At this point the stone's speed is...A.) 2Sqrt(gR) B.) Sqrt(2gR) C. 3Sqrt(gR) or D. Sqrt(3gR)?

Question

A stone of mass m is attached to a stong string and whirled in a vertical circle of radius R. At the exact top of the path, the tenstion in the string is three times the stone's weight.
At this point the stone's speed is...A.) 2Sqrt(gR) B.) Sqrt(2gR) C. 3Sqrt(gR) or D. Sqrt(3gR)
I thought the answer was B but the answer is A. How is this so?

Answer 1

Draw the force diagram on the stone at the top of the path ─ two forces act on the stone, its weight Fg = m x g and the string force, the tension Ft, which is three times the stone's weight, so Ft = 3 x Fg = 3m x g. Both forces act toward the centre of the circle and their resultant is called a centripetal force. According to the second Newton's law, the centripetal force equals:
Fc = m x a, where a is the centripetal acceleration a = sqr(v) / r.
We can write:
Fc = Fg + Ft = m x g + 3m x g = 4m x g and
Fc = m x sqr(v) / r = 4m x g
m is cancelled out of the equation, then we can find the speed:
sqr(v) / r = 4g
sqr(v) = 4gr
and
v = sqrt(4gr) = 2 x sqrt(gr), since sqrt(4) = 2
This is the answer A.

Answer 2

Upward force from rotation = mV²/R
Downward force from gravity = mg
Problem statement: mV²/R - mg = 3mg
Solve for V and you get answer A)