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Constant Angular Acceleration

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Starting from rest at t = 0s, a wheel undergoes a constant angular acceleration. When t = 3.9s, the angular velocity of the wheel is 6.9 rad/s. The acceleration continues until t = 19s, when the acceleration abruptly changes to 0 rad/s^2. Through what angle does the wheel rotate in the interval t= 0s to t= 43s?


    2. Relevant equations

    ω = ωo + αt
    θ = ωot + (1/2)αt^2
    θ = 1/2(ωo +ω)t


    3. The attempt at a solution

    First I found the the angular acceleration using the initial angular velocity. That much I'm sure is right, the rest is where I am having difficulty. I know you have to find theta for each of the different time intervals, but I'm not sure I using the right equations and thus getting the right answers. Anyone willing to help?
     
  2. jcsd
  3. Oct 17, 2007 #2
    First find the angular displacement of the first interval is. θ(0->19) (second equation you listed, ωo = 0)
    Then find the angular velocity at the end of the first interval ω(0->19) (first equation you listed, ωo=0).
    Then find the angular displacement of the second interval θ(19-43) (second equation, ωo=ω(0->19) )
    Now add θ(0->19) and θ(19-43)
     
  4. Oct 17, 2007 #3
    what about the 3.9s?
    or is the 3.9s only needed to find the angular acceleration of the first interval?
     
  5. Oct 17, 2007 #4
    Yup, only needed to find the constant angular acceleration in the first interval.
     
  6. Oct 17, 2007 #5
    I'm confused about one thing. It says at 3.9s angular velocity = 6.9 rad/s, shouldn't that quantity come into play at some time? And also in the last step of the explanation you gave where you say the initial and final angular velocities are equal do you mean the angular velocity I found in the previous step or the given angular velocity?
     
  7. Oct 17, 2007 #6
    Only to determine the angular acceleration during the first interval:
    [tex]\alpha=\frac{6.9 rad/s}{3.9s}[/tex]
     
  8. Oct 17, 2007 #7
    I ended up getting a pretty big number, but I'm not sure if it's right, I got 1636 approximately for my final answer, could you possibly double check for me?

    never mind I just checked it's not right
     
    Last edited: Oct 17, 2007
  9. Oct 17, 2007 #8
    I got something that was a few hundred less than your answer (can't just give it to you :) ) Make sure you're only calculating the constant angular velocity part for a time of (43-19) seconds.
     
  10. Oct 17, 2007 #9
    I've tried everything and I just can't get this right, so I'll walk you through my math step by step. First I found the angular acceleration by dividing the 6.9 by 3.9 and got 1.77 rad/s^2.
    First check point, is my angular acceleration right? I think it is but if isn't let me know.

    Second I did what you said and found the angular velocity at the end of the first interval by multiplying the angular acceleration times the time of 19 seconds since the initial velocity is zero. I got 33.63 from that. Is that right?

    Third I found the angular displacement over the 19 second time interval by multiplying 1/2 times 1.77 times 19^2 and got 319.485. Is that right?

    Fourth I found the angular displacement over the second time interval (24 s) by multiplying 33.63 times 24 and adding that quantity to 1/2 times 1.77 times 24^2 and got the number 1636, which is obviously wrong so at what point did I go wrong?
     
  11. Oct 17, 2007 #10
    anyone still willing to help me out?
     
  12. Oct 30, 2007 #11
    Don't forget that when you're calculating the angular displacement for the 2nd time interval, that the acceleration is 0. so your equation should be:

    θ(19-43) = θ = ωot

    You should get the correct value now after adding the two angular displacements together.
     
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