1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rigid body, torque and moment of inertia

  1. Sep 27, 2014 #1

    it's been quite some years since I took some entry-level mechanics at uni, and my book is no longer with me, so I'm struggling a bit here.

    I'm making a very simple rigid body simulator just for fun, but the angular velocity is not working out right. I suspect I've forgotten something crucial and basic. My setup is as follows: I've got a long rectangular box (stick) with dimensions w x h x d, and which has uniform density. The center of mass of the box is [itex]\vec{x}[/itex], and the orientation is given by the rotation angle [itex]\phi[/itex]. I use simple Euler integration (for now) to find the linear velocity [itex]\vec{v}[/itex] and angular velocity [itex]\omega[/itex].

    I apply a force [itex]\vec{F}[/itex] at some point [itex]\vec{r}[/itex] at one end of the stick. Here both [itex]\vec{F}[/itex] and [itex]\vec{r}[/itex] are in the local (unrotated) frame centered at the center of mass of the box.

    For the linear part I find the rotated force vector [itex]\vec{F}'[/itex] by rotating [itex]\vec{F}[/itex] by [itex]\phi[/itex], and then integrate [itex]\vec{x}_{t + \Delta t} = \vec{x}_t + \Delta t \vec{v}_t[/itex] and [itex]\vec{p}_{t + \Delta t} = \vec{p}_t + \Delta t \vec{F}'[/itex], and recover the linear velocity [itex]\vec{v}_{t + \Delta t} = \frac{1}{m} \vec{p}_{t + \Delta t}[/itex].

    To find the torque I first find [itex]\tau = \| \vec{r} \times \vec{F} \|[/itex]. Then I integrate [itex]\phi_{t + \Delta t} = \phi_t + \Delta t \omega_t[/itex] and [itex]L_{t + \Delta t} = L_t + \Delta t \tau[/itex]. I then recover the angular velocity [itex]\omega_{t + \Delta t} = I^{-1} L_{t + \Delta t}[/itex] for use in the next timestep.

    I calculate the moment as inertia as [itex]I = \frac{1}{12}m(w^2 + h^2)[/itex], where the mass is given as [itex]m = \rho w h d[/itex] and [itex]\rho[/itex] is the density of the box.

    The issue I have is that for a long stick the moment of inertia is so large compared to the mass (due to the [itex]w^2[/itex] [itex]h^2[/itex] terms) that even though I apply a constant tangential force at the end of the stick ([itex]\vec{F}[/itex] stays constant in the local, unrotated frame) the stick hardly rotates. Instead the linear velocity increases and the stick translates as if I had applied the force very close to the center of mass.

    I assume I've done some silly above, and would really appreciate if someone could point it out.
  2. jcsd
  3. Sep 28, 2014 #2
    Ok so stepping back a bit, assume I have a thin rod of uniform density and length [itex]\ell = 2r[/itex] (that is, [itex]r[/itex] is the distance from center of mass to either endpoint).

    For the linear acceleration I have
    [tex]\vec{v} = \frac{1}{m}\vec{p} \\
    \frac{d\vec{v}}{dt} = \frac{1}{m}\frac{d\vec{p}}{dt} \\
    \frac{d\vec{v}}{dt} = \frac{1}{m}\vec{F}[/tex]

    Assuming the force is always tangential to the the length of the rod and that I apply the force at one end of the rod then [itex]\tau = rF[/itex] where [itex]F = \|\vec{F}\|[/itex]. The moment of inertia is [itex]I = \frac{1}{12}m\ell^2 = \frac{1}{3}mr^2[/itex]. Using this the angular acceleration becomes, as far as I can gather, the following:
    [tex]\omega = I^{-1}L \\
    \frac{d\omega}{dt} = I^{-1}\frac{dL}{dt} \\
    \frac{d\omega}{dt} = I^{-1}\tau \\
    \frac{d\omega}{dt} = I^{-1}rF \\
    \frac{d\omega}{dt} = \frac{3}{mr^2}rF \\
    \frac{d\omega}{dt} = \frac{3}{mr}F[/tex]

    Then if the above is correct, this means for a fixed force [itex]F[/itex] the linear acceleration will go as [itex]1/r[/itex] as I increase [itex]r[/itex], but the angular acceleration will go as [itex]1/r^2[/itex]. Thus for a sufficiently long rod the angular acceleration will vanish. So, is this a correct analysis or am I doing it wrong?

    The results feels very counter intuitive to me, but it wouldn't be the first time my gut feeling is wrong when it comes to physics :)
  4. Sep 28, 2014 #3
    A problem with your first post is that you assume you can describe the orientation of rigid body with one angle, and that the change of that one angle is angular velocity. That is not true motion is not planar; you need more angles and their relationship with angular velocity is quite complex.

    In your second post, you fared better, except in two places: you said "tangential" where you should have said "orthogonal", and your assessments of acceleration as functions of ##1/r## are wrong.
  5. Sep 28, 2014 #4
    Which other angles are you referring to? My external forces will be strictly in the xy plane.

    Use of tangential is probably due to my 3D graphics background, but yes I meant orthogonal to the length of the rod. I'd be very happy if you could shed some light on where I'm making the mistake w.r.t. the acceleration.

    Last edited: Sep 28, 2014
  6. Sep 28, 2014 #5
    Are you saying that your motion is essentially planar? That was not evident from your description. If that is the case, then you can use a single angle and all the simplicity coming with that.

    Your final equation relates angular acceleration with force via ##1/r##, yet you spoke of ##1/r^2##, which is definitely wrong.
  7. Sep 28, 2014 #6
    Sorry, my bad. I rewrote the post a couple of times, must have accidentally deleted that part :(

    The rod has uniform density so the mass would increase with r as well, no? But that shouldn't really matter I guess as long as the angular acceleration has "one extra" [itex]1 / r[/itex] factor compared to linear acceleration?

  8. Sep 28, 2014 #7
    Lets go back to your post #1. Since the problem is planar, I suggest that you stop treating your object as a 3D box, but treat it as a rectangular plate, and adjust your equations correspondingly.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook