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Moment of inertia and angular velocity

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=67419&stc=1&d=1394286461.jpg

    (a) Calculate the moment of inertia I of the disc when it rotates about the pivot as shown in the figure.
    (b) If the disc is released from rest, determine the angular speed, ω, of the disc at its lowest point.

    2. Relevant equations

    a) Id = Icm + md^2
    Icm = 1/2*M*R^2
    b) a = r*γ, a=g
    (ωf)^2 = (ωi)^2 + 2*γ*θ

    3. The attempt at a solution

    a) Id = Icm + md^2
    = 1/2MR^2 + 5(0.3^2)
    = 1/2*5*0.3^2 + 5(0.3^2)
    = 0.675 kg/m^2

    b) a = r*γ, a=g, γ = g/r
    θ = ¼*2pi
    (ωf)^2 = (ωi)^2 + 2*γ*θ
    ωf = sqrt(2*g/r*1/4*2pi) = sqrt(2*9.8/0.3*1/4*2pi = 10.1 rad/s

    Am I doing the question correctly?
    Thank you very much.
     

    Attached Files:

  2. jcsd
  3. Mar 8, 2014 #2
    I am not sure what the "relevant equation" for (b) means.
     
  4. Mar 8, 2014 #3
    b) a = r*γ, a=g
    (ωf)^2 = (ωi)^2 + 2*γ*θ

    Oh, I'm sorry for this.
    I was trying to show that

    linear acceleration = radius * angular acceleration,
    where linear acceleration = gravitational acceleration, in this case

    (final angular velocity)^2 = (initial angular velocity)^2 + 2*angular acceleration*angular displacement
     
  5. Mar 8, 2014 #4
    Don't you think angular acceleration should depend on the moment of inertia?

    Besides, why do you think it will be constant?
     
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