Flywheel calculations

  • Thread starter Katy96
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  • #26
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Read the question.
I have, I cant figure it out and by that I mean how do I change it to something I can use units wise
 
  • #27
PeroK
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I have, I cant figure it out
##\omega## is "angular velocity", which must be used in radians per second. The question gives you it, however, in revolutions per something. Just as an aside, you seem to have truncated most of the units when quoting the question.
 
  • #28
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yes its given in revolutions per min, and im unsure of how to change it from revolutions per min to radians per second once I do that I know how to do part b)
 
  • #29
PeroK
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yes its given in revolutions per min, and im unsure of how to change it from revolutions per min to radians per second once I do that I know how to do part b)
How many radians are in a revolution?
 
  • #30
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2Pi/60?
 
  • #31
PeroK
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2Pi/60?
A poor guess. It's ##2\pi##. You could always have googled that!
 
  • #32
SammyS
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so how do I change 48km/h into m/s? never mind I got it now!!
Is editing a response as you type it a lost skill?

I see this so often it's ridiculous !
 
  • #33
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A poor guess. It's ##2\pi##. You could always have googled that!
okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60
 
  • #34
PeroK
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okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60
I see what you did. There are ##2\pi## radians in a revolution, so yes it's ##2\pi /60## to convert from revs per minute to radians per second.
 
  • #35
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I see what you did. There are ##2\pi## radians in a revolution, so yes it's ##2\pi /60## to convert from revs per minute to radians per second.
when I did that I got the right answer so in part c do I use vmax=rωmax or do I use I=0.606mr2
 
  • #36
PeroK
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when I did that I got the right answer so in part c do I use vmax=rωmax or do I use I=0.606mr2
Does that give the right answer?

I'm going offline now. Good luck with the rest of it.
 
  • #37
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Does that give the right answer?

I'm going offline now. Good luck with the rest of it.
thanks.
 
  • #38
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Its just part e i'm struggling on now
 
  • #39
jbriggs444
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e) Determine also the
Its just part e i'm struggling on now
Yes. It is quite the conundrum.
 
  • #40
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Yes. It is quite the conundrum.
sorry it meant to say Determine also the mean frictional torque causing the slowing down in the two minute stop.
 
  • #41
jbriggs444
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You know how much energy it had at the beginning of the two minute stop. You have already used that to calculate its angular velocity at that time.

The question tells you what fraction of the starting energy it has at the end of the two minute stop. How much energy is that? How fast does that mean that it must be rotating at the end of the two minutes?
 
  • #42
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this is what I know, i need to use the change in the KE to find the final angular speed, and hence the angular deceleration, using the equations of angular motion and torque Important to show answer is negative, as result of deceleration
The force providing the centripetal acceleration acts towards the axis so cannot have a torque about it.and flywheel does not stop during the 2 mins.
 
  • #43
jbriggs444
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this is what I know, i need to use the change in the KE to find the final angular speed
Then there is a starting point. What is the change in KE?
 
  • #44
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Then there is a starting point. What is the change in KE?
that's where I get stuck
 
  • #45
jbriggs444
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The problem statement says that four fifths of the flywheel's energy remains after two minutes.
 

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