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Propeller Rotation Problem - Please check

  1. Nov 30, 2008 #1
    Thanks in advance - it went smoothly so I'm hoping I'm right but... usually that's when I've made a big mistake.

    1. The problem statement, all variables and given/known data
    An airplane propeller is 2.08 m (tip to tip) and has a mass of 117 kg. When the engine is first started it applies a constant torque of 1950 N m to the propeller which starts from rest.
    a) What is the angular acceleration of the propeller? (model it on a slender rod).
    b) What is the propeller's angular speed after 5.00 revolutions?
    c) How much work is done during the first 5.00 revolutions by the engine?
    d) What is the average power during the first 5.00 revolutions?
    e) What is the instantaneous power output of the engine the instant it has completed 5.00 revolutions?


    2. Relevant equations
    I think this is all of them
    I = mL2/12
    [tex]\Theta[/tex]f = [tex]\Theta[/tex]i + [tex]\omega[/tex]t + [tex]\alpha[/tex]t2/2
    [tex]\omega[/tex]f = [tex]\omega[/tex]i + [tex]\alpha[/tex]t
    W = [tex]\Delta[/tex]KE = I[tex]\omega[/tex]2/2
    Pavg = [tex]\Delta[/tex]P/2 = [tex]\tau[/tex][tex]\omega[/tex]/2
    Pins = [tex]\tau[/tex][tex]\omega[/tex]


    3. The attempt at a solution
    work that got me to the answers below.
    a) 42.2 b) 53.9 rad/sec2 c) 1.14 kJ d) 52.6 kW e) 10.5 kW


    I = mL2/12 = 117*2.082/12 = 42.2

    a) [tex]\tau[/tex] = 1950 N m = I[tex]\alpha[/tex]
    [tex]\alpha[/tex] = 1950/42.2 = 46.2 rad/sec2

    b) [tex]\Theta[/tex]f = [tex]\Theta[/tex]i + [tex]\omega[/tex]t + [tex]\alpha[/tex]t2/2
    5*2[tex]\pi[/tex] = 0 + 0 + 46.2t2/2

    t = squareroot (20[tex]\pi[/tex]/46.2) = 1.166 seconds

    [tex]\omega[/tex]f = [tex]\omega[/tex]i + [tex]\alpha[/tex]t
    = 0 + 46.2*1.166 = 53.9 rad/sec

    c) W = [tex]\Delta[/tex]KE = I[tex]\omega[/tex]2/2
    = 42.2*53.9/2 = 1.14kJ

    d) Pavg = [tex]\Delta[/tex]P/2 = [tex]\tau[/tex][tex]\omega[/tex]/2 = 1950*53.9/2 = 52.6 kW

    e) Pins = [tex]\tau[/tex][tex]\omega[/tex] = 1950*53.9 = 10.5 kW/s
     
  2. jcsd
  3. Nov 30, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi JWDavid,

    I just wanted to point out here that you seemed to have typed in the wrong value; 42.2 is the moment of inertia; but I see you have the correct answer for the angular acceleration below.

    I don't think this is correct; it looks like you neglected to square the angular speed.

    This also looks incorrect to me. The average power is not just the work done divided by 2; the formula is something like:

    [tex]
    P_{\rm ave} = \frac{\mbox{work}}{\Delta t}
    [/tex]

    Your book may have it written differently, of course. But the point is its the fraction (work done in some time interval)/(time interval).

     
  4. Nov 30, 2008 #3
    thank you,
    You are correct I did forget to square the angular speed.
    But on the average power - it does work both ways. If you note I wasn't taking the work done I was taking the change in Power/2 {(torque*angular speed)-0}/2.

    thanks again for your observations, I would probably have lost points on this had I not asked for help.
     
  5. Dec 1, 2008 #4

    alphysicist

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    Homework Helper

    Sorry about part d; I was so focused on the dividing by 2 that I did not actually read what you wrote down, and just assumed you did the (work)/(time) route!
     
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