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Homework Help: Moment of inertia calculation, factor of radius out?

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok i have the following problem. There is a pulley with radius 7.5 cm and 2 forces acting on it (2 tensions of a string) acting in opposite directions either side. One is 18.2N, the other 7.5N. I know the acceleration is 3.75 meteres per second squared.

    Find the moment of inertia.

    2. Relevant equations

    Change in torque = Ia

    3. The attempt at a solution

    Ok so the torque of each individual force is r x F (F being the force and r the radius in meters). As the forces are acting perpendicular, the torque of each force is rF.

    So i thought the total torque is τ = r(18.2-7.5) = Ia.

    So i thought the moment of inertia would equal would equal...

    I = [r(18.2-7.5)]/a

    However my answer seems to be a factor of r out, i.e the equation [(r^2)(18.2-7.5)]/a sems to yield the correct answer.

    Why is this? I do not undersdtand where the extra r comes from which it is multiplied by.

  2. jcsd
  3. Apr 24, 2010 #2


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    What is a?

  4. Apr 24, 2010 #3
    Sorry a is acceleration
  5. Apr 24, 2010 #4
    Is it perhaps that the forces are implemented in this way? (picture added)

    Because then you would have to add the tensions on the pulley, meaning it's gonna be:

    [itex]t_1+t_2 = a I <=> F_1 R + F_2 R = R (F_1+F_2) = I a[/itex], then check if the result is correct.

    Attached Files:

  6. Apr 24, 2010 #5
    Actually, I just noticed. You do have the acceleration of the points where the forces are applied on the string, because the units are meters/second*second. That means that the pulley is actually turning with a rotational acceleration [itex]a_{rotation} = \frac{a}{R}[/itex]. If you substitute that in the second law for rotational motion you will get the result you want.
  7. Apr 24, 2010 #6
    Ah i think i understand. So the a in the Ia part of the equation refers to angular acceleration but what i had was the linar acceleration (the situation is one weight hanging off a table and the other being pulled along the table, both are moving with acceleration 3.75). I still do not really understand how dividing by the radius gives the angular acceleration though.

  8. Apr 24, 2010 #7
    Yes exactly. The following [itex]a_{center-mass}=a_{angular}*R [/itex] belongs to the theory of my school Physics book. In this situation, you can proove something like it since the string isn't sliding on the pulley. The proof is the following:

    Consider a small change of the angle during the angular motion of the pullet [itex]d\theta[/itex]. The string has moved a distance [itex]ds[/itex] and since [itex]s=\theta R[/itex] we have [itex]ds=d(\theta R)=R d\theta[/itex]

    By definition, we have [itex]\omega=\frac{d\theta}{dt}[/itex] and [itex]u=\frac{ds}{dt}[/itex],so

    [itex]\frac{ds}{dt}=\frac{d\theta}{dt} R \Leftrightarrow a = a_{angular} R[/itex] by differentiating.

    I hope I've done everything right :)
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