angular KE/momentum

Discussion in 'Physics & Math' started by slayerdeus, Jun 21, 2004.

  1. slayerdeus Registered Member

    Messages:
    15
    Five identical particles of mass m = 0.30 kg are mounted at equal intervals on a thin rod of length l = 1.01 m and mass M = 2.0 kg, with one mass at each end of the rod. If the system is rotated with angular velocity = 60 rev/min about an axis perpendicular to the rod through one of the end masses, determine
    (a) the kinetic energy and
    (b) the angular momentum of the system.

    I have KE = 1/2*Irod*w^2 + 1/2*Iparticle*w^2
    = 1/2(1/3*Mass of rod + 1/2*Mass of particles)*L^2*w^2

    What am I doing wrong? I haven't looked at part b yet.
     
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  3. Pete It's not rocket surgery Registered Senior Member

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    10,167
    I will be different for each particle.
    The centre particle, for example, will have I=0.
     
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  5. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    Each particle has rotational inertia mr<sup>2</sup>, where r is its distance from the rotational axis. The rod itself has rotational inertia Ml<sup>2</sup>/3 about the rotation axis (I hope you can work out where this value comes from). These must be added to get the total rotational inertia of the system, I. Once you have that, you can use the usual expressions for kinetic energy and angular momentum.
     
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