Centripetal force help

A ball of mass m, at one end of a string of length L, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. The speed of the ball at the bottom of the circle is:

the answer is Sqrt(5gL). I did this problem and I only got sqrt(gL).
First I did Fy = Ft - Fg = 0
mv^2/r = mg where r=l
v = sqrt(gL)

I think the speed is not meant tp be constant; as the mass falls it loses potential energy and gains kinetic energy.

At the top its speed is given by mv^2/L = mg .as you say.
So its kinetic energy at the top is (1/2)mv^2 = mgL/2

At the bottom it has dropped a distance 2L, so its potential energy has decreased 2mgL. Its kinetic energy has increased by 2mgL.

Total kinetic energy at bottom = mgL/2 + 2mgL = 5mgL/2

At the bottom (1/2)mv^2 = 5mgL/2
v = sqrt (5gL)

the velocity at the top of the string is not zero; we find the velocity at the top of the string by applying newton's laws to the mass at the top:

sum of forces = m a

the sum of forces is tension acting down, gravity down, and these combine to produce a centripetal force of magnitude mv^2/r acting toward the center of the circle, also down

we have

-T - mg = - mv^2/r

for T->0, we have the minimum velocity at the top

mg = m v^2/r or v^2 = rg

now, apply energy methods; the PE at the top is mg (2r) compared to the bottom, so we have

1/2 m v(bottom)^2 = 1/2 mv(top)^2 + mg(2r)

1/2 m v(bottom)^2 = 1/2 m(r g) + 2 m g r

v(bottom) = Sqrt[5 g r]

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