Calculating Tensions in a Vertical Circular Motion Problem

[cy19861126]

Homework Statement

A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path,(a) at the beginning of his motion, (b) at a height of 1.5 m above the bottom of the circular arc, and (c) at the bottom of his arc.

Homework Equations

F = ma
F = m * v^2/r

The Attempt at a Solution

a) Using the equation F = m*v^2/r, F is 0 because v is 0
b) First, I used the energy approach attempting to solve the problem:
mgh = 0.5mv^2 + mgh,
9.8 * 4 = 0.5 * v^2 + 9.8 * 1.5
v = 7m/s.

As I get velocity, I plugged the value into
F = m * v^2 / r
F = 70kg * 7^2 / 4
F = 857N

I do not think this is right, since the value is too large. Also, I found out that I can calculate the angle of the distance he travelled: arccos(2.5/4) = 0.89rad --> pi/2 - 0.89 rad = 0.68 rad, this is the angle which he travelled. But I do not know where to plug this in anywhere.

 

[Hootenanny]

Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?

 

[cy19861126]

Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?

Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all

 

[Hootenanny]

Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all

Yep it looks spot on to me . Conservation of energy is by far the easiest approach to solving circular motion, many students overlook this method initially as they want to use all the new equations (toys) they've been given . They forget their basic principles, well that's what I find anyway...