = mg)

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