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Disk accelerating against drag

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    I'm struggling to formulate the mathematical description of this question;



    Can anyone help me get started with it?

    What's confusing me is that the power is not given a value, but nor does it show up in the final result as P, so I'm trying to work out how to formulate the equation of motion.

    Thank you.
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    (Original post by 99wattr89)
    I'm struggling to formulate the mathematical description of this question;



    Can anyone help me get started with it?

    What's confusing me is that the power is not given a value, but nor does it show up in the final result as P, so I'm trying to work out how to formulate the equation of motion.

    Thank you.
    You know that P=\tau \omega and  \alpha = \frac{d \omega}{dt} = \tau_{total} /I which allows you to set up your differential equation.
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    (Original post by suneilr)
    You know that P=\tau \omega and  \alpha = \frac{d \omega}{dt} = \tau_{total} I which allows you to set up your differential equation.
    Actually, I didn't know those!

    Tau represents the torque, right? r x F
    And is equal to the time derivative of the angular momentum.

    I'm not sure how that gives power. I was thinking of power as work/time, or (force*distance in direction of force)/time.
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    (Original post by 99wattr89)
    Actually, I didn't know those!

    Tau represents the torque, right? r x F
    And is equal to the time derivative of the angular momentum.

    I'm not sure how that gives power. I was thinking of power as work/time, or (force*distance in direction of force)/time.
    Power is work per unit time and the work done by a rotating body is the integral of the torque wrt to theta.
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    I'd think you could experss P in terms of \omega_0 - saves one integral.

    Edit: But it's a little beyond me. :sad:
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    (Original post by ghostwalker)
    I'd think you could experss P in terms of \omega_0 - saves one integral.

    Edit: But it's a little beyond me. :sad:
    There is only the one integral?
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    (Original post by suneilr)
    There is only the one integral?
    Fair enough. I did say it was a bit beyond me.
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    (Original post by suneilr)
    Power is work per unit time and the work done by a rotating body is the integral of the torque wrt to theta.
    Thank you! I think that makes sense to me. Torque is the rotational force, so the angle rotated through times the force gives the work, and power is the rate of work.

    I'm still trying to work out how angular acceleration is given by total torque times moment of inertial though. Acceleration is normally force/mass, but that suggests that moment of inertia is 1/(mass), which doesn't make sense.

    (Original post by ghostwalker)
    I'd think you could experss P in terms of \omega_0 - saves one integral.

    Edit: But it's a little beyond me. :sad:
    Thanks for trying!
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    (Original post by 99wattr89)
    Thank you! I think that makes sense to me. Torque is the rotational force, so the angle rotated through times the force gives the work, and power is the rate of work.

    I'm still trying to work out how angular acceleration is given by total torque times moment of inertial though. Acceleration is normally force/mass, but that suggests that moment of inertia is 1/(mass), which doesn't make sense.



    Thanks for trying!
    woops that should indeed be torque/ I
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    (Original post by suneilr)
    woops that should indeed be torque/ I
    Ahhh, I see! So moment of inertia acts like mass when dealing with rotation, I think that seems logical too.

    Unfortunately, the solution is also troubling me - I have:

    Angular acceleration = (Power - Drag)/I

    Where Power = Angular velocity x torque from power, and Drag = -k x Angular velocity

    So to find the time you integrate the acceleration with regard to time and set the result equal to w, with the constant being zero. But how can you integrate the drag term with regards to time?
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    (Original post by 99wattr89)
    Ahhh, I see! So moment of inertia acts like mass when dealing with rotation, I think that seems logical too.

    Unfortunately, the solution is also troubling me - I have:

    Angular acceleration = (Power - Drag)/I
    That shouldn't be power - drag. They both need to be torques, so it should be P/w - drag torque.
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    (Original post by suneilr)
    That shouldn't be power - drag. They both need to be torques, so it should be P/w - drag torque.
    Oh, sorry, yes. So angular acceleration is (P - kw2)/I ?
    I'm still not sure how to integrate that though, because I don't know what w is as a function of time.
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    (Original post by 99wattr89)
    Oh, sorry, yes. So angular acceleration is (P - kw2)/Iw?
    do you remember how to solve differential equations like dy/dx = y?
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    (Original post by suneilr)
    do you remember how to solve differential equations like dy/dx = y?
    Oh, I've never done that in reverse! :eek:

    I think I see how to do the integral now, treating the w as integral of dw, then rearranging and integrating to get t = Iln(P-kw2).

    The trouble is that the given answer is of a different form. It uses w0, w when the net torque is zero, but doesn't use power. :confused:
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    (Original post by 99wattr89)
    Oh, I've never done that in reverse! :eek:

    I think I see how to do the integral now, treating the w as integral of dw, then rearranging and integrating to get t = -1/2Iln(P-kw2/P).

    The trouble is that the given answer is of a different form. It uses w0, w when the net torque is zero, but doesn't use power. :confused:
    It says that w0 is the velocity at steady state ie when dw/dt = 0 which gives you an expression for P in terms of w0.
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    (Original post by suneilr)
    It says that w0 is the velocity at steady state ie when dw/dt = 0 which gives you an expression for P in terms of w0.
    Er, at the moment I have t = -(I/2k)ln(P-kw2)

    Is that right? I can't get the kw2/P term you suggested.
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    (Original post by 99wattr89)
    Er, at the moment I have t = -(I/2k)ln(P-kw2)

    Is that right? I can't get the kw2/P term you suggested.
    You need to integrate between w=0 and w, so you get the 1/P term in the logarithm as well.
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    (Original post by suneilr)
    You need to integrate between w=0 and w, so you get the 1/P term in the logarithm as well.
    Ah, I'm sorry, I have the right integral now, and I have the final solution.

    Thank you very much for all your help!

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