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physical pendulum

Refer to figure $1$.

Sticking to the convention (counterclockwise is positive) the differential equation for a physical pendulum is $$I\cdot D^2(θ) =-mgd\sin(θ)$$ since the torque of the gravity force is clockwise, so it has a negative sign in front which with the small angle approx becomes $$D^2(θ) + (mgd/I)θ = 0,$$ where $I$ is the moment of inertia, $d$ is the distance to the center of mass, and $D^2$ is the second derivative with respect to time.

No problem there, now referring to figure $2$.

I want to find the equation using clockwise as positive $$I*D^2(θ) = mgd\sin(θ),$$ now the torque has the positive sign, but something must be wrong, because this lead to the equation: $$D^2(θ) - (mgd/I)θ = 0,$$ whose solution won't be the same as in the previous case.

I guess I have to put a “-” in front of $D^2(θ)$, but I don't know how to justify it.

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I have rewritten my answer to address the concerns made in comments by @juancarlosvegaoliver.

Instead of it being a rotation let it be simple harmonic motion in one dimension along the $x$-axis (vales increasing from left to right) and then I will show its relevance to the question asked.

enter image description here

The displacement from $O$ is $\vec x = x \hat x$ where $x$ is the component of the displacement in the direction of $\hat x$.
The velocity is $\dot x \hat x$ and the acceleration is $\ddot x \hat x$.
The force is $-k\vec x = -kx \hat x$ and $-kx$ is the component of the force in the direction of $\hat x$.

At a position like $A$ the displacement $\vec x$ in the direction of $\hat x$ and the direction of the (restoring) force is in the direction of $-\hat x$.
At position $B$ the displacement is in the direction of $-\hat x$ and the direction of the (restoring) force is in the direction of $+\hat x$.

So using $\vec F = m\vec a \Rightarrow -kx\hat x = m\ddot x \hat x \Rightarrow \ddot x = -\frac km x$ for all values of $x$ whether positive or negative.

Switching the direction of the unit vector to $\hat X = - \hat x$ makes no difference because now $\vec x = -x \hat X = -x \,(-\hat x) = x \hat x$ and $\vec a = -\ddot x \hat X = -\ddot x \,(-\hat x) = \ddot x \hat x$.
$\vec F = +kx\hat X = +kx (-\hat x) = -kx \hat x$.
For example, $2 \hat X = -2 \hat x$ and both give the position as $x=-2$

Having the $x$ axis pointing from right to left only results in a change of sign so, for example a position of $x=-2$ would now be $x=+2$.


Diagram 1 is the same as my diagram except that now $\vec x = x\hat x$ etc is replaced by $\theta \hat k$ etc and $\vec F - -k x \hat x$ is replaced by $\vec \tau = - mgd \sin \theta \,\hat k$.
The direction of $x$ increasing is to the right is replaced by the direction of $\theta$ increasing is anticlockwise.

Diagram 2 is just a reversal of the unit vector such that the new unit vector $\hat K = - \hat k$ with the direction of increasing $\theta$ still anticlockwise.
I have shown for the one dimension motion reversing the direction of the unit vector does not change anything and it is the same for the example with rotation.
$\vec \theta = -\theta \hat K = -\theta \,(-\hat k) = \theta \hat k$ and $\vec \alpha = -\ddot \theta \hat K = -\ddot \theta \,(-\hat k) = \ddot \theta \hat k$.
$\vec \tau = +mgd\hat K = +mgd (-\hat k) = -mgd \hat k$.

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  • $\begingroup$ 1 Is the angular unit vector you are using ($\hat \theta_{\rm a}$) the polar angular unit vector? Because if so, since it is not constant in direction, when taking the derivative of the angular position to get the angular velocity and the angular acceleration, shouldn't they also be derivated? $\endgroup$
    – J.C.VegaO
    Jun 23 '19 at 14:27
  • $\begingroup$ 2.According to your solution, $\dot \theta_{\rm a}$ and $\ddot \theta_{\rm a}$ represent just magnitudes of angular velocity and acceleration. When I wrote the equation for figure 2 , I wanted $\ddot \theta_{\rm a}$ to represent the component of the acceleration, that is magnitude and sign, so the sign should be "inside" the variable, following this approach, why don't I get the correct equation? $\endgroup$
    – J.C.VegaO
    Jun 23 '19 at 14:38
  • $\begingroup$ 3 when writting the torque for figure 2, shouldn't the theta inside the sine function have a minus sign?, because the angle as drawn is negative and $d \sin \theta$ is the distance from the pivot point to the direction of the gravity force, and therefore it should be positive, then the torque should be $mgd \sin (-\theta)$ = -$mgd \sin (\theta)$ ,which gives back the wrong equation $\endgroup$
    – J.C.VegaO
    Jun 23 '19 at 14:46
  • $\begingroup$ @juancarlosvegaoliver Perhaps it might have been better that I stated that when the oscillation was in the $xy$ plane the $\pm z$ direction is the unit vector direction which does not change with time? $\endgroup$
    – Farcher
    Jun 23 '19 at 15:10
  • $\begingroup$ yes, I guess you mean that, so it's not the polar unit vector, is it?. $\endgroup$
    – J.C.VegaO
    Jun 23 '19 at 15:23

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