Calculating Energy Required to Rotate a Spiral Spring by 180 Degrees

AI Thread Summary
To calculate the energy required to rotate a spiral spring by 180 degrees, the torque must be understood as a variable dependent on the angle, specifically τ = 10^-5 * θ Nm. The correct approach involves integrating the torque over the angle from 0 to 180 degrees, using the formula for work done in circular motion, W = ∫ τ dθ. The initial misunderstanding stemmed from treating torque as constant rather than variable. After correcting for units and integration, the final energy calculated was approximately 4.93 x 10^-4 J. Proper notation and conversion from degrees to radians are crucial for accurate results.
KiNGGeexD
Messages
317
Reaction score
1
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!
 
Physics news on Phys.org
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!
 
Back
Top