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Moment of Inertia hw problem

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data

    2 metal disks, one with radius R1= 2.5cm and mass M1=.8kg and the other with radius R2=5.00cm and mass M2=1.60kg, are welded together and mounted on a frictionless axis through their common center.

    a)What is the total moment of inertia of the two disks?
    b)A light string is wrapped around the edge of the small disk, and a 1.5kg block, suspended from the free end of the string. If the blook is 2.00m above the floor what is the speed of just before it strikes the floor?


    2. Relevant equations
    I = Σ mi *ri^2
    K = 1/2 *I*ω^2
    W(g)= m*g*h


    3. The attempt at a solution
    a) I = 0.8kg*(2.5cm)^2 +1.6kg(5.00cm)^2 = 45kg*cm = .0045kg*m

    b) m*g*h = 1/2 *I*ω^2
    i) 1.5kg*9.8m/s^2*2m = .5*(.0045kg*m) * ω^2
    ii) ω = sqrt[(1.5kg*9.8m/s^2*2m ) / (.5*(.0045kg*m))]

    This is what i thought i should do, but looking at the solution in the back of the book, i doesnt seem to be correct. Any help is appericated, thanks.
     
    Last edited: Apr 29, 2007
  2. jcsd
  3. Apr 29, 2007 #2
    If you are going to use energy methods, I think you need to account for the ke of the falling block as well. In other words some of its potential energy of the block goes into rotating the flywheel and some into its own motion. Also I think you dropped a factor of 2m in "1)"
     
  4. Apr 29, 2007 #3
    Ture, that was just a copy error though from my paper to the computer.
    In part a. my book says the answser should be have of what i calculated it to be, do you know why that might be?
     
    Last edited: Apr 29, 2007
  5. Apr 29, 2007 #4

    hage567

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    Homework Helper

    You are using the wrong formula for the moment of inertia of a solid cylindrical disk. It is 0.5MR^2, not MR^2, which is what you are using.
     
    Last edited: Apr 29, 2007
  6. Apr 29, 2007 #5
    Thanks i figured it out when i was reading my book, and i figured out why my velocity calcuation was coming out wrong i forgot to add the KE of the weight. Thanks for your help.
     
  7. Jan 27, 2008 #6

    TFM

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    I am doing this question, and I am using Conservation of Energy and the following equation:

    mgh = 0.5*I*(v^2/r^2)+0.5*m*v^2, which I then arrange:

    2mgh = v^2((I/(r^2))+m)

    v^2 = (2mgh)/((I/r^2)+m)

    but cannot get the right answer when I put in the variables. Any Ideas why?



    TFM
     
    Last edited: Jan 27, 2008
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