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Inertia & Torque Calculation Assistance

  1. Mar 8, 2012 #1
    Hi, I would like my workings for inertia and torque checked, and advice on formula if you would be so kind.

    I have an aluminium disc of 300mm diameter and 8mm thick and a mass of 1.58Kg. To work out the inertia I have used the formula:

    I = 0.5m*r^2
    I= 0.79*0.0225
    I= 0.0178 Kg m^2

    Now, the disc is to spin at 30rpm (0.25 secs to move 45 degrees) and I would like to find the torque, so I calculate:

    angular velocity ω1= 0
    angular velocity ω2= ((30rpm/60)*2pi) = 3.1415 rad/s
    α = (ω2-ω1)/t = (3.1415-0)/0.25 = 12.57 rad/s^2

    T= Iα = 0.0178*12.57 = 0.2237 Nm

    How close am I, or have I gone wrong somewhere?

    Also, does anyone know if the formula below is correct:

    M = (Jω)/t

    Thanks in advance.
     
  2. jcsd
  3. Mar 8, 2012 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi Simples!Welcome to PF! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)

    If the question says that the disc starts at 0 rpm, and is uniformly accelerated to 30 rpm after 0.25 seconds and 45°,

    then yes, everything is correct :smile:

    (but I have a suspicion that it doesn't say that :redface:)
    Perhaps I'm being dim :blushing:, but what are M and J ? :confused:
     
  4. Mar 8, 2012 #3

    OldEngr63

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    Gold Member

    I'll take a guess on

    M = (Jω)/t

    I'm willing to bet that J = I = polar mass moment of inertia, ω is system final angular velocity starting from zero, M is average moment acting over the impulse time interval t. Thus, this is a very crude statement of change in angular momentum is equal to the angular impulse, for the particular case where angular momentum is zero and everything is expressed in terms of average values.

    Don't think I've seen this one before!
     
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