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Homework Help: Maximum angular velocity

  1. Sep 9, 2014 #1
    I calculated a, b, and c, I just need help with d.

    1. The problem statement, all variables and given/known data[/b]
    The operation of a "reverse" three-speed automotive transmission is illustrated the in
    figure. Each of the gears rotate about a fixed axis. Note that gears A and B, C and D, E
    and F are in mesh. The radii of each of these gears are shown in the figure. The shaft G
    starts from rest and reaches its maximum angular velocity of ωG = 60 rad/s. The mass
    moment of inertia of the load shaft is given as 5.0 kgm^2
    . The shaft G is driven by a 2 hp motor. Assume that the efficiency of the gear box is 80 %.

    a) If gear A has 45 teeth and if B, C and D, E have the same diametral pitch
    calculate the number of teeth in gears B, C, D, E, F.
    b) Determine the angular velocity of the drive shaft H.
    c) Calculate the torque on the output shaft.
    d) How many turns does the load shaft make to reach its maximum angular velocity.

    2. Relevant equations

    3. The attempt at a solution
    a) Used the ratios to solve, all end up having to be .5, so 45, 15, 15, 25, 35, 30

    b) Multiplied ω*45/15, and repeated for every gear until I reduced it to 127ω
    c)(127/.8) for efficiency, converted 2hp into 1491.5 W. 1491.5/(127/.8)=9.4
    d) Not sure where to start for this... I assume I need something that has rpm in it, and I have to use the moment of inertia somehow.

    Attached Files:

  2. jcsd
  3. Sep 9, 2014 #2

    Simon Bridge

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    (d) that would be correct - though it will help to think through the physics before looking for an equation.
    What determines the mag angular velocity?
    What is the initial angular velocity?
    Does it have a uniform acceleration?
  4. Sep 10, 2014 #3
    The rpm from the motor?
    127ω? Or is that the max? I don't know how the acceleration of this system works
    I think it is constant, but I don't know what it is
  5. Sep 10, 2014 #4

    9.4 N-m=5kg*m^2 * Ac

    Solving for time: ω=ω0+Act

    126 rad/s * 60 / 2*∏ = 1203 rpm
    1203 rpm *(67.02s/60)=1344 revolutions

    Does this look correct?
  6. Sep 11, 2014 #5

    Simon Bridge

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    That looks likt the sort of reasoning I'd use yeah.
    I knew you'd get there :)
  7. Sep 12, 2014 #6
    According to my solution the answer is 667 rev, should I be using the 60 rad/s anywhere? Or is there any mistake you see?
  8. Sep 12, 2014 #7

    Simon Bridge

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    Of course... That's the max speed.
    I didn't go through your work in detail. Where did you get the figure you used?
  9. Sep 12, 2014 #8
    Never mind
    Last edited: Sep 12, 2014
  10. Sep 12, 2014 #9
    Never mind I figured it out.
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