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Simplifying slope-angular acceleration vs torque

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    In a lab we conducted in class we graphed angular acceleration vs torque, and found a positive linear correlation-easy enough. However I am getting rather stuck in simplifying the units of slope in terms of Kg, m, and s... I am fairly sure this will give me that angular acceleration equals torque times inertia but I have trouble with simplifying the slope AND with even getting torque and angular acceleration simplified to kg m and s. I would really appreciate some help! thanks!!

    2. Relevant equations

    Torque=NM
    NM=Kg*a*m
    kg*a*m=kg*(m/s^2)*m
    =kg*m^2/s^2

    Angular Acceleration= rad/s^2
    rad/s^2=at/r
    at/r=m/s^2/r
    m^2/s^2


    3. The attempt at a solution
    (kg*m^2/s^2)/(m^2/s^2)
    =kg
     
    Last edited by a moderator: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2

    vela

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    Your mistake is here. You have
    $$\left[\frac{a_t}{r}\right] = \frac{\text{m}/\text{s}^2}{\text{m}} = \cdots$$ You shouldn't get m2/s2.


     
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