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Moment of Inertia help pls

  1. Dec 23, 2004 #1
    A pulley of moment of inertia 0.021kg-m^2 and radius 0.12m is acted upon by a force which varies in time as F= 0.23t + 0.12t^2 where F is in Newton and t is in second. suppose that the pulley is initially rotating at 0.18 r/s and the force acts tangentially to the pulley. Find the magnitude of the angular velocity of the pulley two seconds after the force began to act on the pulley.

    I was thinking if I can use F(tan) r = I (alpha)

    alpha = (0.12m(0.23t+0.12t^2))/0.21kgm^2 to get the accelration and then plug it in

    w = wi + (alpha)t to get the velocity?

    Is my reasoning correct?

    What if I integrate the Force to get the work W, and then W = Kf - Ki?
     
    Last edited: Dec 23, 2004
  2. jcsd
  3. Dec 23, 2004 #2

    Doc Al

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    Good!
    That equation assumes constant alpha, which is not the case here. You have to integrate to find the final omega.
     
  4. Dec 23, 2004 #3

    You mean intergrate in the Force to get K = 1/2 Iw^2? I'm a bit lost! What do you mean?
     
    Last edited: Dec 23, 2004
  5. Dec 23, 2004 #4

    Doc Al

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    You have [itex]\alpha (t)[/itex]. So realize that [itex]d\omega/dt = \alpha (t)[/itex], so:
    [tex]\Delta \omega = \int_{0}^{2} \alpha(t) dt[/tex]
     
  6. Dec 23, 2004 #5

    dextercioby

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    Nope,as Doc said,u need to integrate this relation

    [tex]d\omega=\alpha dt [/tex]

    between corresponding limits.

    Well,you figure it out.Is it possible without finding [itex] \theta (t) [/itex] and reversing it.?? :rolleyes:
    [tex] \frac{I}{2}(\omega_{f}^{2}-\omega_{i}^{2})=\int_{\theta_{0}}^{\theta_{f}} F(t(\theta))r d\theta [/tex]

    Daniel.
     
  7. Dec 23, 2004 #6
    Do I need to go through this?


    Can't I just use [tex]\Delta \omega = \int_{0}^{2} \alpha(t) dt[/tex]
    But do I need to subtract the initial omega that is given from the final omega?

    One more question please: when I did multiply the force equation by the radius, should I do the dot product or just multiply it commutatively. the reason I'm asking is the units when multiplying 0.12m by 0.23t + 0.12t^2 of course its m/t where t is in seconds. but after that I have to divide the answer by the moment of inertia which is kg m^2. Logicaly I know the answer should be in m/s^2 since its accelration and I also know that accelration could be written in N/kg, but here I have m(t)/kg m^2.
     
    Last edited: Dec 23, 2004
  8. Dec 23, 2004 #7

    Doc Al

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    [tex]\omega_f = \omega_i + \Delta \omega[/tex]

    Also realize that the units of "0.23t + 0.12t^2" are Newtons (when t is in seconds), and the units of torque are N-m. The units of [itex]\omega[/itex] will be radians/sec.
     
  9. Dec 23, 2004 #8
    I keep thinking about rotation in terms of motion. Yes, radian/sec not m/s.
    I really appreciate all your help guys. Thank you very much.
     
  10. Dec 23, 2004 #9

    dextercioby

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    Yes,it's the only way to get the answer.


    U can.Doc gave you the integral formula,i gave u the differential formula.

    Nope,correct integration will give the result.

    The problem gives the force without units.It should be Newtons.But the constants are given numerically:they should be measured,first in Newtons per second,and the second in Newtons per second squared.

    Daniel.
     
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