## Homework Statement

A stuntman swings from the end of 4 m long rope along the arc of a vertical circle. If his mass is 70kg, find the tension in the rope required to make him follow his circular path, assuming he starts from rest when the rope is horizontal:
a) at the beginning of his motion

b) at the bottom of his arc

## Homework Equations

centripetal acceleration = v^2/r

F = mv^2/r

## The Attempt at a Solution

For part a) at the beginning of his motion, is the tension equal to zero? The tension of the rope is pointing to the center and therefore the equation is F = mv^2/r. However, v is equal to 0 because he starts from rest. is that right?

For part b) at the bottom of the arc, tension in the rope and mg causes acceleration and therefore the equation is F - mg = mv^2/r. But what's v???

Doc Al
Mentor

For part a) at the beginning of his motion, is the tension equal to zero? The tension of the rope is pointing to the center and therefore the equation is F = mv^2/r. However, v is equal to 0 because he starts from rest. is that right?
Right.

For part b) at the bottom of the arc, tension in the rope and mg causes acceleration and therefore the equation is F - mg = mv^2/r. But what's v???
You'll need to figure out the speed at the bottom. (What's conserved?)

Right.

You'll need to figure out the speed at the bottom. (What's conserved?)

Kinetic energy

Doc Al
Mentor

Kinetic energy
Total mechanical energy is conserved as man swings, not just kinetic.

Total mechanical energy is conserved as man swings, not just kinetic.

Ok. In that case:

KEmiddle + PEmiddle = KEbottom +PEbottom
1/2mv^2 +mgh = 1/2mv^2 +mgh
0 + mgr = 1/2mv^2 + 0
mgr = 1/2mv^2
(70)(9.8)(4) = 1/2(70)v^2

therefore, v = square root off 2gr = 8.9m/s

F1 - mg = mv^2/r
F1 = mv^2/r +mg
= (70)(8.9)^2/4 + (70)(9.8)
= 2.07 x 10^3 N

Can you confirm if this is right? THANKS

Doc Al
Mentor

Looks good! (Don't round off until the end.)