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Determining the power of a rotating system

  1. Dec 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Force of 200 foot pounds of torque is being exerted on a axle of a disc which is turning at 1.57 radians. The disk has a moment of inertia of 234 kg per meter squared and it's 1.5 inch diameter axle has a prony brake installed applying a load of 200 foot pounds of torque at the axle. How much work and power is being created by this situation

    .
    2. Relevant equations
    W=fd
    Moi
    Power = f x d /time

    3. The attempt at a solution

    1.57 rad/sec x 234 kg /m2 = 367 Nm
    How does the prony brake get accounted f
     
  2. jcsd
  3. Dec 2, 2017 #2

    CWatters

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    Power (in Watts) = torque (in Nm) angular velocity (rads per second)

    200 foot pounds is 271 NM so the power going in is..

    271 * 1.57 = 425W

    If the brake is also applying a torque of 271Nm then the same power is being dissipated into the brake.

    Energy cant be created only moved or converted. In this case 425W is moving from the motor or whatever is turning the flywheel through the flywheel and into the brake. From there it leaves in the form of heat.
     
  4. Dec 2, 2017 #3

    scottdave

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    One thing to note about your formula: the units/dimensions do not correspond, so that can tell you that something is wrong with your formula
     
  5. Dec 2, 2017 #4

    haruspex

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    Wrong dimensions. Moment of inertia is mass times square of distance, not divided by square of distance.
     
  6. Dec 2, 2017 #5

    CWatters

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    And anyway if it's going at a constant angular velocity the moment of inertia is irrelevant to the question of power flow.
     
  7. Dec 4, 2017 #6
    Yes you are right. I typed in the wrong symbol. Thank you.

    But you guys bring up a good point. The MOI x angular velocity = angular momentum, correct? Or in my case angular momentum = 367 Nm. What us thia number used to determine? How is it relevant to power or work?

    So right now we know there is 425 watts being accomplished by the system. If i want to know total power of the system, what else should be added to the 425 watts? The power it took to go from 0 rad/sec to 1.57 rad/sec?
     
  8. Dec 4, 2017 #7

    CWatters

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    Angular momentum is only relevant when the angular velocity is changing. For example...

    If you put in more energy than you take out the energy stored in the flywheel increases eg it goes faster.
    If you take out more energy than you put in the energy stored in the flywheel reduces eg it goes slower.

    There are several way to express this but the most common is probably...

    Net torque = MOI * angular acceleration

    Where net torque is the sum of all torques acting on the wheel taking into account the direction in which they act. Note that if the wheel is going at a constant angular velocity the acceleration is zero so the net torque is zero eg

    Driving torque + brake torque + friction torque = zero
     
  9. Dec 4, 2017 #8

    CWatters

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    Power is the rate of doing work. It's not really meaningful to add rates like that.

    You can calculate the energy/work that was added to the flywheel using

    Energy = 0.5 * MOI * angular velocity^2

    Think of this energy as like the energy stored in a battery. You can only get it back if you discharge the battery eg slow down the flywheel. How much power you can extract just depends on how fast you discharge the battery eg how fast you slow down the flywheel. There is no way to get more energy out than you put in.
     
  10. Dec 4, 2017 #9

    CWatters

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    Suppose a flywheel was spun up until it contained 100 joules of energy, then you could choose how much power to extract from it as follows...

    0w by not slowing it down
    1w by stopping it in 100 seconds
    10w by stopping it in 10 seconds
    100w by stopping it in 1 second
    1000w by stopping it 0.1 seconds

    So it's up to you how much power you extract and that determines how long the flywheel can deliver it. If you don't want the flywheel to slow down you have to apply energy at the same rate you extract it eg at the same power.
     
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