# Work done to move a spring

1. Dec 15, 2013

### KiNGGeexD

A spiral spring exerts a restoring torque on an axis proportional to the angle through which the axis is turned. If it provides a torque of 10-5 Nmrad-1, find the energy required to turn it through 180degrees from its relaxed state?

My solution was simple

Work done = torque* the angle theta but I seemed to get the wrong answer!

2. Dec 15, 2013

### tiny-tim

Hi KiNGGeexD!

(try using the X2 button just above the Reply box )

I think you're misunderstanding the question …

the torque isn't constant, it's 10-5*θ Nm, depending on the instantaneous angle θ.

3. Dec 15, 2013

### KiNGGeexD

So how would I go about solving the problem?:(

4. Dec 15, 2013

### tiny-tim

linear work done = ∫ F·ds

circular work done = ∫ τ dθ

5. Dec 15, 2013

### KiNGGeexD

Ah so I need to integrate τ dθ for 0-180 degrees

6. Dec 15, 2013

### tiny-tim

yup!

7. Dec 15, 2013

### KiNGGeexD

Haha cheers friend!

8. Dec 15, 2013

### KiNGGeexD

Maybe I'm getting confused but would the answer not just be the same

9. Dec 15, 2013

### tiny-tim

no

10. Dec 15, 2013

### KiNGGeexD

I'm clearly integrating wrong haha

W= τ dθ

From 0-180 ok I'm not 100% lol

11. Dec 15, 2013

### tiny-tim

12. Dec 15, 2013

### KiNGGeexD

W=τ dθ

So

W= τ*180 + c

13. Dec 15, 2013

### tiny-tim

an integral should have an ∫ in it

(and limits)

and what is τ ?​

14. Dec 15, 2013

### KiNGGeexD

τ= Iα ?

And yea I know about the imetrgral sign and limits I just can't do it on my phone:(!

15. Dec 15, 2013

### tiny-tim

try typing two #, then \int, then two more #

16. Dec 15, 2013

### KiNGGeexD

For a non constant torque

W= ταθ

17. Dec 15, 2013

### KiNGGeexD

That tau was supposed to be I, moment of inertia

18. Dec 15, 2013

### KiNGGeexD

I am integrating

τ from 0-180

Or τθ from 0-180?

19. Dec 15, 2013

### KiNGGeexD

If so

W= τ*θ^2 all divided by 2?

20. Dec 15, 2013

### tiny-tim

hmm … what you mean is correct, but that's certainly not the correct way to write it