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Moment of Interia. Help please.

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data
    How long does it take for the hanging mass to fall the same distance 31.5 cm?
    I don't know how to solve for T with everything I have. Help please.

    2. Relevant equations

    mgh=(1/2)I(2h/tr)2 + (1/2)m(2h/t)2

    3. The attempt at a solution

    Okay, previous questions are[All correct]
    1.
    Two masses of 150. g are suspended from a massless rod at a distance of 9.0 cm from the center. What is the moment of inertial of the two-mass system about the center of the rod?

    I=m1*R^2
    since they are two masses that weigh the same
    I= 2(.15kg)(.09m)^2=2.43E-03 Correct


    2.
    If the masses rotate with an angular velocity of 2.45 rad/s, what is the rotational kinetic energy of the system?

    KEr=(1/2)Iw^2 = (2.43E-03 )(2.45 rad/s)^2 = 7.29E-03 J Correct


    3.
    Consider the setup shown in the lab manual but with the large masses removed from the support rod. If the hanging mass is 100. g and drops a distance 31.5 cm in a time of 6.1 s, what is the moment of inertia of the support rod and shaft? The radius of the shaft is 0.50 cm.

    I=mr^2(gt^2/2h - 1) = 1.446E-03 kg m^2 Correct


    4.
    Now two masses each of 200 g are placed on the rod at a distance of 11.0 cm from the point of rotation. What is the TOTAL moment of inertia of the masses plus rod and shaft?
    Yes, Computer gets: 6.286E-03 kg m^2

    I=2MR^2+Io is what you get from #3
    I=2(.2kg)(.11m)^2 + Io = 6.286E-03 kg m^2 Correct
     
  2. jcsd
  3. Nov 12, 2007 #2
    BUMP if anyone can help out.
     
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