🎉 The Study-to-Win Winning Ticket number has been announced! Go to your Tickets dashboard to see if you won! 🎉View Winning Ticket

Problem 34

Center of mass of a curved wire A wire of density $\delta(x, y, z)=15 \sqrt{y+2}$ lies along the curve $\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+$ $2 t \mathbf{k},-1 \leq t \leq 1 .$ Find its center of mass. Then sketch the curve and center of mass together.

Answer

\documentclass{article}

\usepackage[utf8]{inputenc}

\usepackage{amsmath}

\usepackage{pgfplots}

\pgfplotsset{every axis/.append style={

axis x line=middle, % put the x axis in the middle

axis y line=middle, % put the y axis in the middle

axis line style={<->,color=blue}, % arrows on the axis

xlabel={$y$}, % default put x on x-axis

ylabel={$z$}, % default put y on y-axis

}}

\title{Density and shape of a curved wire}

\begin{document}

\maketitle

To begin, recall that finding the center of mass of a wire A requires finding the following centroids:

\\

\vspace{5mm}

\\

$\bar{x}=\frac{\int_Ax\delta(x,y,z)ds}{\int_A\delta(x,y,z)ds}$

\\

\vspace{5mm}

\\

$\bar{y}=\frac{\int_Ay\delta(x,y,z)ds}{\int_A\delta(x,y,z)ds}$

\\

\vspace{5mm}

\\

$\bar{z}=\frac{\int_Az\delta(x,y,z)ds}{\int_A\delta(x,y,z)ds}$

\\

\vspace{5mm}

\\

where $\delta(x,y,z)$ is the mass density function

\\

\\

Note that the total mass M(A) is given by $M(A)=\int_A \delta(x,y,z)ds$. Also note A is parameterized by,

\\

\[ \begin{cases}

x(t)=0\\

y(t)=t^2-1\\

z(t)=2t

\end{cases}

\]

Recall that $ds=|v(t)|dt$.Thus,

\\

$|v(t)|=\sqrt{(\frac{dx(t)}{dt})^2+(\frac{dy(t)}{dt})^2+(\frac{dz(t)}{dt})^2}$

\\

\indent \indent$=\sqrt{(0)^2+(\frac{d(t^2-1)}{dt})^2+(\frac{d(2t)}{dt})^}$

\\

\indent \indent$=\sqrt{(2t)^2+2^2}$

\\

\indent \indent$=\sqrt{4t^2+4}$

\\

\indent \indent$=2\sqrt{t+1}$

\\

\hline

\vspace{1 cm}

\\

Now we can sub $x(t), y(t), z(t)$ into $\delta(x,y,z)$ to obtain$,

\\

\\

$\delta(x,y,z)=15\sqrt{y+2}=15\sqrt{(t^2-1)+2}=15\sqrt{t^2+1}$

\\

\\

Thus $\int_A \delta(x,y,z)ds=\int_a^b\delta(0,t^2-1,2t)|v(t)|dt$

\\

\indent\indent $=\int_a^b\delta(0,t^2-1,2t)|v(t)|dt$

\\

\indent\indent

$=\int_{-1}^{1}15\sqrt{t^2+1}*2\sqrt{t^2+1}dt$

\\

\indent\indent

$=30 \int_{-1}^{1}t^2+1 dt$

\\

\indent\indent

$=30[\frac{t^3}{3}+t]^{1}_{-1}$

\\

\indent\indent

$=30[\frac{1}{3}+1-(-\frac{1}{3}-1)]$

\\

\indent\indent

$=30[\frac{8}{3}]$

\\

\indent\indent

$=80$

\\

\hline

$\vspace{2 mm}$

\\

Now for the numerators of our respective centroids:

\\

\vspace{5 mm}

\\

$\int_Ax\delta(x,y,z)ds=\int_{a}^{b}x(t)\delta(0,t^2-1,2t)|v(t)|dt$

\\

\indent\indent

$=\int_{-1}^{1}0*15\sqrt{t^2+1}*2\sqrt{t^2+1}dt$

\\

\indent\indent

$=\int_{-1}^{1}0dt$

\\

\indent\indent

$=0$

\\

Thus $\bar{x}=\frac{0}{80}=0$.

