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Moment of inertia problem please help

  1. Nov 23, 2007 #1
    Moment of inertia problem...please help !!!

    Hi,
    I have been sat in front of this computer all day trying to work this problem out.

    A turbine rota accelerates to 60rpm from rest in 9 revs then the force is removed and it continues to rotate for a further 20 revolutions.

    Basically I know the torque is 260Nm
    I know the acceleration is 0.35 rads/s^2
    I know the deceleration is 0.16 rads/s^2

    I need to find the moment of inertia in kgm^2

    And i need to find the bearing friction in Nm

    Please help, think my head is going to explode
     
  2. jcsd
  3. Nov 23, 2007 #2
    Did you try just use this formula

    [tex]\tau=I\alpha[/tex]

    where [tex]\tau[/tex] is torque and [tex]\alpha[/tex] is angular acceleration.
     
  4. Nov 23, 2007 #3
    What i didnt explain is I know the answer is 509.8Kgm^2 and I cant get anywere near it with any equation
     
  5. Nov 23, 2007 #4
    I don't know,but I also cannot get the answer.

    Here's the my solution.

    [tex]\tau_f=I\frac{d\omega}{dt}=I\frac{d\omega}{dt}\frac{d\theta}{d\theta}[/tex]

    [tex]\int_{0}^{20\times 2\pi}d\theta\tau_f=I\int_{0}^{1 rad/s}\omega d\omega[/tex]

    [tex]40\pi \tau_f=I\frac{\omega^2}{2}[/tex]

    when we have [tex]\tau_0=260 Nm[/tex]

    [tex]\tau_0-\tau_f=I\frac{d\omega}{dt}\frac{d\theta}{d\theta}[/tex]

    [tex]d\theta(\tau_0-\tau_f)=I\omega d\omega[/tex]

    [tex]9\times 2\pi\left(\tau_0-\frac{\omega^2}{80\pi}\right)=I\frac{\omega^2}{2}[/tex]

    [tex]I=\frac{18\pi\tau_0}{\frac{\omega^2}{2}+\frac{9\omega^2}{40}}[/tex]
     
  6. Nov 23, 2007 #5
    ermm at a glance
    torque=I*alpha + friction*omega

    if it max's out at 60rpm (2pi rads/s) then alpha is 0
    so 260nm=friction* 2pi

    friction= 260/2pi nm/rad/s ??? roughly 40nm/rad/s ??
    dunno if thats any help?
     
  7. Nov 23, 2007 #6
    also when you say that

    Basically I know the torque is 260Nm
    I know the acceleration is 0.35 rads/s^2
    I know the deceleration is 0.16 rads/s^2

    i'm pretty sure that the acc/deelertaion is NOT constant if you have friction which is usually proportional to velocity
     
  8. Nov 23, 2007 #7
    also, i dont know how to do that fancy maths typing!!
    but

    using o for theta
    w for omega

    T=I*Alpha +D*w

    T=I*(d2o/dt2)+D(do/dt)

    in laplace(if you know it?)

    T=Ios^2 + Dos

    lol i should really have put this allin one post, i'll look at it again if i've time later!

    also its
    i have been sitting,
    and a turbine rotar
     
    Last edited: Nov 23, 2007
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