Tension of a string rotating a ball in a circular motion? 
A ball of mass 5kg is attached to a string of length 120 cm and
  rotating vertically at a speed of 10 cm/s. what is the tension of the string when
  the ball is farthest to the right from the center? (neglect both the
  string's mass and air resistance)

I tried applying Newton's second law that says $$\Sigma force= ma => T-mg = ma$$
but that doesn't give the right answer and i don't know why. 
can somebody please help me, thanks
 A: The centripetal force is not a "separate" force. I think it's best not to think of centripetal forces, but just centripetal acceleration. An object with circular motion means that net sum of all the forces acting on the object results in circular motion... meaning the net acceleration towards the center of the circle is $\dfrac{v^2}{r}$
In your situation there are two forces acting on the ball. The tension in the rope and gravity. (there's no extra centripetal force).
$\Sigma F_{towards center} = m_{ball}a_{towardscenter} => T = m_{ball}\dfrac{v^2}{r}$ 
So gravity does not play a role here because gravity acts downward, and the direction towards the center of the circle is to the left.
Suppose the ball was at an angle of 45 degrees to the right of the upward direction. Then you'd have to consider the tension in the rope and the component of gravity acting towards the center. Specifically you'd get $T+m_{ball}gcos(45) = m_{ball}\dfrac{v^2}{r}$
But anyway, for your question $T = m_{ball}\dfrac{v^2}{r}$
A: The motion of a mass tied to  a string in a vertical circle includes following  mechanical concepts. 
It must satisfy 
(i) availability of centripetal force to remain in a circular path
(ii) satisfy  conservation of energy 
If we take a situation that the ball just reaches the topmost position with velocity equal to zero then the the tension in the string will be such that it is just taut.
Therefore the   gravitational pull must be providing a force equal to magnitude of centripetal force (m.v^2)/r  so one can get the value of speed   and the total energy is known.
In other cases also where the body can reach the top and can cover it with some finite velocity the total energy conservation can be applied with due consideration of change in potential energy of the body.
I will advise you to take the topmost velocity to be  say v(3) the bottom point as V(1) and at horizintal midway V(2) and relate the energies K.E. +P.E. at the three points to be equal.
you have a info that at top point the only force is mg acting downward  providing the centripetal force. that will  facilitate with the value of V(3).
Then you can calculate  v(1) and then naturally  V(2) can be easily computed.
The Tension in the string  at the horizontal point where the speed of the ball is v(2)  T=  m(v(2))^2/r  as the mg force is perpendicular to the string and not contributing to the tension.
