Uniform Circular Motion: Solving for String Tension

[QCHabsQC]

Homework Statement

A ball of mass 4 kg on a string of length 1.5 m is being swung in a circle in the vertical plane at a constant speed of 10 m/s. Find the tension in the string at the bottom and top of the balls path

Homework Equations

ƩF = m(v^2/r)

The Attempt at a Solution

This is how I attempted to solve this problem, but my answer is not correct.
Is it correct that the tension in the string at the bottom is: T = m(v^2/r) + mg
And at the top: T = m(v^2/r) - mg

 

[Simon Bridge]

Please show your working - not just some answers.

Consider:
T = m(v^2/r) + mg

This is saying that, at the bottom of the path, the tension has to overcome gravity and still have enough left over to give a centripetal force... the net force points upwards.
Does that make sense?

 

[QCHabsQC]

Yes; so the tension at the bottom would be, T = (4*(10)^2)/1.5 + 9.8*4 = 217 N. Is this correct?

 

[Simon Bridge]

I don't check arithmetic.
But you appear to have answered your own question.