1. Jul 24, 2010

### mizzy

1. The problem statement, all variables and given/known data
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

2. Relevant equations

centripetal acceleration = v^2/r

F = mv^2/r

3. 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???

2. Jul 24, 2010

### Staff: Mentor

Right.

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

3. Jul 24, 2010

### mizzy

Kinetic energy

4. Jul 24, 2010

### Staff: Mentor

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

5. Jul 24, 2010

### mizzy

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

6. Jul 24, 2010

### Staff: Mentor

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

7. Jul 24, 2010