Flywheel RPM when Power Returns: 477.628992424 rad/s

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The discussion focuses on calculating the flywheel's angular velocity when power returns after a 32-second outage. Initially spinning at 450 RPM, the flywheel slows due to friction, completing 170 revolutions during the power loss. The correct angular acceleration is determined to be -0.857617188 rad/s², leading to a final angular velocity of approximately 19.656 rad/s when power is restored. Additionally, it is calculated that the flywheel would have taken about 54.9 seconds to stop completely if power had not returned. The final number of revolutions made during this time is confirmed to be approximately 205.849.
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At what rate is the flywheel spinning when the power comes back on?
A high-speed flywheel in a motor is spinning at 450rpm when a power failure suddenly occurs. The flywheel has mass 44.0kg and diameter 77.0cm. The power is off for 32.0s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.

1. At what rate is the flywheel spinning when the power comes back on?

2. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on?

3. How many revolutions would the wheel have made during this time?

So it seems like if I get the first part the rest will be easy. This is what I have for the first part but it doesn't seem right. 170 revs = 1068.141502221 rad
450 rpm=47.1 rad/s
t = 32.0 sec

1068.141502221 = 47.1 * 32 +0.5 * α * 32^2
So α = 0.528992424 rad/s?
 
Last edited:
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oldspice1212 said:
At what rate is the flywheel spinning when the power comes back on?
A high-speed flywheel in a motor is spinning at 450rpm when a power failure suddenly occurs. The flywheel has mass 44.0kg and diameter 77.0cm. The power is off for 32.0s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.

1. At what rate is the flywheel spinning when the power comes back on?

2. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on?

3. How many revolutions would the wheel have made during this time?

So it seems like if I get the first part the rest will be easy. This is what I have for the first part but it doesn't seem right.


170 revs = 1068.141502221 rad
450 rpm=47.1 rad/s
t = 32.0 sec

1068.141502221 = 47.1 * 32 +0.5 * α * 32^2
So α = 0.528992424 rad/s?

You solved that last equation wrong. You got a positive value of α. That would say it's speeding up. Try it again.
 
α is the angular acceleration. You need to calculate the rpm at t=32 s. ehild
 
ehild said:
α is the angular acceleration. You need to calculate the rpm at t=32 s.


ehild

True. I probably shouldn't have stopped there. Finding α is only a step along the way. And you have the units of α wrong. Should be rad/s^2.
 
How do I get it to be negative unless it should be -.5? Oh right it is rad/s^2.
 
oldspice1212 said:
How do I get it to be negative unless it should be -.5?

Show how you got α = 0.528992424 rad/s^2.
 
I divided right from left

Formula I used: θ = ωi * t + ½ * α * t^2
 
oldspice1212 said:
I divided right from left

Formula I used: θ = ωi * t + ½ * α * t^2

No complaints about the equation you derived from but in numbers that becomes: 1068.1 =α 512.0 + 1507.2. How does "divided right from left" give you a positive α?
 
α = (1068.1 - 47.1 * 32) / ( 0.5 * 32^2)

Mhm not sure what I was doing before but now I did this and got negative; -0.857617188 rad/s^2

To get it to rad/s I use ωf = ωi + α * t

Therefore it's (47.1 + ( -0.857617188 * 32) = 19.656 rad/s
 
  • #10
oldspice1212 said:
α = (1068.1 - 47.1 * 32) / ( 0.5 * 32^2)

Mhm not sure what I was doing before but now I did this and got negative; -0.857617188 rad/s^2

That's better. Can you get the final angular velocity now?
 
  • #11
Sorry I edited the one beforeTo get it to rad/s I use ωf = ωi + α * t

Therefore it's (47.1 + ( -0.857617188 * 32) = 19.656 rad/s
 
  • #12
oldspice1212 said:
Sorry I edited the one before

edit:

To get it to rad/s I use ωf = ωi + α * t

Therefore it's (47.1 + ( -0.857617188 * 32) = 19.656 rad/s

Ok, so the rest is easy, yes?
 
  • #13
I'm doing it right now, I'll post my answers just to be sure. Btw thanks a lot!
 
  • #14
2. ωf = ωi + α * t

0 = 47.1 -0.857617188 * t

So 47.1/ ( 0.857617188 ) = 54.9 seconds3. θ = ωi * t + ½ * α * t^2

41.7 * 54.9 + 0.5 * -0.857617188 * 54.9^2

2289.22 - 1292.4 = 996.82

996.82 radians / 2 * pi = 158.64

It says number 3 is wrong?
 
Last edited:
  • #15
oldspice1212 said:
2. ωf = ωi + α * t

0 = 47.1 -0.857617188 * t

So 47.1/ ( 0.857617188 ) = 54.9 seconds3. θ = ωi * t + ½ * α * t^2

41.7 * 54.9 + 0.5 * -0.857617188 * 54.9^2

You made a mistake (in red).

ehild
 
  • #16
Blah human blunder, lol thanks a ton.

2585.79 - 1292.4 = 1293.39

1293.39/ (2 * pi) = 205.849 revs

Thanks again!
 
  • #17
You could have saved lot of work by using the formula θ=0.5(ωiif)t, which is the same as

number of revolutions= 0.5(fi+ff)t (f is revolutions/s).

fi=450/60=7.5 s-1
170=0.5(7.5+ff))32--->ff=3.125 s-1, ωf= 19.6 rad/sehild
 
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