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Homework Help: Why does I*a = TR?

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I am first learning about angular motion and came across this formula while doing a homework problem. Can anyone explain to me why ɪa = TR? That is the moment of inertia * angular acceleration = tension in rope * radius.

    For reference, in the problem it's a stunt guy jumping off a building with rope tied around his waist connected to a cylindrical drum with rope wound around it.

    I'm only interested in knowing why those two quantities equal each other. Thanks in advance!

    2. Relevant equations

    torque = ɪ

    torque = TR

    ɪa = TR

    3. The attempt at a solution

    I already solved the problem, just need clarification.
  2. jcsd
  3. Nov 19, 2016 #2


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    TR is the torque exerted on the drum by the rope. Do you know Newton's second law of motion:

    Force = mass x acceleration

    Well its rotational analog is:

    Torque = Moment of inertia x angular acceleration.

    Each item in that equation is the angular analog of the corresponding element in the Newton law. The second equation can be derived from the first.
  4. Nov 19, 2016 #3


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    Just for info all the equations of motion (for example SUVAT) have a rotational analog.
  5. Nov 19, 2016 #4
    Hey, I'm also working on rotational motion in physics right now as well. Whenever I try and fit things into equations I think of the units involved.

    I = MR^2 = Kg * m^2

    T = N* Meters = Kg * m^2/(s^2)

    a= rad/ s^2 (angular acceleration)

    I = Ta

    Kg * m^2 = (Kg * m^2/(s^2)) * rad/ s^2

    radian has no dimensions so the left side and right side of the above equation end up being the same.

    Basically do the same with the given equation and you should see why the equation works out.
  6. Nov 20, 2016 #5
    I understand that T = Ia is the rotational analog of F = ma, but why does the tension * radius = Ia? Why does TR = Ia... why are both Ia and TR both equal to torque?
  7. Nov 20, 2016 #6


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    Did you read your textbook? It's very likely explained in there. Here's a derivation for a point mass.
    \vec{F} &= m\vec{a} \\
    \vec{r}\times\vec{F} &= m(\vec{r}\times\vec{a}) \\
    \lvert \vec{r}\times\vec{F} \rvert &= m\lvert \vec{r}\times\vec{a} \rvert \\
    rF_\perp &= mra_\perp \\
    rF_\perp &= mr(r\alpha) \\
    rF_\perp &= mr^2 \alpha \\
    rF_\perp &= I\alpha
  8. Nov 20, 2016 #7


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    In plain English... Torque equals force multiplied by radial distance.

    Depending on the problem the force might be provided by "tension" and the distance might be the "radius". However this isn't always the case.

    It sounds like you are working on a particular problem/example. I suggest you post the details.
  9. Nov 20, 2016 #8
    Luke, I think a picture would be very helpful. However from the description I suspect this question is all about the drum. The radius R is the radius of the drum. The rope comes off of the drum at a tangent. The tension on the rope is also the force the rope is applying to the drum. Force applied tangentially at a distance from the axis of rotation is the definition of torque. The rope applies a torque TR to the drum.
  10. Nov 21, 2016 #9
    Okay wow that makes a lot more sense. I didn't know that was the exact definition. I thought it was just some sort of angular force. Makes a lot more sense now, I understand. Thanks a lot!
  11. Nov 21, 2016 #10
    You're welcome
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