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# A wire of constant density $\delta=1$ lies along the curve$\mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+(2 \sqrt{2} / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq 1$Find $\bar{z}$ and $I_{z}$.

Integrals

Vectors

Vector Functions

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Okay, folks. So now we're gonna talk about problem number 40. Um, we have a wire of constant density lying along curve, and we're gonna be finding the Z component of the center of mass and then the Z component of the moment of inertia. So, for for the Z component of the center of mass, which is which is denoted C bar. Um, that's really just one over him z d m. And that's equal. Well, first of all, let's look for em. First Mass is integral d m, which is integral Delta DS. Because that's what GM is. Um, this could be written as, um Delta, which is a constant. So I'm just gonna leave it out because that's one, um ds, on the other hand, is just x dot squared plus why dot squared plus z dot squared pt where x dot means dx DT. Same as y in Z. Okay, when you take the time derivative, we take the time derivative of of X and y and Z, and you square them and you add them up. It's really messy. It's You don't have to use change rule, and you have to use some trig identities like coastlines agree. Plus why wise Excuse me? Like Osan Square plus signs were give you one and all that. Um but when you have done all of that, you're going to get hopefully one plus t squared plus two teeth mortified by DT. Okay, um and this thing is just t plus one squared and and the square root cancels the square, so we have t plus one t t. This is equal to t squared over two plus t between zero and one, which is 3/2. So that's total Maris. Um as for zd em, it's a similar idea. DDM is the same thing as the adult a. D s. Because this thing is what the M is and Z as a function and t is to root two over three t to the power 3/2 that Z and um D s as we have obtained before. The S is T plus one and B T and integration limit is between zero and one. And so now we have to route to over three t to the power of 55 have plus T to the power of 3/2 multiplied by d. T. And between zero and one. Of course. And when you crank it out when you crank it out and you do your algebra here, you're gonna have This is not even eligible anymore. This is just pure arithmetic. 16 route to over 35. Okay, that's what Z Bar is. Z bar, which is the Z component of the center of mass vector. This is to over three 16. Route to the over 35. Okay, um, where I got this to over three from is by. You know this thing I have calculated to be 16 through to over 35 and one over M while ems 3/2 right here. One over AM is's to over three. That's where I got this to. Over three from anyway, I can write this as 32. Route to over 105 So that Z bar and I'd see it's just delta R squared minus disease worried ds. And this thing is just 01 Ah t squared t plus one 80. And, um Well, first of all, in case you don't already know, r squared means x squared plus y squared plus z squared. But now, when you subtract all Disease Square. You just have expiry plus y squared. But X's t co sign t. And why is t signed t when you squirm? You? Yeah, the mom you get t squared. Okay, so that's where that's where you got this t square from? Um, this is just a simple integral to, ah, to evaluate. So we have t fourth over four plus t cute over three between their own one, and that's just seven over. Well, so that's the Z component of the moment of inertia. And we're done for this video. Thank you for watching.

University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp