Mechanics> rotational motion...

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.

harsh gurung , 14 Years ago

Grade 12

3 Answers

Chetan Mandayam Nayakar

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))

Approved

Last Activity: 14 Years ago

harsh gurung

bingo

Last Activity: 14 Years ago

Kavan Gondalia

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

Last Activity: 7 Years ago

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Mechanics> rotational motion...

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.

harsh gurung , 14 Years ago

Chetan Mandayam Nayakar

harsh gurung

Kavan Gondalia

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