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Angular Velocity

  1. Jan 20, 2008 #1
    [SOLVED] Angular Velocity

    1.
    The angular acceleration is given [tex]\alpha[/tex]=-2w^2 rad/s^2 where [tex]\omega[/tex] is the angular velocity in rad/s. When [tex]\theta[/tex]=30 deg the angular velocity is 10 rad/s. What is the angular velocity when [tex]\theta[/tex]=60deg?


    2.
    Used [tex]\alpha=d\varpi/d\theta * \varpi[/tex]


    3.
    [tex]\int-2 d\theta=\int1/\varpi d\varpi[/tex]

    after integration I got

    [[tex]\(-2)*theta[/tex]]=[[tex]ln\varpi[/tex]]

    limits 0-60 for [tex]\theta[/tex] and 10 to [tex]\varpifor \varpi[/tex]

    I think I'm missing a step somewhere. The book gives an answer of w=3.51. With my calculations I get w=1.14. Can you advise?
     
    Last edited: Jan 20, 2008
  2. jcsd
  3. Jan 21, 2008 #2

    Shooting Star

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    Redo the calculation. Book is correct. The limits of theta should be from pi/6 to pi/3.
     
  4. Jan 21, 2008 #3
    Using pi/6 to pi/3 I get

    -2pi/3 + pi/3 = lnw-ln10

    -pi/3 + ln10 = lnw

    From here I think it is

    1/[e^(-pi/3+ln10)] = w

    Can you check me on this part?
     
    Last edited: Jan 21, 2008
  5. Jan 22, 2008 #4
    1/[e^(-pi/3+ln10)] = w

    e^(-pi/3+ln10) = w [ln a = b ie e^b = a]
     
  6. Jan 22, 2008 #5

    Shooting Star

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    ln 10/w = pi/3. Take antilog of pi/3, use a calulator (or something) => w = 3.51.

    (By antilog, I meant, 10/w = e^pi/3. You have done everything correctly.)
     
    Last edited: Jan 22, 2008
  7. Jan 22, 2008 #6
    I see where I was incorrect.

    Thanks for the help.
     
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