\\

\vspace{5 mm}

\\

$\int_Ay\delta(x,y,z)ds=\int_{a}^{b}y(t)\delta(0,t^2-1,2t)|v(t)|dt$

\\

\indent\indent

$=\int_{-1}^{1} (t^2-1) *15\sqrt{t^2+1}*2\sqrt{t^2+1}dt$

\\

\indent\indent

$=30\int_{-1}^{1}(t^2-1)(t^2+1)dt$

\\

\indent\indent

$=30\int_{-1}^{1}t^4-1dt$

\\

\indent\indent

$=30[\frac{t^5}{5}-t]^{1}_{-1}$

\\

\indent\indent

$=30(\frac{1}{5}-1-(-\frac{1}{5}+1))$

\\

\indent\indent

$=30(-\frac{8}{5})$

\\

\indent\indent

$=-48$

\\

Thus $\bar{y}=\frac{-48}{80}=-\frac{3}{5}$.

\\

\vspace{5 mm}

\\

$\int_Az\delta(x,y,z)ds=\int_{a}^{b}z(t)\delta(0,t^2-1,2t)|v(t)|dt$

\\

\indent\indent

$=\int_{-1}^{1} 2t *15\sqrt{t^2+1}*2\sqrt{t^2+1}dt$

\\

\indent\indent

$=60\int_{-1}^{1}t(t^2+1)dt$

\\

\indent\indent

$=60\int_{-1}^{1}(t^3+t)dt$

\\

\indent\indent

$=60[\frac{t^4}{4}+\frac{t^2}{2}]^{1}_{-1}$

\\

\indent\indent

$=60(\frac{1}{4}+\frac{1}{2}-(\frac{1}{4}+\frac{1}{2}))$

\\

\indent\indent

$=60(0)$

\\

Thus $\bar{z}=\frac{0}{80}=-\frac{3}{5}$.

\\

\vspace{5 mm}

\\

Thus $(\bar{x},\bar{y},\bar{z})=(0,-\frac{3}{5},0).$

\\

\hline

$\vspace{2 mm}$

\\

From the parameterization of the curve A, we may conclude that since,

\\

$y=t^2-1$ and $z=2t$,

\\$y=(\frac{z}{2})^2-1 \Rightarrow \frac{z^2}{4}-1=y$

\\

\hline

$\vspace{2 mm}$

Note that $x=0$ and $\frac{z^2}{4}-1=y$ tells us that our wire has zeros at the coordinates (0,0,2), (0,0,-2), and (0,-1,0). Thus taking into account

\\

$-1\leq t \leq1$, yields the following the curve:

\\

$\vspace{2 mm}$

\\

\begin{tikzpicture}

\begin{axis}[

xmin=-2,xmax=2,

ymin=-3,ymax=3,

grid=both,

]

\addplot [domain=-2:2,samples=20]({x^2/4-1},{x});

\end{axis}

\end{tikzpicture}

\end{document}

You must be logged in to bookmark a video.

...and 1,000,000 more!

