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Spinning/Falling Rod

  1. Jul 23, 2004 #1
    Having some trouble understanding why the conclusion I am arriving at is not true. It has to do with showing that it is easier to balance a pool stick with finger on the tapered end.

    Summary of the problem:

    Rod of mass M with clay of mass M attached to one end. The clay represents the thick end of pool stick. Find the angular acceleration of the system if the rod is placed vertical on a frictionless tabletop and tipped a small angle theta (for both clay touching and not touching surface). The conclusion should be that the angular acceleration is smaller when the clay is not in contact with the surface.

    My analysis:

    Take as a frame of reference the center-of-mass of the system and assume that the axis of rotation in this frame is fixed. The external torque is equal to the inertia times ang acceleration. From this it should become mostly algebraic.

    [tex] \alpha_\text{thick} = \frac{\tau}{I} = \frac{N \sin \theta \cdot L/4}{10/48 ML^2} = \frac{6}{5} \frac{N \sin \theta}{M L}

    [tex] \alpha_\text{taper} = \frac{\tau}{I} = \frac{N \sin \theta \cdot 3L/4}{10/48 ML^2} = \frac{18}{5} \frac{N \sin \theta}{M L}

    From this it appears that the angular acceleration with the tapered end down is greater, which means it would be harder to balance.

    edit: i see now that the book did its calculations with the axis of rotation passing through where the rod touches the surface. but i don't understand how this could be correct because that axis moves while not passing thru the center-of-mass. therefore i don't see how torque = inertia x ang. acc.
    Last edited by a moderator: Jul 23, 2004
  2. jcsd
  3. Jul 24, 2004 #2
    If the rod falls without slipping, then the rotation axis goes through the point of contact between the rod and the surface. Note that the rotational inertia of the system (rod and clay) will not be the same in both situations...I think.
  4. Jul 24, 2004 #3

    Doc Al

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    Staff: Mentor

    The center of mass frame is not convenient for analyzing this problem as it is accelerating. Also realize that the normal force N is not the same in each configuration.

    As e(ho0n3 explained, the way to look at this problem is to treat the contact point as the axis of rotation and then compare the rotational accelerations for each case.

    With the clay part on top:
    [itex]I = \frac{4}{3}ML^2[/itex]

    [itex]\tau = \frac{3}{2}MgL sin\theta[/itex], thus

    [itex]\alpha = \frac{9}{8}\frac{g}{L} sin\theta[/itex]

    With the clay part on the bottom:
    [itex]I = \frac{1}{3}ML^2[/itex]

    [itex]\tau = \frac{1}{2}MgL sin\theta[/itex], thus

    [itex]\alpha = \frac{3}{2}\frac{g}{L} sin\theta[/itex]
  5. Jul 25, 2004 #4
    Yes, but i'm not so sure that the rotational dynamics formula applies because the contact point is moving across the surface. I would be happy if you could show me why the formula does apply.

    Here is my reasoning:

    [tex] \vec{L} &= \sum \vec{r}_i \times m_i \dot{ \vec{r}_i } [/tex]
    \dot{ \vec{L} } &= \sum( \dot{ \vec{r}_i } \times m_i \dot{ \vec{r}_i } + \vec{r}_i \times m_i \ddot{ \vec{r}_i } ) = \sum \vec{r}_i \times m_i \vec{a}_i

    Unless we can conlude that the contact point travels at a constant speed, it is not an inertial frame of reference. Therefore the angular momentum does not appear to equal the torque as we cannot substitute F=ma. I guess you could try substituting a fictitious torque to get things moving, as in

    \frac{d \vec{L}}{dt} &= \sum \vec{r}_i \times m_i ( \vec{a}_{i/I} - \vec{a}_{f/I} ) \\ &= \sum ( \vec{r}_i \times m_i \vec{a}_{i/I} - \vec{r}_i \times m_i \vec{a}_{f/I} ) \\
    &= \sum \vec{r}_i \times m_i \vec{a}_{i/I} - ( \sum m_i \vec{r}_i ) \times \vec{a}_{f/I} \\
    &= \vec{ \tau }_\text{ext/f} - R \times M\vec{a}_{f/I} \\

    where f stands for the frame and I stands for an inertial frame. Anyways, unless the frame is the center of mass, then the torque does not equal the deriv of ang momentum. (I suppose you know all this already, but it's helps me to say it cuz i'm still just getting a hang of it.) My guess is that because the problem is only considering a small instant near the beginning of the motion, then the contact point frame is approximatly stationary?
  6. Jul 25, 2004 #5

    Doc Al

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    Staff: Mentor

    balancing a pool stick?

