Work Done by Spring at A: Calculating from Position 2 to Position 3

[maniac90]

Homework Statement

The torsion spring at A applies the counterclockwise couple C(θ)= − 10θ N · m to the uniform bar.The angle θ (in radians) is measured counterclockwise
from position 1, where the spring is undeformed.

Calculate the work done by the spring on the bar as it rotates from position 2 to position 3.

http://img849.imageshack.us/img849/1696/182z.jpg

Homework Equations

U_1-2 = ∫C dθ

U_1-2 = C(θ2 − θ1)= C Δθ

The Attempt at a Solution

U_1-2 = ∫− 10θ dθ = -10∫θ dθ

= -10 (1/2) θ²

50° = 5π/18 rad

U_1-2 = -3.8

I am not sure if that is the correct way of solving it, any guidance would be helpful

 

[WJSwanson]

Looks right to me.

 

[maniac90]

I just checked the book. The answer says it is −8.38 N·m
I can't figure out what I am doing wrong

 

[WJSwanson]

Oh, here's the issue. Sorry I didn't check more closely before. I think your limits of integration were off. It should be

$W = \int^{\theta = 4\pi/9}_{\theta = \pi/6} -10\theta d\theta$

because the equilibrium position of the system is at position 1, where $\theta = 0$. You want to find the work done by the spring moving from position 2 ($\theta = \frac{\pi}{6}$, relative to position 1) to position 3 ($\theta = \frac{4\pi}{9}$, with respect to position 1).

 

[WJSwanson]

So the moral of the story is that it's highly important to take into account your equilibrium conditions and how the interactions occur relative to the equilibrium conditions. It's also important for people who answer questions on PF not to get ahead of themselves and forget (like I did here) to check that the approach the OP takes is duly considerate of the equilibrium conditions and their relation to the conditions stated as the bounds of the problem.

 

[maniac90]

That helped me a lot. Thank you very much