OR

## Discussion

## Video Transcript

all right, So the problem presented years to find the center of Mass. Given the mass density function, Delta X Y Z is equal to 15 times a square root of lifeless, too. Along the curve, R of T is equal to t squared minus one J plus two t k Pretty varies from negative 1 to 1. Now we're trying to find a century of mass. We won't remind yourselves with a couple key formulas. First of this. Go away. It's long click in my notes. Okay, so forth. One first member are the formulas for our central it specifically. If want center of Mass, you want to think of the center of mass in terms of X, y and Z So these were just the interest of master Centrowitz with, ah long perspective access So X y z? No. So they're all the same. Only thing that varies is the variable in the numerator. Where on each of these were that Delta X y Z's air mass density function and, well, I want to fight first. No sign of the denominator is the same of each of these Central. It's sobering to solve this problem. We want to kind of go about this sufficiently as possible. The first thing we can do is find the job mater of each function so no to the total mass. And I use dissertation in May is given by the total mass on A is our wire We're going along a wire A This is just a line integral along some curve wire is the curve eloping this time Some book you see it, see? But we labeled it a So this is our total mass along our wire, eh? With this density mass function Now we go back to the prom presented a number that is privatized over Artie equal to the square. Miles wanted shapeless to take to t k. So from this, since there's no eye, so we get from this is we know that there's there's actually a plus zero. Hi. We could add to this statement And that means that if we try to think of our parameters in terms of X Y and Z 50 is equal to zero. Why response to R J vectors that was our T squared minus one, and Z would be in the K direction of the Ark duty now when It's also no girl. We're gonna be putting everything in terms of tea. Story means we're going to change d s to be in terms of tea as well. Now, one way to understand this formula. So the former D s is gonna be the magnitude of the tea kinds DT. And the way you think about that thinking my pen to work that I have some curve I want to go from here Time. I will go from there too. They're all right. This changing my curve. G ass has a length to it We could think of, uh I can't think of a normalized unit vector in our X y Z plane, and they were going to take that normalize the vector, and we're gonna take it along some amount of time. And so we're gonna have a unit vectors of Thompson like the time necessary, and provide the length of this little bit If we imagine this bit being some incontestably small little change. So way we find the magnitude of the normalized specter would be to take the magnitude of ah, the square root. We'll make two beauty which is the square root of the derivative of X er t squared plus the derivative of Why, If he squared plus the derivative of the city square this week was our magnitude our length of our unit vector and it's going to the calculations so driven a zero is still zero. That's not gonna do anything derivative. T squared minus one is to t and the derivative of two tea is too. So I see that this becomes a square of two t squared plus two squared, just just 40 square plus four which becomes some latitude and spirit of tea plus one. So this is the length of our unit vector And then we will just be taking that length and multiplying it times some change in time DT. And that will result in this overall change along my curve. Yes, All right. So we now have that bit. So we have the d s part of cynical. Now we want to take care of the rest of it so we can sub r, x, y and Z and two are probably mass function Unclear My screen. So it's that way. There we go. But it's over. Call from the beginning. The problem. We know a podcast function is 15 10 square to whiteness to, and we know that why, as a function, tea is just tea squared minds once we're going to substitute that in here and then simplify So we get 15 to the square of P squared plus one. So my total mass is integral line integral along my curve a not the X y z d s. It's really gonna be a curve from a to B of Was playing in my X y and Z here. So x zero, why was to spread mus one z was to t times d s rewritten in terms of tea. Now you've solved for all these bits we know that this delta function is in fact, now would be 15 t school party scrip plus one. We know they're ti's range from negative 1 to 1 has presented the problem and we know that this magnitude the tea is really just two times square root tea script was one. So this now my new integral in terms of teeth. Now you just superb. I so freaking has two is 30. We call it the constant swearing swearing of spirits from go away because the power becomes one such a snotty scripless one. We apply the integral suite at the anti derivative of T squared T cubed over three and one is tea and we're gonna evaluate that from negative one. What? Plugging in my one and negative one. We get 30 times 1/3 plus one minus. Don't fear Princess is really the missus out by forgetting Quincy's minus negative on third minus one. These are both on our terms, So these remain negatively Beginnings of one and 13 plus one is 4/3 minus and negative. So plus wondered on one again for third. So 8/3. So 30 times 1/3 8 So the total dropped here it's the total mass of my wire. Is he? Now we're gonna burn the point. This is we need to find our individual central. It's so we now need to find senior of mass respect, Excellency. Two. So we're gonna go with these one. At times, we'll start X So numerator writes in Troy X Is it a girl along any of ex Delta X y z? Yes, we privatized everything in terms of tea. So again, Delta is just felt function to see before. Now this is D s in terms of tea and then X is really exit e will exit e was just zero, as I remember from beginning there. So this is just zero times things you've seen before Zero times anything is zero and integral of zero is still zero. So the central oId for exes, actually. Nice, because we get the numerator to be zero and from our work right here, we know that the denominator is he. So x bar, which is the notation we used for central zero. Now we'll do the same thing for why so again, we only know this nominator. We're fucking out of the end. We are the numerator, so take the inner roll longer curve. Why Tim's apartment mass function. Pramuk tries everything in terms of tea ever been seen before? The right part of this is the same as we've seen before. 15 to the Square T script was one time two times Square T script was one T and their wives t was really t squared minus one. So door new cram occurs in 15 times two is 30. So get back to that out. Swear roots certify to next 11 strike T scruples one have t squared minus one. This is a difference of squares. So we know that this is actually the This is a unf actor for Misty of the fourth minus one. So I'm just a guy waiting is integral 30 times in a roll of negative 1212 year the fourth must won t t applying and a derivative t the fourth becomes to you This is five native one because t evaluated for negative 1 to 1 hugging in again. These are odds bucking the negative. It's going to stay the negative inseventy will be preserved. So playing the one I get 1/5 minus one looking in. Negative one negative one sis minus negative wants a plus one and 1/5 minus one is negative. 