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Angular Acceleration Of A Wheel

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A wheel rotating with a constant angular acceleration turns through 22 revolutions during a 9 s time interval. Its angular velocity at the end of this interval is 15 rad/s.

    What is the angular acceleration of the wheel? Note that the initial angular velocity is not zero.

    2. Relevant equations

    w = omega
    a = alpha

    wf = wi + at
    theta = wit + 1/2at2

    3. The attempt at a solution

    wf = 15
    wi = ?
    a = ?
    theta = 2pi * 22 = 128.23
    t = 9

    So I have two unknowns and two equations but I can't seem to solve for the two equations. Both equations have two unknowns in them which is throwing me off. Any help is appreciated.

    Thanks
     
  2. jcsd
  3. Oct 16, 2009 #2

    rl.bhat

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    The first equation can be written as
    wf*t = wi*t + a*t^2and subtract it from the second equation.wi*t get canceled out.Then solve for a.
     
  4. Oct 16, 2009 #3
    How does this look?

    wft = wit + at2
    theta = wit + 1/2at2

    wft - theta = at2 - 1/2at2

    wft - theta / t2 = a - 1/2at2

    wft - theta / t4 = a - 1/2a

    2wft - theta / t4 = a

    Thanks for the help.
     
  5. Oct 16, 2009 #4

    rl.bhat

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    wft - theta = at2 - 1/2at2

    wft - theta / t2 = a - 1/2at2
    This step is wrong. It should be
    wft - theta = 1/2*at2
     
  6. Oct 16, 2009 #5
    How come the at2 can just be dropped?

    Following your correction:

    wft - theta = 1/2*at2

    wft - theta / t2 = 1/2a

    2wft - theta / t2 = a
     
  7. Oct 16, 2009 #6

    rl.bhat

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    It should be
    2[wft - theta] / t2 = a[
     
  8. Oct 18, 2009 #7
    Here is what I got:

    2((15*9)-138.23) / 92 = -.0798

    Does that mean that the disc has a negative acceleration?

    Also, I still do not understand how you just dropped the at2:

    wft - theta = at2 - 1/2at2

    wft - theta / t2 = a - 1/2at2
    This step is wrong. It should be
    wft - theta = 1/2*at2
     
  9. Oct 18, 2009 #8

    rl.bhat

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    Also, I still do not understand how you just dropped the at^2
    I am not dropping at^2. It is a simple algebra.
    wft-θ = at^2 - 1/2*at^2
    = 2*1/2*at^2 - 1/2at^2....= ? [Just like (2x - x) = ?]
     
  10. Oct 18, 2009 #9
    Ah, okay. I see it now.

    I am unsure if my answer is in rad/sec2 which the question specifies for. I multiplied 15 rad/s times 9 sec - 138.23 rad all divided by 9 sec. The terms seem correct but the negative is still throwing me off.
     
  11. Oct 18, 2009 #10

    rl.bhat

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    See whether the angular velocity increasing or decreasing. If it is decreasing then the acceleration is negative.
     
  12. Oct 18, 2009 #11
    22 rev / 9 sec = 2.44 rev/sec = 15.36 rad / sec. At the end of the time interval it is at 15 rad/s, so I guess the acceleration is negative.
     
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