# Rope,potential energy

1. Dec 6, 2011

### Silken

Hi everyone

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

See picture in attachment.

2. Relevant equations
-

3. The attempt at a solution

well, the solution should be:

$$V=\rho g[(h-h_{2}) \frac {h+h_{2}} {2}+(h-h_{1}) \frac {h-h_{1}}{2}]$$

where rho is the density per length of the rope. I don't understand where (h+h2)/2 and (h+h1)/2 come from. When I tried to solve it, my guess for the potential energy (just for one 'side' let'say say) was: $$\rho g (h-h_{1})*h$$ but that's wrong. Can anyone help me out?

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2. Dec 6, 2011

### Staff: Mentor

Hint: Where's the center of mass of the left hand side?

3. Dec 6, 2011

### grzz

For each side use PE = mgh were h is the position of the centre of mass of m.

4. Dec 6, 2011

oops!

5. Dec 6, 2011

### Silken

There was a mistake in my formula:

$$V=\rho g[(h-h_{2}) \frac {h+h_{2}} {2}+(h-h_{1}) \frac {h+h_{1}}{2}]$$

Thanks for the help. You both say I shall use the center of mass on both sides. I guess that's where the 1/2 comes from. My problem is here, if I want the center of mass of one side, why do I have to add them like (h1+h)/2. I thought If I want to center of mass of the rope on one side I'd had (h-h1)/2. Where's my mistake?

6. Dec 6, 2011

### Staff: Mentor

You want the location of the center of mass measured from the ground.

7. Dec 7, 2011

### Silken

Yes thank you.I found me error in reasoning. :)