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# If a wire with linear density $\rho(x, y, z)$ lies along a space curve $C$, its $\textbf{moments of inertia}$ about the $x$-, $y$-, and $z$-axes are defined as $$I_x = \int_C (y^2 + z^2) \rho(x, y, z) ds$$ $$I_y = \int_C (x^2 + z^2) \rho(x, y, z) ds$$$$I_z = \int_C (x^2 + y^2) \rho(x, y, z) ds$$Find the moments of inertia for the wire in Exercise 35.

## $$\begin{array}{c}{I_{x}=I_{y}=4 \sqrt{13} k \pi\left(1+6 \pi^{2}\right)} \\ {I_{z}=8 \sqrt{13} k \pi}\end{array}$$

Vector Calculus

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Okay, so we computed on these conscious moment of inertia three dimensions and, uh and tha This part's getting hindered by this part. I ask you to computer the using the information from problems thirty five, which is thiss and we compute the yes, which is the exit he square prostate. What if he square prostituted his score, which is square with thirteen. So anyway, all these probably just be square Those thirteen our times of constant did he and similarly so we just have to compute those into gross. In fact, it might be easier if we can just compute in the growth X square Y score and Z square separately. So integral from zero to two pi x square iss Ah, this will be by bye Ah, one more for my thesis. And this will be for Piper hose. The first thing that was for a second term the entire the ripped scientist t the importance of Sim a scientist fear to pines here are the same. So the second term in the group goes to zero and similarly Ah, you integrate Why square we get the same answer And if winter crazy square which is night, he square. That was three t cubits and I derivative. So she had twenty. Problem to pi twenty power. Hi. Q is now fourteen before pie. And then after that we can just write all these answer. We have squared off thirteen K. Why? School process square. So for Piper first honey for fight you and this one should be the same X square, plus y square s O. Here we should get a pie.

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Vector Calculus

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