- #26

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I have, I cant figure it out and by that I mean how do I change it to something I can use units wiseRead the question.

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- #26

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I have, I cant figure it out and by that I mean how do I change it to something I can use units wiseRead the question.

- #27

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##\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.I have, I cant figure it out

- #28

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- #29

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How many radians are in a revolution?

- #30

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2Pi/60?

- #31

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A poor guess. It's ##2\pi##. You could always have googled that!2Pi/60?

- #32

SammyS

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Is editing a responseso how do I change 48km/h into m/s? never mind I got it now!!

I see this so often it's ridiculous !

- #33

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okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60A poor guess. It's ##2\pi##. You could always have googled that!

- #34

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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.okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60

- #35

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when I did that I got the right answer so in part c do I use vI 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.

- #36

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Does that give the right answer?when I did that I got the right answer so in part c do I use v_{max}=rω_{max}or do I use I=0.606mr^{2}

I'm going offline now. Good luck with the rest of it.

- #37

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thanks.Does that give the right answer?

I'm going offline now. Good luck with the rest of it.

- #38

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Its just part e i'm struggling on now

- #39

jbriggs444

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e) Determine also the

Yes. It is quite the conundrum.Its just part e i'm struggling on now

- #40

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sorry it meant to say Determine also the mean frictional torque causing the slowing down in the two minute stop.Yes. It is quite the conundrum.

- #41

jbriggs444

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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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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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Then there is a starting point. What is the change in KE?this is what I know, i need to use the change in the KE to find the final angular speed

- #44

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that's where I get stuckThen there is a starting point. What is the change in KE?

- #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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