I don't understand why in this example the gravitational force is not considered in the sum of the external torques, when computing the angular momentum principle. Why is that?

enter image description here

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I was thinking I need to start from the equation:

$\dot{H}_O = M_O^{ext}$

The problem I see is on the right side. If we consider the starting time to be 0, we have

$\underline{r} = (c + v_0 t) \underline{i} + d \underline{j} \\ \underline{F} = -mg\underline{j}$.


$M_O^{ext} = \underline r \times \underline F = -mg(c+v_0 t) \underline{k}$

Integrating the principle of angular momentum, and defining the time the string becomes taut as $t_f$, we get

$ H_O(t_f) - H_O(0) = -mgct_f - 0.5 mgv_0 t^2_f$

The left side of the last equation is already computed on the book as you can see in the picture. Why is the right side not there?

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    $\begingroup$ The question doesn't state that the y-coordinate is vertical (i.e. in the direction of gravity). x-y could be the coordinates on a frictionless table. $\endgroup$ Aug 1, 2017 at 10:11
  • $\begingroup$ @JulianHelfferich I hadn't thought of that, maybe that is the reason of the derivation in the book. In the case y was the direction of gravity, would my analysis be correct then? $\endgroup$
    – charles
    Aug 1, 2017 at 10:17
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    $\begingroup$ Adding gravity makes the problem much more complicated. In particular, finding the time and the x-y coordinates when the string becomes taut is not trivial. That said, I can not find a flaw in your reasoning. $\endgroup$ Aug 2, 2017 at 5:13
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    $\begingroup$ At no time is it mentioned that the experiment is occurring under an inertial system. It is left out intentionally, in order to simplify the problem, which is complicated enough. No other "external" forces need be considered. $\endgroup$
    – Guill
    Aug 2, 2017 at 7:27