Calculating Energy Required to Rotate a Spiral Spring by 180 Degrees

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Homework Help Overview

The discussion revolves around calculating the energy required to rotate a spiral spring by 180 degrees from its relaxed state, focusing on the relationship between torque and work done in rotational motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the concept of torque as a function of the angle and discuss the need for integration to find the work done. Questions arise regarding the correct setup of the integral and the interpretation of torque in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, offering guidance on the need to integrate torque over the specified angle. There is a mix of attempts to clarify misunderstandings about the integration process and the nature of the torque involved.

Contextual Notes

Some participants express confusion about the integration limits and the conversion of degrees to radians, indicating potential gaps in understanding the mathematical framework required for the problem.

KiNGGeexD
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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!
 
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Hi KiNGGeexD! :smile:

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

I think you're misunderstanding the question …

the torque isn't constant, it's 10-5*θ Nm, depending on the instantaneous angle θ. :wink:
 
So how would I go about solving the problem?:(
 
linear work done = ∫ F·ds

circular work done = ∫ τ dθ :wink:
 
Ah so I need to integrate τ dθ for 0-180 degrees?:)
 
yup! :biggrin:
 
Haha cheers friend!
 
Maybe I'm getting confused but would the answer not just be the same?:)
 
no

show us your integral​
 
  • #10
I'm clearly integrating wrong haha

I had

W= τ dθ

From 0-180 ok I'm not 100% lol
 
  • #11
show us your integral! :rolleyes:
 
  • #12
W=τ dθ

So

W= τ*180 + c
 
  • #13
an integral should have an ∫ in it :confused:

(and limits)

and what is τ ?​
 
  • #14
τ= Iα ?

And yea I know about the imetrgral sign and limits I just can't do it on my phone:(!
 
  • #15
KiNGGeexD said:
And yea I know about the imetrgral sign and limits I just can't do it on my phone:(!

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

tiny-tim said:
the torque isn't constant, it's 10-5*θ Nm, depending on the instantaneous angle θ. :wink:
 
  • #16
For a non constant torque W= ταθ
 
  • #17
That tau was supposed to be I, moment of inertia
 
  • #18
I am integrating

τ from 0-180

Or τθ from 0-180?
 
  • #19
If soW= τ*θ^2 all divided by 2?
 
  • #20
hmm :redface: … what you mean is correct, but that's certainly not the correct way to write it
 
  • #21
Yea I know sorry :(But is that correct?
 
  • #22
if it means what i think it means, yes

what result do you get?​
 
  • #23
1.62 j
 
  • #24
you forgot to convert from degrees to radians
 
  • #25
Haha ok in that case4.93*10^-4 :)
 
  • #26
-4 ? :confused:
 
  • #27
Haha sorry where do you get this notation from? Lol!
 
  • #28
[NOPARSE]type -4[/NOPARSE] :wink:
 
  • #29
Is my wrong answer right though? Lol
 
  • #30
Wait never mind the dimensions are spot on haha!
 

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