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Introductory Physics Homework Help
Rotational inertia of square about axis perpendicular to its plane
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[QUOTE="ChiralSuperfields, post: 6855283, member: 731016"] Thank you for your reply [USER=334404]@haruspex[/USER] ! Sorry, let me try to explain it better. [ATTACH type="full" alt="1676500671426.png"]322345[/ATTACH] (sorry about the tilted coordinate system, the ##\hat i## axis should be horizontal and ##\hat j## axis should be vertical) ##\vec r_1 = \sqrt {x^2 + y_1^2} ## since it has a ##x\hat i## and ##y_1\hat j## component. However since ##y_1 = x## in this special case then ##\vec r_1 = \sqrt {2x^2} = R ## For this other vector their y-component is not ##x## ##\vec r_2 = \sqrt {x^2 + y_2^2} ≠ R## (cannot equal R since ##\vec r_2 < \vec r_1 ##) ##\vec r_3 = \sqrt {x^2 + y_3^2} ≠ R## So what I was saying is that we should consider the general case where, ##\vec r = \sqrt {x^2 + y^2} ## Where x is a constant equal to ##\frac {\sqrt {2}}{2}R## and y is a variable position vector component in the y-direction. However, if we choose a different coordinate system, such as what [USER=700856]@erobz[/USER] has done in post #43 [ATTACH type="full" alt="1676500849267.png"]322346[/ATTACH] Then now y a constant equal to ##\frac {\sqrt {2}}{2}R## and x is variable position vector component in the y-direction. I think the confusing thing about this problem is choosing the right side to find the moment of inertia for and defining the right coordinate system. What do you think? How would you solve this using the parallel axis theorem out of curiosity? Many thanks! [/QUOTE]
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Rotational inertia of square about axis perpendicular to its plane
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