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Values for mass

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Compact discs and long-playing records are made from similar materials. The former have a diameter of about 12 cm, and the latter, about 32 cm. When in use, records spin at 33.333 rev/min, and compact discs spin at, say, 405 rev/min. Ignoring the holes in both objects and assuming that a compact disc has half the thickness of a record and 0.90 of its density, what is the ratio of the angular momentum of a compact disc in use to that of a record?
    mc= mass of cd
    mR= mass of record
    Wrec = 33 1/3 rev/min = 3.49 rad/s
    Wcd = 405 rev/min = 42.4 rad/s
    rcd= 6cm = .06m
    rR = 32cm = .16m
    Height R = 1????
    Height cd = .5????
    density R = 1????
    density CD = .9???


    2. Relevant equations
    m= d*v
    v= 2pi*r^2*H
    rev/min = 2pi/60 rad/s
    I = 1/2mr^2

    3. The attempt at a solution

    i have all the work written out im just kind of stuck on what to give the height and density values. So ill show you guys the symbol math and hopefully you can help me figure out what numerical values to give them.
    mR = p*2pi*rR^2*HR
    mcd = .9p*2pi*rcd^2*Hcd

    Lcd/LR = Icd*Wcd/IR*Wr
    = [(1/2mcd*rcd^2)Wcd]/[(1/2mR*rR^2)WR]
     
  2. jcsd
  3. Mar 17, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Since what they want is a ratio just develop the mass as a ratio to begin with.

    For instance Mc/Mr = (6/16)2*(1/2)*(.9)

    That should get you almost the whole way there, because otherwise

    Lc/Lr = Ic * ωc / Ir * ωr

    And ω/ω = 405/33.3 because here radians, revs work out the same.
     
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