Question

A 56-kg student runs at 4.2 m/s , grabs a hanging 10.0-m-long rope, and swings out over a lake (Figure 1) . He releases the rope when his velocity is zero.
What is the angle θ when he releases the rope?
What is the tension in the rope just before he releases it?
What is the maximum tension in the rope during the swing?

Answer 1

find his KE when he grabs the rope.
then his max PE will equal the initial KE
draw the figure.
his mg is down, Tension is up. so Tension*cosTheta=mg solve for Theta

PE=initKE
mgh=1/2 mv^2 solve for h.

Max tension. At the bottom
tension=mg+mv^2/r