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Angular acceleration: round 2

  1. Jan 5, 2010 #1
    Ok, so...

    [tex]\tau = \frac{dL}{dt}[/tex]

    and

    [tex]L\equiv I\omega[/tex]

    so

    [tex]\tau = \dot{I}\omega + I\dot{\omega} = \dot{I}\omega + I\alpha[/tex]

    and i don't believe it'd be wrong to write (shouldn't be...right?)

    [tex]\alpha = \frac{\tau-\dot{I}\omega}{I}[/tex]

    ---
    the situation i'm looking at is that of a disc (of radius 'R') falling straight down due to gravity.

    [tex]I_{cm} = \frac{mR^2}{2}[/tex]

    and then I with respect to any arbitrary point...

    [tex]I_{arb} = I_{cm} + mr^2 = \frac{mR^2}{2} + mr^2[/tex]

    (where 'r' is the distance from the center of mass to the arbitrary reference point)

    [tex]\dot{I}_{arb} = 2mr\dot{r}[/tex]

    (as R doesn't change with time, but 'r' does, as the disc is moving with respect to the point)


    assuming that all is right...then the following is true (everything with respect to a non-moving arbitrary point)

    [tex]\alpha = \frac{\tau-2mr\dot{r}\omega}{\frac{mR^2}{2} + mr^2}[/tex]

    the m's can be factored out leaving

    [tex]\alpha = \frac{\tau-2r\dot{r}\omega}{\frac{R^2}{2} + r^2}[/tex]

    This is very troublesome IMO...
    This says that [tex]\alpha[/tex], something that is just based on kinematics (it is terms of the relative position function and its derivatives) and not anything else, depends on 'R'...the radius of the disc. I'm sure this isn't true...so where have I failed in my reasoning?...
     
  2. jcsd
  3. Jan 6, 2010 #2

    ideasrule

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    Homework Helper

    First, you can't factor out the "m" because the torque is by itself; it isn't multiplied by "m".

    Also, try to express both the torque and the angular acceleration in terms of the disc's linear (downward) acceleration, R, and r. The R should cancel out.
     
  4. Jan 6, 2010 #3
    oopsie..right 'm' stays...the 'm' leaves when the torque is due to gravity which would be mg*r*sin(angle)
    ......

    for the case where the disc is being accelerated by gravity...

    [tex]\alpha = \frac{(mg)(r)cos(\theta) - 2mr\dot{r}\omega}{\frac{mR^2}{2}+mr^2}[/tex]

    where [tex]\theta[/tex] is the angle that [tex]\vec{r}[/tex] makes with the positive x-axis..so that when [tex]\theta=0[/tex], the torque is just [tex](mg)(r)[/tex].

    and so, when you remove the [tex]m[/tex] you get

    [tex]\alpha = \frac{(g)(r)cos(\theta) - 2r\dot{r}\omega}{\frac{R^2}{2}+r^2}[/tex]

    or even more simply

    [tex]\alpha = \frac{(g)(r)cos(\theta) - 2r\dot{r}\omega}{\frac{I_{cm}}{m}+r^2}[/tex]

    and the problem persists. An object with the same mass, moving in the same way, relative to the same point, would have an [tex]\alpha[/tex] (of the center of mass) depend on its shape.
    Perhaps i'm interpreting something wrong?
     
  5. Jan 6, 2010 #4
    point is...the [tex]\alpha[/tex] should just be derived from a coordinate transformation (cart->polar)...the cartesian position function has absolutely nothing to do with the shape of the object..only the mass. [tex]\alpha[/tex] should be the same
     
  6. Jan 6, 2010 #5
    after thinking about it...the only issue i can find is this

    perhaps the torque due to gravity isn't just the force (mg) applied at the center of mass, crossed with the [tex]\vec{r}[/tex]

    but even if this is the case, and even if the torque due to gravity was some function of R...it doesn't seem physically possible to algebraically take the R out of the picture.

    Tomorrow, i'll try to apply it to a simple homogeneous rod, or even just a discrete (perhaps 2 or 3) point masses that make up a structure.
     
  7. Jan 6, 2010 #6
    nope..checked it out..
    the torque due to gravity is the same if you treat it moving through the center, or sum the torques on every element of the structure all about the same point.

    [tex]\alpha[/tex] for each element depends on the shape, as that would change it's position (such as the length of a rod)...however, the shape, (radius in this case) doesn't change the position of the center, nor does it (or should it) change the way the object falls due to gravity.
     
  8. Jan 6, 2010 #7
    nevermind...i figured it out
    the moment of inertia with respect to the arbitrary point is just mr^2..without the added moment of interia about the center of mass - because the object isn't rotating!



    this forum isn't as active as i thought it'd be
     
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