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Homework Help: Flywheel and Gearbox Coupling Help

  1. Jan 8, 2016 #1

    I've been working my way through the following question but I am a little bit stuck at this part:

    A gearbox and flywheel are as shown in FIGURE 4. The output shaft rotates in the opposite direction to the input shaft at 5 times its speed. The gearbox has an efficiency of 92%.

    If the flywheel is solid, has a mass of 50 kg, a diameter of 1.5 m and is to accelerate from rest to 300 revs min–1 in 1 minute:

    Calculate the magnitude and direction of the torque required to hold the gearbox stationary (holding torque Th).

    2. Relevant equations

    Input shaft rotating clockwise.
    Output shaft rotating anti-clockwise.

    Input Angular Velocity W1 = 0 RAD/s
    Final Angular Velocity W2 = (300*2π)/60
    Final Angular Velocity W2 = 31.42 RAD/s

    Angular Acceleration α = (W2 - W1)/t
    Angular Acceleration α = (31.42 - 0)/60
    Angular Acceleration α = 0.52 RAD/s

    Moment of inertia for solid disc I = 1/2*MR2
    Moment of inertia for solid disc I = 0.5*50*0.752
    Moment of inertia for solid disc I = 14.06 KG/m2

    Torque of output shaft T2 = I*α
    Torque of output shaft T2 = 14.06*0.52
    Torque of output shaft T2 = 7.3Nm
    Regarding gearbox efficiency T2 = (7.3/100)*92
    Therefore T2 = 6.716Nm

    Torque of input shaft T1 = T2 * 5
    Torque of input shaft T1 = 6.716 * 5
    Torque of input shaft T1 = 33.58Nm

    3. The attempt at a solution

    I'm struggling because a lot of the worked examples already have input or output power, which is what I need to calculate the holding torque TH. Could I calculate the output power using T2 * W2? Thus 6.716 * 31.42 = 211Kw and work backwards from there?

  2. jcsd
  3. Jan 8, 2016 #2


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    I haven't checked your arithmetic but you seem to be doing ok so far.

    Take a look at your worked examples again. How do they get from input power to holding torque?
  4. Jan 8, 2016 #3
    By using the angular velocity Wi: input power = Ti * Wi.

    I'm guessing to get Wi that you divide the output speed by 5, so 300/5 = 60. Then convert to RAD/s to get 6.283 RAD/s.

    Then placing that into input power formula: 33.58 * 6.283 = 210.98 Watts.

    Is that correct?
  5. Jan 8, 2016 #4
    Don't worry, I've figured it out. The above post is incorrect anyway. Thanks for the hint :)

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