| 6 comments ]

Assume the spring consists of N small springs, where N>>1. Hence, relaxed length of the small spring is .
The large spring equals to the small springs in series, so






From the equation above, we can conclude that multiply between spring constant of a part and its relaxed length is constant. And we can write



differentiate it, and we get




The total length of unwinding spring is L. If the cylinder rotates by small angle , a small part of the spring with total length will wind the cylinder, while the other part deformed to length L.
Relaxed length of the small part is,



Because the small part will wind the spring, the change of relaxed length of unwinding spring is,





Then, from equation (1) and (3), we can get



Integrate the equation, we will get




When the unwinding spring have relaxed length l, it will give force to the cylinder,



and give torque,




Thus, the change of energy when it rotates by small angle is





Substitute kl from equation (1),




Substitute k from equation (4),


Integrate two sides of equation,



And now, substitute and , and we get



Substitute for uniform solid cylinder.





That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

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6 comments

Anonymous said... @ April 16, 2009 at 9:44 AM

em...
I would luke to ask...
why don't we use energy???

so the initial energy will be the energ of rotation, and the last energy will be the spring potential energy.
the result is not the same, but I don't know why..
can you tell me why I can't use this formula?


By: indonesian boy.
^^

Anonymous said... @ April 28, 2009 at 11:23 PM

Good solution ^^ The best part of the problem is how to find the formula of k.
Why don't you make it a little more difficult, for example, let the cylinder move freely, not just rotate only?

#Indonesian boy: In my opinion, the method of energy conservation is still ok. Did you notice that the potential energy of the spring was partly in the winded part?

Anonymous said... @ May 4, 2009 at 11:22 PM

The integration to find E_o is wrong. It should be {e^(Rq/L) - 1 - Rq/L}KL^2, where q is theta. You forgot to substitute 0 for q in e^(Rq/L).

akrthegreat said... @ March 18, 2010 at 11:41 AM

Would you please explain why did you take relaxed length of small part as (Rdq)l/L? That's just the only thing I'm confused about in this wonderful solution.

Anonymous said... @ June 13, 2010 at 4:04 PM

why spring force of already winded spring is not taken into cosideration as it will also act tangentially to the disc at each point

Sanchar said... @ June 19, 2010 at 11:54 PM

agreed to anonymous, the tension in the spring should be same throughout. you have assumed that the tension in the winded part is different than the unwinded one. explain.

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