4/5 minus. This is the same A minus one minus. Want this 4/5. Made negative is also negative. 4/5. So I have negative 4/5 minus forfeits his total negative eight fits and 30 times negative Ate. This reduces to negative 48. Those are centrally, Inspector Y is Where are y Cintron? I should say is negative. 40/80 which reduces to negative three fits and a real doozy. So again, same thing we've done twice. Now we have our delta had our delta function and D s with prime tries. In terms of tea received t was actually to t so 2 to 15 tends to That's gonna 60 t is gonna stay in the inn a grand. I swear it's gonna go away. So I have t tempts the times. Princes T Square plus one distribute the tea through its becomes t cubed plus t apply them to derivative T to the fourth of a four plus t squared over two. Again, we're evaluating negative 1 to 1 note that here the degrees are even so Negatives drink function will become positive. So I end up with 1/4 plus 1/2 minus one for us would have these are the same or they become zero because this side is so positive. Three force minus three force. So 0 60 times zero is zero. And again, we could plug in 80 for the John Matrix. We did that scratch work already, and this is a This should the that this woman right here ignore that draw. There's my drama. Yeah, this that's the type of obviously a zero divided. My 80 is equal to zero, not three. Let's do this. The latex. So I must have acted, copied and pasted the same equation there. Oh, yeah, actually, zero. I have put it to clear the solar will follow me along as we go. Just keep in mind that is actually zero. Okay, So like calculating the centers of massive respect each access we now know the X bar was equal to zero. Why? Mar was because the Nega 3/5 and Z bar equal zero. Thus, as this ridiculous expert on Whitewater, clumsy bar is equal to zero call a negative 3/5 comma zero. So that is a center of mass. Now, the second part of this question was then to sketch the curve. Well, if y is equal to T squared minus one and Z is equal to t notes that we can substitute, we can solve this equations equals to tea in terms of tea. Simply divide both sides by two. When we get t is equal to Z over too. You could draw that look like this guy. Give us that Z over too equals T We just take this, you plug it in right there. So we get wise equals zero. Choose squared minus one, which is really the same as C squared. Over four. Minus one equals. Why now this We can kind of think about this A couple ways of you understand how to draw this. Make clear some sap away. No note that in the problem R r T director Artie Artie, Marty is curved to find over y and Z where? X zero So we know that we're in the We're always in X equals zero plane. So that's gonna make this nice, because we can actually draw this simple curve. If I just take what I have here this equation and we try to graph it. You try to put that in there correctly, you change it. I would type of go away. Okay. Right, right. So I must fly forth to multiply it for two everything. Let me make one quick note years. This makes more sense if I switch to it for you. This is the same as e squared minus for equal to for Why was more for everything before we get this equation here, which is the equation that I decided to craft right here. C squared minus four. Oh, and I'm Joe Or so note, this is to save my minus four white of this side and that ad for dinner. At this question, these the same question. I just typed you done differently when I put it right here. You know, if we look at this graph Woop, it's going to look like cheeks for recording the red. The red is my wire. Now the red is not quite my wire. Because remember that in the original pressurization, the curve exes always zero. So if I add the equation, X equals zero note that therefore always next equals aeroplane that's gonna cut this. We're gonna be kinda hard to get perfect, but we're gonna be tracing this shape. So I see that this equation my Z squared minus four equals four is this curve My center of mass that we found was zero comma *** freefest, Come. Zero was right here. And if we think about this, it's curve. So my curve of the wire makes sense of the center of mass would be right here and know this curve isn't actually quite as actors. We're doing this in Judge Breyer, so it grafts over an infinite domain. We be restricted the domain between negative one and one for tea. So we we'll need to actually cut this off just a little bit. And if you go back to our notes here, waits for X equals zero. Another excuse heroes were a two dimensional plane. Here's the two dimensional plane. I just cut it down. Now, here have kind of given away already because you see that I cut it off at negative to a two year. But that's supposed to die. We're still looking at the infinite graph. When I looked at this function without looking at a graphic software, I know that flooding in plus or minus two from easy access will result in 4/4 or one and one minus 1 to 0 for another fucking in to a *** to is gonna give This is zero. So I know it's gonna look kind of like, and I know that if Z is, your illness would be negative one. So in terms of my y Z plane, well, my explain and looking at X Y and Z I know that I'm gonna have this intercept here where wise negative one and everything else is zero. And then these two points, where's the use? Plus or minus to an ex wire zero. And then to see that this line does indeed not keep going. Note that if we think about our original parameters, let's draw those real quick. So laxity is equal to zero. So we're just in a two dimensional playing around and three dimensional. Why of tea two to T square mice, one of leave and see t is equal to two. Okay, so it's like a negative one. I get negative to plug in one I get to So we see that Zeze indeed going from native choose to end up looking negative one your negative one here I would get well, negative one. I get zero. So a T is negative one when he gets one way we could do this. Is this something that helps to dio table? And we know that our key values here he's going from negative one toe once we're in a negative one zero and one and we'll do why and see rather be talking this problem a little easier to see. That's supposed to be a Kama region. That's common. Okay, Tee's negative. One negative. One squares, one almost from zero. So zero comma, two times negative. One is negative, too. That is indeed this point. And if you plug zeroes in well, zero minus one is negative. One to 10 00 So when do you get this point and looking? One one minus 10 And she does want is to. So we do indeed get this point. So we see that the curve does indeed stop here at Z equals. Let's remind us to like I craft besides the shape of our curve, I believe you should be able to see that I uploaded the late tech code What I did here, in case you want to read through any of the notes, I've uploaded yourself. We're playing a little bit of this craft, and that's it. Thanks

