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A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .

  1. May 24, 2012 #1
    A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .
    The shaft continues to accelerate at this rate for the next 8 sec . Thereafter the shaft rotates with a uniform angular speed . Find the total time to complete 400 rev .

    my answer is ...

    10 rev/s----------------18rev/s-------------------
    4s
    W0 = 62.83 rad/s
    Wf = 113.0
    total time at 400 rev ========400/2Pi
    so Q = 2513 rad
    w = w0 + at
    133.0 = 62.83 + at
    a = 12.4 so this is niforml accelerated
    now I will take the velocity at 12 sec (4+8 = 12 )
    so
    w = w0 + w
    w = 113.0 + 12.4 X (12)
    =261.8

    ------------ now I will take time after 12 s which the a = 0 because of angular speed ----

    Q = w0 +1/2 at^2
    2513=261.8t+1/2(0)^2 t
    t= 9.5 s

    so Total time = 12 + 9.5 = 21.5 s



    t = 2513/261.8 =
     
  2. jcsd
  3. May 24, 2012 #2

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    I think you mean "113.0" Not "133.0." I'm guessing that's just a typo.
    Even with using "113.0" your calculation is off by a significant figure. See my hint below on how to be more exact on some of these calculations.
    If you are looking for the final velocity after t = 12 seconds using that formula, you need to start with initial velocity at 12 seconds previous (t = 0). You used the initial velocity figure at 8 seconds previous (i.e. t = 4 s).

    Here's a hint that might help avoid rounding errors as you solve this problem. Keep things in terms of π.

    For example, when you convert the initial angular velocity from 10 rev/s to units of radians per second, leave it as ωi = 20π rad/s. Similarly, when solving for the acceleration, leave things in terms of π.

    (36π [rad/s] - 20π [rad/s])/(4 ) = 4π [rad/s2]


    You can convert the π to its numerical value at the very end, if it doesn't happen to completely cancel out before then. In the mean time, you'll have to carry πs around, but you won't have to worry about any rounding errors, and you might not even need to use a calculator at all! :wink:

    [Edit: another potential option is to stick with revolutions instead of radians all along. As long as you never need to convert to linear distance/velocity/acceleration, it will still work.]
     
    Last edited: May 24, 2012
  4. May 25, 2012 #3
    thanks sir ,, but know you don't show to me where I do the mistake and
     
  5. May 25, 2012 #4

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    I'll try again, and highlight in red.
    I think you meant 113.0 [rad/s], not 133.0. I'm guessing that's just a typo though.
    There's a rounding error involved with that. The exact angular acceleration is 4π [rad/s2] ≈ 12.56637 [rad/s2]
    The 113.0 [rad/s] value was calculated at t = 4.0 . Yet your equation assumes that the acceleration continues for 12 after that. Either you need to use the ω0 = 62.83 [rad/s] *OR* change Δt to 8 . (Either one will work -- just do one or the other, not both.)

    As far as my previous hint goes, if you keep 'π' as a symbol, rather than a number, it can help avoid rounding errors, if you wish to work everything in units of rad, rad/s, and rad/s2.

    And there's nothing wrong with converting everything to rad, rad/s, and rad/s2 if you are more comfortable working that way. But for this particular problem, it's actually not necessary. You can work directly with θ in units of rev, ω in units of rev/s, and α in units of rev/ss, if you want to. The choice is yours, but just be sure to keep track of your units and stay consistent, whichever way you choose. :wink:
     
  6. May 25, 2012 #5
    Ok I will do next time the anwer with pi ... but let me complete in this Q ^_^

    W0 = 62.83 rad/s
    Wf = 113.0
    total time at 400 rev ========400/2Pi
    so Q = 2513 rad
    w = w0 + at
    113.0= 62.83 + a(4)
    a = 12.5 rad/s^2

    now I will take the velocity at 12 sec (4+8 = 12 )
    so
    w = w0 + w
    w = 62.8+ 12.4 X (12) (12)
    w = 213.13 rad/s

    Q = w0t +1/2 at^2
    2513=12.5t+1/2(0)^2 t
    t= 11 s

    so Total time = 12 + 11 =
    so total time now I got 23.7 s

    please check my answer now ...
     
  7. May 25, 2012 #6

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    That's fine. :smile:
    'Looks good! :approve:
    That's roughly what I got (to the first two or three significant figures anyway). :approve:
    That's not quite right. I see where the 2513 comes from. That's
    3513 [rad] ≈ (2π [rad/rev])(400 [rev])​
    so that's the total number of radians (2π revolutions) that the problem statement is asking you to find.

    But you've calculated the time it takes to make 3513 [rad] at a constant angular velocity of 12.5 [rad/s].

    A couple of things:
    1. The angular velocity is never at a constant 12.5 [rad/s]. You seem to be confusing angular velocity with the 12.5 [rad/s2] angular acceleration.
    2. Not all of the total 400 rev (~3513 [rad]) happen at a constant angular velocity. A large part of them happen when the shaft is accelerating, and only the rest happen at a constant angular velocity.

    Here's another hint: What's the angular displacement (i.e "how many radians are turned") after the first 12 seconds, when the shaft is accelerating? (i.e. what is θ at t = 12 [sec], knowing that the shaft accelerates from time 0 < t < 12 [sec]):wink:
     
  8. May 25, 2012 #7
    sorry little i have confusing anyway

    Q = w0 + 1/2at^2
    Q = 213.13 + 1/2(12.5)X(12)^2
    Q = 1113.3 rad

    now what .... ?
     
