# Supplementary Prob 1 Conservation of Energy.

1. Mar 18, 2004

### cristina

A girl of mass m ties a rope of length R to a tree branch over a creek and starts to swing from rest at a point that is a distance R/2 lower than the branch. What is the minimum breaking tension for the rope if it is not to break and drop the girl into the creek?

2. Mar 18, 2004

### ShawnD

Her potential energy will change into kinetic energy.

$$\frac{1}{2}mv^2 = mgh$$

$$v = \sqrt{2gh}$$

substitute values into that

$$v = \sqrt{19.6(\frac{R}{2})}$$

The tension in the rope is based on the girl's centripetal acceleration AND the force of gravity (am I right now?)

$$F = ma (gravity) + ma (centripetal)$$

$$F = m(9.8) + m(\frac{v^2}{R})$$

$$F = 9.8m + m\frac{(\sqrt{19.6\frac{R}{2}})^2}{R}$$

$$F = 9.8m + m\frac{19.6\frac{R}{2}}{R}$$

try to simplify from there

Last edited: Mar 18, 2004
3. Mar 18, 2004

### cristina

May you please explain more on why The tension in the rope would be based on the girl's centripetal acceleration?

4. Mar 18, 2004

### HallsofIvy

Staff Emeritus
Because force is mass times acceleration!

5. Mar 18, 2004

### ShawnD

I think I made a mistake, I did not factor in gravity. Re-read my previous post now that I have fixed it.

Last edited: Mar 19, 2004
6. Mar 18, 2004

### cristina

Ok true, I start getting the concept!
thank you very much

7. Mar 18, 2004

### cristina

I am lost now, where you didn't factor gravity? the minimum tension would be 19.6m.

Last edited: Mar 19, 2004