    Perhaps I am unclear as to the actual problem. If the problem is "explain how it's easier to balance a pool stick with the tapered end on one's finger", then I would not not model it as something sliding on a frictionless surface. If there's no friction, good luck balancing that stick! That's how I was looking at the problem anyway, in my answer in post #3. I assumed that the point of contact was stationary.

    On the other hand, if the problem is to find the angular acceleration of the stick when displaced from vertical while resting on a frictionless surface--that's a different problem. For that problem, the only issue I have with your solution in post #1 is that you seem to be assuming that the normal force N is the same in both cases. I don't think so.
  7. Jul 25, 2004 #6
    your right, the normal forces can be different, but i just pretended they would be nearly equal so i could get somewhere, otherwise i couldn't even compare the results.

    the problem itself has nothing to do with balancing pool sticks, but the book uses it as a reason to show that pool sticks are easier to balance on the tapered end. i think the connection is not correct.

    a more proper analysis seems to show that the angular acceleration (about the contact point) is actually higher when the the clay piece does not touch the surface.

    here is how i figure it goes: by considering that the rod/clay system is release from rest after being tipped a very small angle theta allows one to eliminate the normal force. also note that the angle the contact point makes with the normal line to the surface is the same as the angle through which the rod has turned about its center-of-mass. their angular accelerations then are equal as well. as always, one can begin by writing the general dynamics equations. If [tex]R[/tex] represents the distance of the center-of-mass from the contact point and [tex]\theta[/tex] is the angle discussed above then it follows that

    [tex] N - Mg = Ma [/tex]
    [tex] N R \sin \theta = I_\text{cm} \alpha [/tex]

    but this brings the problem to a dead stop unless a direct relationship between [tex]a[/tex] and [tex]\alpha[/tex] can be found. this can be done by writing

    [tex] y = R \cos \theta [/tex]

    and subsequently taking derivatives to obtain

    [tex] a = -R [ \alpha \sin \theta + \omega \cos \theta ] [/tex]

    if [tex]\theta[/tex] is much less than one, we can apply [tex] \sin\theta = \theta [/tex]. also, if the rod is tipped by [tex]\theta[/tex] and then released, it will be starting from rest, so we can take [tex]\omega[/tex] to be zero. this leaves the clear relationship of

    [tex] a = -R\alpha\theta [/tex]

    that we want. substiuting this into the dynamics equations and eliminating the normal force leaves the equation

    [tex] (Mg - MR\alpha\theta)R \sin\theta = I_\text{cm} \alpha [/tex]

    letting [tex] \sin\theta = \theta [/tex] again and then solving for [tex]\alpha[/tex] yields

    [tex] \alpha = \frac{M g R \theta}{M R^2 \theta^2 + I_\text{cm}} [/tex]

    finally, letting (R = 3/4L or R = 1/4L) and [tex] I_\text{cm} = 10/48ML^2 [/tex], one can resolve the instantaneous angular acceleration after the rod has been tipped a small angle theta with the clay on either end. This leads to

    [tex] \alpha_\text{H} = \frac{ 36 g \theta }{10 L + 27 L \theta^2 } [/tex]
    [tex] \alpha_\text{L} = \frac{ 12 g \theta }{10 L + 3 L \theta^2 } [/tex]

    where H stands for when the clay is "high" and L stands for when the clay is "low". the ratio of the accelerations in the high position to the low position is greater than one for small angles no matter what the length is. this goes against the conclusion arrived at by treating contact point as a frame of reference.
    Last edited by a moderator: Jul 25, 2004
  8. Jul 25, 2004 #7

    Doc Al

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    Staff: Mentor

    But I see you were perfectly capable of doing the full (correct) analysis.

    Right. The book got it backwards. To show the connection with balancing pool sticks, you can't assume a frictionless surface.

    Exactly right. And I agree with your analysis. :smile:

    Note that with a frictionless surface, the center-of-mass falls straight down.


    Of course. It's a different problem. (Can't trust those books!)
  9. Jul 25, 2004 #8
    thanks for reading. saying things on the forums forces one to be clear and that helps me understand what i'm saying myself more too.
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