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Rotational Inertial and Torque question

  1. Nov 14, 2005 #1
    Can anyone help me out? :smile:

    The top has a moment of inertia
    of 0:0002 kg.m2 and is initially at rest. It is
    free to rotate about a stationary axis AA'. A
    string, wrapped around a peg along the axis
    of the top, is pulled in such a manner as to
    maintain a constant tension of 5.22 N in the string.
    If the string does not slip while wound
    around the peg, what is the angular speed
    of the top after 82.1 cm of string has been
    pulled off the peg? Answer in units of rad/s


    I have a problem with this question. I don't know how to calculate the linear distance given to an angular distance. I found the acceleration first using Torque=(I*acc). But then I don't know what to do next after this. My acceleration turned out to be 26100 rad/s^2
     
  2. jcsd
  3. Nov 14, 2005 #2

    Chi Meson

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    in order to check your torque, we knweed to know the radius of the peg.
    edit:

    aha! by reverse engineering I have discovered that you have used the force as the torque (unless the radius of the peg is 1 meter, which I doubt).

    If the radius of the peg (that the string is wrapped around) is not given, then there is not enough information to solve this problem.
    [tex]\tau = I \alpha[/tex]
    [tex] \tau = Fr [/tex]
    [tex]Fr = I \alpha [/tex]
     
    Last edited: Nov 14, 2005
  4. Nov 14, 2005 #3
    Try square root of [(2*Tension*String Length)/(Moment of Inertia)] .
     
  5. Nov 14, 2005 #4
    also watch your units... (i.e. make sure length is in m NOT cm)
     
  6. Nov 14, 2005 #5
    thanks.it worked.
     
  7. Nov 15, 2005 #6

    Chi Meson

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    D'oh!

    "i've made a huge mistake."
     
    Last edited: Nov 15, 2005
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