Timeline for How should the differential equation of a physical pendulum be written using clockwise rotation as positive?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 23, 2019 at 18:46 | vote | accept | some_math_guy | ||
| Jun 23, 2019 at 18:40 | comment | added | some_math_guy | $\vec x = -x \hat X $ --> my concern is :since x is a variable,I can call it as I like, why not calling it just x, why is it necessary to write the - sign? it's like solving a equation and calling the unkown "-a", I might as well called it "a", and it would be more natural, because the answer will be direct instead of solving for a and later having to substitute the value of a in the expression "-a" | |
| Jun 23, 2019 at 18:10 | comment | added | Farcher | @juancarlosvegaoliver From your diagram you have defined the direction of increasing $\theta$ to the right/clockwise. It is like having the direction of increasing $x$ to the right and the unit vector pointing to the left. | |
| Jun 23, 2019 at 17:56 | comment | added | some_math_guy | My only concern is why we start off with a minus sign $\vec \theta = -\theta \hat K $. Is it because the angle in diagramm is anticlockwise, and the defined positive direction is clockwise?Can't it just be $\vec \theta = \theta \hat K $ and consider the sign to be "inside" the variable(like if I called $ -\theta = \alpha $ )? and then since variables are just name I change it back to $ \theta $) | |
| Jun 23, 2019 at 16:54 | history | edited | Farcher | CC BY-SA 4.0 |
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| Jun 23, 2019 at 15:37 | comment | added | some_math_guy | all right, what about the issues in comments 2 and 3? | |
| Jun 23, 2019 at 15:26 | comment | added | Farcher | @juancarlosvegaoliver I have edited my answer as a result of your comment. | |
| Jun 23, 2019 at 15:25 | history | edited | Farcher | CC BY-SA 4.0 |
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| Jun 23, 2019 at 15:23 | comment | added | some_math_guy | yes, I guess you mean that, so it's not the polar unit vector, is it?. | |
| Jun 23, 2019 at 15:10 | comment | added | Farcher | @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? | |
| Jun 23, 2019 at 14:46 | comment | added | some_math_guy | 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 | |
| Jun 23, 2019 at 14:38 | comment | added | some_math_guy | 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? | |
| Jun 23, 2019 at 14:27 | comment | added | some_math_guy | 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? | |
| Jun 23, 2019 at 13:10 | history | answered | Farcher | CC BY-SA 4.0 |