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Rotational motion of a disc

  1. Nov 9, 2006 #1
    One more problem on a tough worksheet...I have tried it for a while, but can't find an equation(s) suitable for the problem...

    A constant force of 12N is applied to the axle of a disc rolling along a flat plane. The disc has mass m=22kg, diameter D=.50m, and rotational inertia I=.688kgm^2. What is the linear speed of the center of the disc, V, after it has rolled for 12m?

    I drew a free body diagram, and I know that the sum of the moments is equal to I times alpha, and I know the good ol' F=ma. But I can't seem to be able to find an acceleration with what I am given. A hint would be wonderful.
  2. jcsd
  3. Nov 9, 2006 #2
    Oh boy...I went somewhere, but I don't know if it was right. Here is what I did:

    I used m*a=f-f(friction) and friction*radius = I*alpha. Alpha is equal to a*r, so I plugged that in to make friction times radius = I*a*radius.

    Plugging that in to m*a=F-f(friction) i got:


    This is rather confusing, but I think it is somewhat correct. Any suggestions or gross errors on my part?
  4. Nov 9, 2006 #3


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    Right idea, but alpha does not equal a*r. alpha and a are certainly related and you need that relationship, but that is not it.
  5. Nov 10, 2006 #4
    it is....acceleration=radius x alpha
  6. Nov 10, 2006 #5


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    [tex]a_{T}=\alpha \cdot r[/tex], i.e. tangential acceleration equals angular acceleration times radius.
  7. Nov 11, 2006 #6
    I hope this can help you:

    total K = (1/2)*m*v^2 + (1/2)*I*w^2

    actually you don't have to use I = 0.688 which is given if you know the I of the disk is (1/2)*m*r^2.

    plug all in, find out K = 3/4 * m * v^2


    the work does on the wheel also equals to the total K above. W = F*s = 12*12 = 144J

    solve for v.

    Good luck.

    Minh T. Le
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