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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.

Chetan Mandayam Nayakar
312 Points
9 years ago

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))

harsh gurung
20 Points
9 years ago

bingo

Kavan Gondalia
11 Points
2 years ago
According to Energy Conservation:-
mgR=1/2 (I of disc)(omega)² + 1/2 mv²
mgR=1/2 (mR²/2)(v/R)² + 1/2mv²
2mgR= 3/2 mv²
4/3 mg = v²/R
4/3 mg = a

Let Normal force be N
N - mg = ma
N - mg = 4/3 mg
N = 4/3 mg + mg
N = 7/3 mg