Thread: Moment of Inertia and Angular Velocity

  1. #1
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    Moment of Inertia and Angular Velocity

    Hello all,

    (Edit: "Angular Velocity" should read "Angular Acceleration". Looking for the wrong thing explains why I couldn't find it. I think I already answered the question myself after figuring out I was looking for the wrong term.)

    As a pet project I'm working on a simple 2D physics engine. There's one problem: my physics is rather rusty.
    Now I've given the object a moment of inertia (according to wikipedia: kg*m^2), and there are some forces working on the object (these objects work on different places of the object). I want to find out the angular acceleration.

    For instance, let's say I have a shape of moment of inertia of 5.0 kg*m^2 (I'm not even sure if that's a realistic number, but it doesn't matter for the question. Now there's, say, a gravitational force of 9.81N working on the center of mass on the object, and a force of 20N upward, 2m left from the center of mass.

    I believe the only force that is important for the angular velocity is the 20N upward (as the gravity, in this case, works on the center of mass and doesn't cause any angular velocity). This is equal I believe, for this purpose, to a 10N upward force that causes some angular acceleration.

    So I'm left with a force (kg*m/s^2) and a moment of inertia (kg*m^2). And I just can't see how I'd get to angular acceleration (rad/s^2) from these two.
    I think I'm missing something here... But what is it?

    Edit 2:
    Okay, looking for the right names helped a lot, I think I understand now. The 10N force upwards is called a "torque" or "moment" (of course this relates to distance of the center of mass and the angle of the force). And then, the angular acceleration is simply the torque divided by the moment of inertia, so in this case that would be "a = T/I = 10/5.0 = 2 rad/s^2"
    Am I right?

    Thanks in advance
    Last edited by EVOEx; 03-10-2011 at 09:04 AM.

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    Registered User NeonBlack's Avatar
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    That sounds pretty much right, but I sense some confusion.

    "The 10N force upwards is called a "torque" or "moment" (of course this relates to distance of the center of mass and the angle of the force). "

    This is not entirely true. The 10N force is not a torque itself, but can impart a torque (depending on the distance from COM and angle, as you correctly state). The units for Torque in SI is newton*meters. So if your force is 20 newtons directly upwards, and orthogonal to the axis of rotation, then your torque is 40 N*m. (Gravity as you said, affects all parts of the object equally, so has no affect on torque.)

    Also, if it helps, you can view torque as the rotational equivalent of force with the follow analogs:
    Force -> Torque
    mass -> moment
    linear acceleration -> angular acceleration

    So,
    Force = Mass * acceleration
    ->
    Torque = Moment * angular acceleration
    I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo

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    Quote Originally Posted by NeonBlack View Post
    That sounds pretty much right, but I sense some confusion.

    "The 10N force upwards is called a "torque" or "moment" (of course this relates to distance of the center of mass and the angle of the force). "

    This is not entirely true. The 10N force is not a torque itself, but can impart a torque (depending on the distance from COM and angle, as you correctly state). The units for Torque in SI is newton*meters. So if your force is 20 newtons directly upwards, and orthogonal to the axis of rotation, then your torque is 40 N*m. (Gravity as you said, affects all parts of the object equally, so has no affect on torque.)

    Also, if it helps, you can view torque as the rotational equivalent of force with the follow analogs:
    Force -> Torque
    mass -> moment
    linear acceleration -> angular acceleration

    So,
    Force = Mass * acceleration
    ->
    Torque = Moment * angular acceleration
    Okay thank you for your answer. I did take into account the distance to the center, though I have to admit I made a mistake in my post: I wrote that 20N upwards 2m to the left of the center of mass (that is, 90 degrees, I made the example easy for my self and still did wrong) would be 10N upwards 1m to the left of the mass (and I didn't even write the 1m, so it was even two mistakes), but of course that should have been 40N (as the further away it is from the center of rotation, the less force needed to make the rotation).
    But yes, it was only equal to the torque because of the easy figures I choose; the rest is simple vector maths (in fact, I was surprised how easy the final formula for the torque was; "c[0]*f[1] - c[1]*f[0]" where c is the distance from the center of mass and f the force).

    Thanks for those equivalents, I didn't realize that yet...

    Thanks,
    EVOEx

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    You may find An Introduction to Physically Based Modeling by David Baraff an interesting read, in particular the parts on rigid body dynamics. This is geared towards full 3D, but should easily be reduced to the 2D that you're after.
    Last edited by MWAAAHAAA; 03-10-2011 at 04:10 PM.
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