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# Moment of inertia of wire hoop A circular wire hoop of constant density $\delta$ lies along the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ in the $x y$ -plane.Find the hoop's moment of inertia about the $z$ -axis.

## $I_{z}=2 \pi a^{3} \delta$

Integrals

Vectors

Vector Functions

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

okay. What we want to do is we're told we have a a circular wire. Um, hoop of constant density that lies along a circle represented by X squared plus y squared equal to a squared. And we want to find the hoops moment of inertia about the Z axis. Okay? And so we know that the moment of inertia about the Z axis is represented by the integral over the curve of X squared plus y squared, um, times the density, um, function t s webs. Okay, So what we want to do is now to find Oliver Terms. And so we know that d s is, um, given by the magnitude of this V of tea times T t Well, the magnitude of u of T v ft is dependent upon the derivative of are empty. And so what we needed to is to define that parametric equation are of tea or Parametric Victor. And so this is a circle. And so we know that Parametric Equation, um, is going to be a co sign of tea. I plus a sign of tea j. And so the derivative of our ved, he is gonna give me a negative eh? Sign of tea. I plus in a co sign of tea. J And so that magnitude of the tea is the square root a negative, a sign of tea. And we're gonna square that, plus a co sign of tea, and we're going to square that. And so this actually just becomes Mm, because sign square Plaza co sign squared, gives me a one, and we end up with a square times they won The square root of a squared is a and therefore D s now is a TT. And so that moment of inertia about the Z is going to go from, um, zero to two pi. Um, it's, um, to pie because we have a complete circle in the X Y plane. And so this becomes, um, Now, um, X squared. Plus y squared is a squared. So this is becomes an A squared, um, times a constant density. So it's just gonna be a constant number and density. It would have been, uh, a function, but because we were told that the constant density is just gonna be that constant density times t s, which is a DT. Um, and so this becomes, um, a cubed times the density times t evaluated at two pie and it zero. So it becomes too pie a cube times that constant density. So there is that moment of inertia about the sea axis.

University of Central Arkansas

#### Topics

Integrals

Vectors

Vector Functions

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp