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A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) witha negligibly small velocity. Find the height H where the bead will reverse direction.
why is the answer 2R and not 5R/2 bcoz the min velocity needed for it to follow the path makes it work out to be 5R/2 (considerning min velocity for req centripetal force and initial potential energy)

Shirish Chandra Srivastava , 9 Years ago
Grade 12
anser 1 Answers
Kanha singhania
Let the potential energy at A be 0, then potential energy at B = -mg(2R-H)On applying "conservation of energy" ,KE(initial) + PE(initial) = KE(final) + PE(final)0+0 = 0 -mg(2R-H) (As velocity at the point B = 0, KE at point B=0)2R-H=0H= 2R
Last Activity: 8 Years ago
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