At the lowest point, the tension in the string is given as \(2W\). The forces acting on the stone are the tension \(T\) and the weight \(W\). At this point, the centripetal force required for circular motion is provided by the difference between the tension and the weight:
\[
T_{\text{lowest}} - W = \frac{mv^2}{r}
\]
Given \(T_{\text{lowest}} = 2W\), we have:
\[
2W - W = \frac{mv^2}{r} \implies W = \frac{mv^2}{r}
\]
At the highest point, the tension \(T_{\text{highest}}\) and the weight \(W\) both act downwards, providing the centripetal force:
\[
T_{\text{highest}} + W = \frac{mv'^2}{r}
\]
Assuming energy conservation, the speed at the highest point \(v'\) can be related to the speed at the lowest point \(v\). The potential energy difference is \(2W\) (since the height difference is \(2r\)):
\[
\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + 2W
\]
Substituting \(W = \frac{mv^2}{r}\) from Step 1:
\[
\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + 2\left(\frac{mv^2}{r}\right)
\]
Solving for \(v'^2\):
\[
\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + \frac{4mv^2}{r}
\]
\[
\frac{1}{2}mv^2 - \frac{4mv^2}{r} = \frac{1}{2}mv'^2
\]
Using the expression for \(v'^2\) and substituting back into the centripetal force equation:
\[
T_{\text{highest}} + W = \frac{mv'^2}{r}
\]
Since \(v'^2\) is less than \(v^2\), the tension \(T_{\text{highest}}\) will be less than \(T_{\text{lowest}}\). Solving the energy equation gives:
\[
T_{\text{highest}} = 0
\]
The tension when the stone is at the highest point is:
\[
\boxed{0}
\]
Answered
