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Consider the following problem:

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My first attempt of a solution was taking moments about Point O and solve for variables by setting the final torque to zero. After consulting the text book and quoting my teacher "the torque can't be set to zero because this is an accelerating system". Setting the final torque equal to ma(L/2)sin30° solved the problem completely.

But the reason I wanted a zero final torque is because obviously the bar stays in place during this motion and if there is a resulting nonzero torque caused by F=ma, then why doesn't the bar OB tip over? Should it be assumed air resistance, as in free fall, creates an opposite torque, resulting in a zero final torque? (Newton's 3rd law). In free fall however, the drag force is smaller than the gravitational force at first, but grows proportionally to the square of the velocity until terminal velocity is reached. So is there an nonzero torque before the system reaches its terminal velocity? Shouldn't the bar want to tip over before reaching terminal velocity?

It's intuitive that the bar does stay in place and only tips over if the system stops, but in my solution, the mathematics says there is a nonzero torque about point O caused by the driving force F=ma acting on the mass center of the bar. And yet this nonzero torque doesn't make the bar tip over.

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  • $\begingroup$ Which forces did you use when calculating the torque? $\endgroup$ – Gabriel Golfetti Dec 19 '18 at 8:54
  • $\begingroup$ I used the horizontal F=ma at the center of mass. This is equal to the sum of the horizontal components of the normal force at A and the hinge force at O. Vertical components equals the weight. So for the torque about O we have that the sum of the gravitational torque and the torque due to force on A equals ma(L/2)sin30 $\endgroup$ – TheMercury79 Dec 19 '18 at 11:05
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Just like the equation for force and momentum $$\vec F=\frac{d}{dt}\vec p$$ there is an equation for torque and angular momentum $$\vec{\tau}=\frac{d}{dt}\vec L$$ In this case the system moving with velocity $v$ has linear angular momentum about point O, since it's motion is displaced by a height $(L/2) \sin 30^\circ$ $$L=mv r_\perp=mv\frac{L}{2}\sin 30^\circ$$ And you see your net torque is exactly equal to $dL/dt$, as it should be, even though in the accelerating frame nothing is rotating.


The way most people would think about this problem, by the way, is to go to the accelerating frame and treat the acceleration $a$ as an extra force $ma$ acting left at the center of mass (much like $mg$). You can see that this gives the same result, since now the extra $ma$ force produces a torque about O.

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