- #1
PhMichael
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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.
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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!
* 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!