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Physics Help!

I have been doing this problem for a while and want to see if my working out for it is fine?

Here is the question itself:

A cylindrical flywheel of radius 0.93 m and mass 54 kg is used to store energy by spinning.

a) Calculate the rotation frequency (in Hz or s-1) required for it to store 5.6 MJ.
(The moment of inertia of a solid cylinder rotating about its centre is given by ½mr2 )

b) Calculate the time taken for a 412 N force acting tangentially at the flywheel’s radius to accelerate the flywheel to 1911 rpm from rest.

IMG_3232.jpg

IMG_3234.jpg

Much appreciated!
IMG_3232.jpg502.3KB
Reply 1
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Yatayyat
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Reply 2
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mqb2766
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Does it say anything more about the flywheel? The inertia is only
I = mr^2
if all the mass is located at the edge of the wheel. If its uniform, the inertia should be 1/2 of that.

Edit - you have it in brackets that its mr^2/2, so your first page answers are missing this 1/2 factor.
(edited 4 years ago)
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Yatayyat
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Original post by mqb2766
Does it say anything more about the flywheel? The inertia is only
I = mr^2
if all the mass is located at the edge of the wheel. If its uniform, the inertia should be 1/2 of that.

Edit - you have it in brackets that its mr^2/2, so your first page answers are missing this 1/2 factor.

Yes I see, so in fact my first page answer should be because it's missing a 1/2 factor, the answer would be '77.938... Hz * 1/2 = 38.969... Hz', so my new value is 38.97 Hz (2 dp)?
Reply 4
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mqb2766
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Original post by Yatayyat
Yes I see, so in fact my first page answer should be because it's missing a 1/2 factor, the answer would be '77.938... Hz * 1/2 = 38.969... Hz', so my new value is 38.97 Hz (2 dp)?


Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?
Reply 5
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Yatayyat
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Original post by mqb2766
Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?

Wait doesn't less inertia imply that the wheel rotates faster when spinning? Since it has less resistance to its change in motion when a torque (rotational equivalent of a linear resultant force) is applied
Reply 6
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Yatayyat
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Original post by mqb2766
Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?

And sorry, I just thought that since I was missing a 1/2 factor in my moment of inertia, if I multiplied my final answer by half would fix the mistake but that isn’t the case here

I redone the problem again and I’m now getting frequency to be 110.2 Hz

CF57415B-E727-47A8-9413-EF8E96A35A4B.jpeg
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mqb2766
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Original post by Yatayyat
And sorry, I just thought that since I was missing a 1/2 factor in my moment of inertia, if I multiplied my final answer by half would fix the mistake but that isn’t the case here

I redone the problem again and I’m now getting frequency to be 110.2 Hz

CF57415B-E727-47A8-9413-EF8E96A35A4B.jpeg

Seems ok? Its sqrt(2) larger than before?
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Yatayyat
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Original post by mqb2766
Seems ok? Its sqrt(2) larger than before?


79.938 * sqrt(2) gives 113.049 so close to mine of 110.2

Does my part b look ok?
Reply 9
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mqb2766
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Original post by Yatayyat
79.938 * sqrt(2) gives 113.049 so close to mine of 110.2

Does my part b look ok?

Sure, all i was saying is that if the inertia is halved w^2 must be doubled for the same energy. So new angular speed is sqrt(2) * old one. Any difference is a numerical from you on the either the old or new calculation.

Checking b now.

Edit: b) appears fine.
(edited 4 years ago)
Reply 10
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Yatayyat
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Original post by mqb2766
Sure, all i was saying is that if the inertia is halved w^2 must be doubled for the same energy. So new angular speed is sqrt(2) * old one. Any difference is a numerical from you on the either the old or new calculation.

Checking b now.

Edit: b) appears fine.

Thank you so much! In future I’d try to do it that way as it saves much more time then having to go back and replugging the numbers

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