Maximum angle before rope breaks

[man_in_motion]

Homework Statement

Walking by a pond, you find a rope attached to a tree limb 5.2 m off the ground. You decide to use the rope to
swing out over the pond. The rope is a bit frayed but supports your weight. You estimate that the rope might break if
the tension is 80 N greater than your weight. You grab the rope at a point 4.6 m from the limb and move back to swing
out over the pond. (a) What is the maximum safe initial angle between the rope and the vertical so that it will not break
during the swing?

Homework Equations

U=mgy
$$F_a=m \frac{v^2}{r}$$

The Attempt at a Solution

$$\Sigma F = ma_c=m\frac{v^2}{r}=T-mg$$
r=5.2-4.6=0.6

 

[willem2]

Your equation for $\Sigma F$ is correct. Now find v at the bottom at the swing with
conservation of energy. Kinetic energy at this point is equal to the difference in potential
energy at the start of the swing and at the bottom.
Also use the fact that the maximum tension is 80N + the weight of the person.
I think r should be 4.6 m. This point is 0.6m above the ground when the rope is vertical.