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Torque Problems

  1. Oct 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 2.0 m diameter and a mass of 270 kg. Its maximum angular velocity is 1500 rpm.

    A) A motor spins up the flywheel with a constant torque of 58 N\cdot m. How long does it take the flywheel to reach top speed?

    B) How much energy is stored in the flywheel?

    C) The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 s. What is the average power delivered to the machine?

    D) How much torque does the flywheel exert on the machine?

    2. Relevant equations
    T = r x F
    a = Tnet/I
    I = [tex]\Sigma[/tex]mr^2
    Krot = .5Iw^2
    Period = 2pi/w

    3. The attempt at a solution

    I first apologize for the lack of units, I'm not very solid with units while doing rotational work and get confused as to what they should be.

    I = .5(270)(2)^2 = 540kg*m^2
    1500rpm = 2pi/w = 25 rounds/seconds
    w = 39.59
    a = (58/540) = .107 (theta)/s^2

    Actually, I just solved the first part when typing that out at 370s. The rest of the problems seem pretty linear, but I can't figure them out. I'm just plugging it into the Krot=.5Iw^2 forumla, and the answer comes out wrong.

    I've tried this problem for a couple hours and asked tenish people from my physics class how to do it, but they couldn't figure it out either.
  2. jcsd
  3. Oct 28, 2010 #2


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    Homework Helper

    ω should be in rad/s (radians per second)

    and α would be in rad/s2 (radians per second squared)

    This is just asking for the rotational kinetic energy.

    Power is simply energy/time

    Another way to write power for rotational motion is P=Tω
  4. Oct 28, 2010 #3
    So it looks like if I can get part B done then I can get the rest by just using that answer.

    I'm doing Krot=.5Iw^2=(.5)(540)(39.69)^2=425000, but this is comming out wrong in Mastering Physics. I've tried 425329 and some others, and now am down to 1 more guess. Any help?
  5. Oct 28, 2010 #4


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    Sorry, I forgot to point out your error. You converted 1500 rpm incorrectly.

    rpm means revolution per minute

    1 revolution = 2π radians
    1 min = 60 seconds

    Hence 1 rpm = 2π/60 rad/s

    so 1500rpm = 1500(2π/60) rad/s
  6. Oct 28, 2010 #5
    Okay, i got 1.7*10^6 for B which is correct, and 3.8*10^5 for C, which is also correct.

    For D, I'm doing 3.8*10^5 = Tω
    (3.8*10^5)/w = T

    I'm not sure how to get w. I do not believe it is my original ω, because using ω = 25*2pi gave me the wrong answer when use as (3.8*10^5)/(25*2pi). I'm not sure how to come up with the new ω to get the new T.
  7. Oct 29, 2010 #6


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    It would be the same angular velocity with half of the power.
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