Find the difference between the magnitude of the tension?

Question

A ball of mass is attached to a string of length . It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. To avoid confusion, take the upward direction to be positive throughout the problem. At the top and bottom of the vertical circle, label the ball's speeds Vt and Vb, and label the corresponding tensions in the string Tt and Tb*Tt. and Tb have magnitudes Tt and Tb.


Find Tb - Tt the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle.
Express the difference in tension in terms of and . The quantities and should not appear in your final answer.

Answer 1

I assume that when you say "enough speed so that the string remains taut" that you really mean just enough speed so that the string does not go slack at the top of the trajectory. For that to happen, the centrifugal force (mv^2/r) must equal the pull of gravity (mg). You are given the speed at the top of the trajectory, so the centrifugal force there is mVt^2/r therefore

mVt^2/r = mg

r = Vt^2/g

and the tension in the string at that point is 0

At the bottom of the trajectory you have two forces adding to produce the string tension, centrifugal force mVb^2/r and gravity, mg. The tension is then

mVb^2/r + mg.

The r is the same at both points and it is Vt^2/g, so

∆T = mg(Vb^2/Vt^2 + 1)

EDIT: Note the answer is 2mg only if the top velocity and bottom velocity are equal.

Answer 2

The answer is 2mg.
Let T be the magnitude of the tension that would be required to keep the object moving in a circle (ignoring gravity).
Then add the forces due to gravity.
So Tb = T + mg and Tt = T - mg.
Tb - Tt = (T+mg)-(T-mg) = 2mg