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Condition for a table to flip

  1. Sep 29, 2014 #1
    I'm arguing with a friend regarding the condition for a table to flip over assuming that the right contact point of its right leg with the ground (denoted hereafter by O) is held fixed i.e. the table cannot slip. The upper horizontal plate is a sort of a rod with mass m and both legs are massless.

    * See the attached figure.

    * [tex] \hat{F} [/tex] denoted an impulsive force.

    * From the balance of angular momentum about the right fixed contact point of the table with the ground over the short period of the application of the impulsive force it follows that the angular velocity of the table is given by

    [tex] \omega = \frac{\widehat{F} h}{m I_{O}} [/tex]

    * Since energy is conserved it can be shown that the angular velocity of the table when its center of mass is right above the fixed contact point O is given by

    [tex] \Omega ^{2}=\omega^{2}+2g \big( \frac{h-\sqrt{h^2+l^2/4}}{I_O} \big) [/tex]

    * Finally, my condition for the table to flip is that the centripetal acceleration at that instant (when the CM is right above O) would be equal to or larger than gravity

    [tex] \Omega ^{2} \sqrt{h^2+l^2/4} \geq g [/tex]

    Soving these last two equation would give the value of the impulsive force for table to flip.


    My friend claims that the initial kinetic energy should be larger than the potentail energy and that the kinetic energy when when the CM is right above O is zero. With the resulting equation, he can solve for the impulsive force. I think that it's wrong.

    What do you think?

    Thank you!

    Attached Files:

  2. jcsd
  3. Sep 29, 2014 #2
    I would tend to agree with the idea of enough (rotational) kinetic energy to get the CM right above O...at that point, it is a coin flip as to which way the table will come down as it is an unstable equilibrium position...anyway, a bit more kinetic energy and flips.

    First, nobody in their right mind would apply a horizontal force in an attempt to flip a table...one would apply a force in a direction tangential to the expected trajectory making more effective use of F by maximizing the perpendicular distance between F and O, thus maximizing the torque ##\tau##

    Having said that, I am thinking one needs to calculate the resulting torque at all times that results from the combination of the applied force and the opposite one produced by the center of mass...this second one reduces in magnitude as the table rotates, the one applied by a horizontal force that is allowed to move up and remain in contact with the table would increase...after some angle theta, you should be able to de-active F and have enough rotational kinetic energy to overcome the opposing/decreasing torque produced by the center of mass through the remaining angle (90-theta) to get CM right above O.

    my 2 cents
  4. Sep 29, 2014 #3


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    I'm inclined to agree. What happens if the initial KE is insufficient to raise the CM to the max required height (max PE)? I would suggest that this at least puts a lower limit on the initial KE required.
  5. Sep 30, 2014 #4
    I think that this problem is very similar to the classical problem which is attached in the figure below. Here, you should find the min. value of h for which the particle would reach the point A, which is above the center of the hoop C, and pass it. In order to do that, you apply the conservation of energy equation between the initial state and the "A" state, namely

    [tex]\frac{1}{2}mv_{A}^{2}+mg(2r-h)=0 \thinspace\Rightarrow\thinspace v_{A}^{2}=2g(h-2r)[/tex]

    Moreover, the condition that the particle would reach A and pass it is that its centripetal acceleration at that instant would be equal to or larger than gravity

    [tex]v_{A}^{2} \geq gr \thinspace\Rightarrow\thinspace h \geq \frac{5r}{2}[/tex]

    Now, in the flipping table case (except that it is a rigid body and not a particle) you have an initial angular velocity due to the application of the impulsive force and so one should apply that same procedure as in the particle example (which I did in my initial post) to obtain the minimum value of the impulsive force.

    Don't you think?

    Attached Files:

  6. Sep 30, 2014 #5


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    Do you actually mean "impulse", instead of "impulsive force"?
  7. Sep 30, 2014 #6
    [tex]\hat{F}=\int_{t_1}^{t_2}{Fdt} \thinspace,\thinspace t_{2} \to t_{1}[/tex]
  8. Sep 30, 2014 #7


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    Yep, that's "impulse".
  9. Sep 30, 2014 #8
    Yeah, I meant a "linear impulse" and that's why I indicated it as an impulsive force.

    But what do you think about my problem? Isn't what I had written correct?
  10. Sep 30, 2014 #9


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    That condition makes no sense. The angular velocity (and thus the centripetal acceleration) when the CoM goes over the support can be
    arbitrarily small. So you should set it to zero to find the limiting value for the impulse.

    What is wrong with that?
  11. Sep 30, 2014 #10
    Well, this problem looks to me very similar to the particle problem which I presented previously and thus the logic should be the same.
    If not, then why?
  12. Sep 30, 2014 #11


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    "Linear impulse" and "impulsive force" are different things. "Impulsive force" is just a name used for that F within that integral when t2-t1 -> 0.

    Maybe if the ball was to roll over a hill, it would be similar. You should try to solve the problem at hand, and do a sanity check on the conditions you are stating for that particular problem.
  13. Sep 30, 2014 #12
    I don't understand your point.

    You can't assume that the kinetic energy of the particle at A is zero in order to solve for the minimum value of h to reach that point because you'll get nothing out of this. Similarly, you shouldn't assume that the kinetic energy of the table is zero when its CM is right above O.

    Why is this claim wrong for the table problem but holds for the particle problem? and why is the condition on the centripetal acceleration wrong for the table problem but holds for the particle problem?
  14. Sep 30, 2014 #13


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    Because they are different problems. The ball on the top of the hoop has no support from below, so it needs a minimal speed to avoid falling down. The table's CoM over the support is supported from below, so it can move arbitrarily slow.
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