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A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.
let r be radius of solid sphere and R be radius of cup
at bottom, mgR= (1/2)(2m/5)(r^2)((v/r)^2)+((m/2)(v^2)), solve for v
normal force at bottom= (mg)+((m/R)(v^2))
bingo
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