Question

A particle of mass m attached to the end of string of length l is released from the horizontal position. the particle rotates in a circle

Answer 1

A particle of mass m attached to end of the string of length l is released from horizontal position.
then, energy at horizontal position of particle = energy at lowest position of particle.

Let at lowest position speed of particle = v
now, mgl = 1/2 mv²
v = √2gl

now we should find speed of particle at highest position . Let v' be the speed of particle at highest position .
so, mg(2l) + 1/2mv'² = 1/2 mv²
2mgl + 1/2mv² = 1/2mg(2gl) = mgl
1/2mv² = -mgl
v² = -2gl , here v is imaginary number .
means this situation is not possible . if we released particle from horizontal position , particle doesn't complete circle.

Answer 2

V> √5r(l-h)

Bottom

V =√2gl

Equate both

√2gl =√5g(l-h)

2l=5l-5h

3l =5h

H>or equal 3/5 l