  9. May 25, 2012 #8

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    You forgot one of the 't's in the formula. See in red below.

    θ = θ0 + ω0t + ½αt2

    Also, 213.13 [rad/sec] is the final angular velocity that you calculated earlier, corresponding to the end of acceleration (at t = 12 [sec]). The formula above calls for the initial angular velocity at the moment the acceleration began (at t = 0).
     
  10. May 25, 2012 #9
    Q = 3457.56

    Now what ?
     
  11. May 25, 2012 #10

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    Try that one again.
    ω0 ≠ 213.13 [rad/s]

    213.13 [rad/s] is the final angular velocity, not the initial.
     
  12. May 25, 2012 #11
    sorry ...

    Q = 62.83 + 1/2(12.5)X(12)^2
    = 962.83 rad

    Now wha
     
  13. May 25, 2012 #12

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    'Forgot the t again. (θ = θ0 + ω0t + ½αt2)
     
  14. May 25, 2012 #13
    Q = 1653.96 rad
    we want get Q at 12 s what you mean by Q = Q0 + w0t+1/2t^2
    what is Q0 ??
     
    Last edited: May 25, 2012
  15. May 25, 2012 #14

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    Allow me to start with a summary of what we've done so far.

    You've determined the acceleration, using the interval from 0 < t < 4 .

    ωf = ωi + αt

    18 [rev/s] = 10 [rev/s] + α(4 )

    8 [rev/s] = α(4 )

    (8 [rev/s])/(4 ) = α

    α = 2 [rev/s2]

    (Although it's not necessary, you could convert that to radians per second squared by multiplying by 2π [rad/rev], such that (2π [rad/rev])(2 [rev/s2]) ≈ 12.566 [rad/s2)

    You've calculated the final angular velocity, after accelerating for 12 .

    ω12s = ωi + αt

    = 10 [rev/s] + (2 [rev/s2])(12 )

    = 10 [rev/s] + 24 [rev/s]

    ω12s = 34 [rev/s]

    (Although it's not necessary, you could convert that to radians per second by multiplying by 2π [rad/rev], such that (2π [rad/rev])(34 [rev/s]) ≈ 213.628 [rad/s])

    And you've calculated the angular displacement at t = 12 .

    θ12s = θ0 + ωit + ½αt2

    = 0 + (10 [rev/s])(12 ) + ½(2 [rev/s2])(12 )2

    = 120 [rev] + 144 [rev]

    θ12s = 264 [rev].

    (And again, although not necessary, that's approximately 1658.76 [rad/sec])

    The first 12 seconds used up 264 revolutions. The problem statement asks for 400 revolutions, so how many revolutions are left to go?

    You know that the remaining revolutions take place at the final angular velocity of ω12s. So how long does it take to rotate through the remaining revolutions at a constant angular velocity of ω12s?
     
    Last edited: May 25, 2012
  16. May 25, 2012 #15
    thaaaaaaaaaaaaaaaaaaaaaaanks good summary that useful for me

    Q = 400 - 260 = 140 revolutions
     
  17. May 25, 2012 #16

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    Hold on, θ12s = 264 [rev], not 260.

    After making that correction, you have one (or maybe two) more calculation(s) to go and you're finished! :smile:
     
    Last edited: May 25, 2012
  18. May 25, 2012 #17
    Draw a revolutions/sec- time diagram.
    You have a line with a point (a sec, 10rev/sec) and another point at (a+4sec, 18rev/sec)
    Since this is a constant acceleration, the gradient of the line is the acceleration.

    Now continue the line until time=(a+4+8)sec. This point is ((a+4+8)sec, X rev/sec)
    You should be able to find the value of X

    As you should know the area under the line and time axis is the revolutions covered within that period of time

    If less than 400rev add then another area(rev) with constant rev/sec.
    The value of this constant velocity is equal to X as in para. 2.
     
  19. May 25, 2012 #18
    In fact I don't know which correction you mean now from 0 to 12 we have 264 revolutions so this the displacement so remaining revolutions Q = ( 400 - 264 ) = 136 rev
     
  20. May 25, 2012 #19

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    Yes, that's right! :approve:

    Almost there. :smile:

    How long does it take a shaft to rotate 136 revolutions, if it is rotating at the constant velocity ω12s? (You've already calculated what ω12s is, btw.)
     
  21. May 26, 2012 #20
    that why I face problem in like this Q because I don't know how to draw
    just I do the period as a line like this
    4s--------------12s-----------------ts(400rev)
    a = 12.5-----w =213.623
    w0 =13-------Q = 1658.76

    ---------anyway ........ I complete my answer ---------

    How long does it take a shaft to rotate 136 revolutions
    Q = 854.8 rad
    w = 213.623
    t = 12
    a = 0

    so after that I use this two equation ..
    Q = W0 + 2at^2
    136 = W0(12) + 2 X (0) ( 12^2)
    w0 = 11.3 rad/s

    from this Q we get time at 136 rev ...

    w = w0 +at
    t = 213.62 - 11.3/0 OMG ........ sure is wrong !>>
     
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