## Recommended Questions

A wire of density $\delta(x, y, z)=15 \sqrt{y+2}$ lies along the curve $\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+$ $2 t \mathbf{k},-1 \leq t \leq 1 .$ Find its center of mass. Then sketch the curve and center of mass together.

Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve $\mathbf { r } ( t ) = t \mathbf { i } + 2 t \mathbf { j } +$ $( 2 / 3 ) t ^ { 3 / 2 } \mathbf { k } , 0 \leq t \leq 2 ,$ if the density is $\delta = 3 \sqrt { 5 } + t$

Find the center of mass of a thin wire lying along the curve $\mathbf{r}(t)=t \mathbf{i}+2 t \mathbf{j}+$ $(2 / 3) t^{3 / 2} \mathbf{k}, 0 \leq t \leq 2,$ if the density is $\delta=3 \sqrt{5+t}$.

Find the mass of a thin wire lying along the curve $\mathbf{r}(t)=\sqrt{2} t \mathbf{i}+\sqrt{2} t \mathbf{j}+\left(4-t^{2}\right) \mathbf{k}$

$0 \leq t \leq 1,$ if the density is (a) $\delta=3 t$ and (b) $\delta=1$.

Find the mass of a wire that lies along the curve $\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, 0 \leq t \leq 1,$ if the density is $\delta=(3 / 2) t$.

Center of mass and moments of inertia for wire with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve

$$

\mathbf { r } ( t ) = t \mathbf { i } + \frac { 2 \sqrt { 2 } } { 3 } t ^ { 3 / 2 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k } , \quad 0 \leq t \leq 2

$$

if the density is $\delta = 1 / ( t + 1 )$

Mass of a wire Find the mass of a wire that lies along the curve

$\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, 0 \leq t \leq 1,$ if the density is $\delta=(3 / 2) t$

Mass of a wire Find the mass of a wire that lies along the curve $\mathbf { r } ( t ) = \left( t ^ { 2 } - 1 \right) \mathbf { j } + 2 t \mathbf { k } , 0 \leq t \leq 1 ,$ if the density is $\delta = ( 3 / 2 ) t$

Mass of wire with variable density Find the mass of a thin wire lying along the curve $\mathbf { r } ( t ) = \sqrt { 2 } t \mathbf { i } + \sqrt { 2 } t \mathbf { j } + \left( 4 - t ^ { 2 } \right) \mathbf { k }$ $0 \leq t \leq 1 ,$ if the density is (a) $\delta = 3 t$ and (b) $\delta = 1$

Find the center of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve

$$\mathbf{r}(t)=t \mathbf{i}+\frac{2 \sqrt{2}}{3} t^{3 / 2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}, \quad 0 \leq t \leq 2$$

if the density is $\delta=1 /(t+1)$.