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    Need some help
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    comoving_frame
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    (Original post by TeeEm)
    Need some help
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    comoving_frame
    Am I being stupid?

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    The mass has lost 4gh of PE when it comes to rest, which can only be converted into the work against the couple. This turns through 2h so C \times 2h = 4gh \Rightarrow C=2g
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    (Original post by atsruser)
    Am I being stupid?
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    The mass has lost 4gh of PE when it comes to rest, which can only be converted into the work against the couple. This turns through 2h so C \times 2h = 4gh \Rightarrow C=2g
    have I mistyped?
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    (Original post by atsruser)
    Am I being stupid?
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    The mass has lost 4gh of PE when it comes to rest, which can only be converted into the work against the couple. This turns through 2h so C \times 2h = 4gh \Rightarrow C=2g
    there is energy lost as the string gets taut and the pulley is set in motion
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    (Original post by TeeEm)
    there is energy lost as the string gets taut and the pulley is set in motion
    Yes, you're right. You can't use conservation of energy here - it's analogous to a completely inelastic collision of two particles, since the "separate" mass and pulley turn into a single system after the string becomes taut. I think I see how to get started though(*), but I'd still need to think about the subsequent motion. Tricky problem.

    (*)
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    conservation of AM to get initial \omega of pulley
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    (Original post by TeeEm)
    Need some help
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    comoving_frame
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    C=\frac{5mg}{9} ?
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    (Original post by atsruser)
    Yes, you're right. You can't use conservation of energy here - it's analogous to a completely inelastic collision of two particles, since the "separate" mass and pulley turn into a single system after the string becomes taut. I think I see how to get started though(*), but I'd still need to think about the subsequent motion. Tricky problem.

    (*)
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    conservation of AM to get initial \omega of pulley
    correct plus the speed of the particle post jerking
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    (Original post by atsruser)
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    C=\frac{5mg}{9} ?
    I got different fraction but I could easily have made a mistake.
    (I get free proof reading here)
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    what about work-energy principle?
    Assuming that the magnitude of the opposing couple is constant (thus its torque), how could we determine its work?
    Is the initial rotational kinetic energy zero? What about the final total kinetic energy? What's the total work done on the system?
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    (Original post by depymak)
    what about work-energy principle?
    See the comments above already. It's not a perfectly elastic process.
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    (Original post by TeeEm)
    I got different fraction but I could easily have made a mistake.
    (I get free proof reading here)
    Spoiler:
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    I get the velocity of the mass after the jerk to be

    v_1 = \frac{\sqrt{2gh}}{3}

    and the equation of motion for the pulley after the jerk to be:

    2C-mg = 2ma \dot{\omega} with initial angular velocity \omega_1 = \frac{v_1}{a}

    The deceleration of the mass is \dot{v}=\frac{g}{18} via SUVAT and we have \dot{\omega} = \frac{\dot{v}}{a}

    I'll leave you to check this and compare to your version.
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    (Original post by atsruser)
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    I get the velocity of the mass after the jerk to be

    v_1 = \frac{\sqrt{2gh}}{3}

    and the equation of motion for the pulley after the jerk to be:

    2C-mg = 2ma \dot{\omega} with initial angular velocity \omega_1 = \frac{v_1}{a}

    The deceleration of the mass is \dot{v}=\frac{g}{18} via SUVAT and we have \dot{\omega} = \frac{\dot{v}}{a}

    I'll leave you to check this and compare to your version.
    we have to wait for the Mechanics master-kid .
    (Hopefully I catch him !!!, as he looks unbeatable at the moment)

    I get 4/3 mg (I could be wrong)
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    my proposal is as follows:
    KFINAL-KINITIAL=WNET
    WC=2aCΔθ
    Δθ=2h/a
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    (Original post by depymak)
    my proposal is as follows:
    KFINAL-KINITIAL=WNET
    WC=2aCΔθ
    Δθ=2h/a
    my proposal is to do a full solution, as you see it, take a photo and post it
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    (Original post by TeeEm)
    my proposal is to do a full solution, as you see it, take a photo and post it
    not sure though but ..in a while
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    (Original post by depymak)
    not sure though but ..in a while
    no worries
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    let me know your thoughts.
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    (Original post by depymak)
    let me know your thoughts.
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    overly simplified

    there are three parts
    Before getting taut
    The instant that it gets taut
    the motion after
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    (Original post by TeeEm)
    we have to wait for the Mechanics master-kid .
    (Hopefully I catch him !!!, as he looks unbeatable at the moment)

    I get 4/3 mg (I could be wrong)
    Shouldn't the dimensions be Nm and not N?
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    (Original post by A Slice of Pi)
    Shouldn't the dimensions be Nm and not N?
    put up a solution and will check it ...
    I might have missed an a when I typed
 
 
 
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