Find the center of mass of a thin wire lying along the curve r(t)=t 𝐢+2 t 𝐣+ (2 / 3) t^3 / 2 𝐤,

step 1 Okay, folks, so let's take a look at problem number 36. We're going to be looking for the center

Find the center of mass of a thin wire lying along the curve...

Key Concepts

Center of Mass

The center of mass of a continuous object is the point at which the weighted relative position of the distributed mass sums to zero. For a one-dimensional object like a wire, it is determined using integrals that balance the moments of mass along each coordinate direction.

Parameterized Curves

Parameterized curves are used to describe the position of points on a geometry as functions of a parameter, usually denoted by t. This allows the curve to be represented in a form that is conducive to integration and analysis in space.

Line Integrals

Line integrals allow the computation of quantities like mass and moments over a curve by integrating a function along the length of the curve. This is particularly useful when the density is not constant, and each segment of the wire contributes differently.

Arc Length Differential

The differential element of arc length, often denoted as ds, represents an infinitesimal segment along the curve. It is calculated from the derivative of the parameterized curve and is essential for converting integrals from a parameter domain to actual length along the curve.

Density Function

The density function specifies how mass is distributed along the wire. In problems involving center of mass, the density function multiplies the arc length element to yield an element of mass, affecting both the total mass and the computed moments.

Moment of Mass

Moments of mass are calculated by integrating the product of the position vector and the differential mass element. These moments are used to calculate the weighted averages of the coordinates, which directly yield the center of mass.

Transcript

00:01 Okay, folks, so let's take a look at problem number 36.

00:05 We're going to be looking for the center of mass of a thin wire lying along a curve, and we're given the density of the wire.

00:14 Okay, there's a few preliminary things that i think you should know.

00:21 First of all, center of mass is a vector, and it's got, you know, a formula for it, which i'm assuming that you already know.

00:29 Center of mass is really just a weighted sum of the positions of all of the particles that make up the object that you're looking for the center of mass for divided by the total mass of course so what i mean by that is we have integral r dm okay and we're integrating over the the entire volume or area of of this mass okay and because r is a vector um r is just, you know, x had plus x had x plus y had y plus z had z.

01:12 So i'm going to write this thing as a sum over i multiplied by a fraction, x, i, d, m multiplied by e, d, m multiplied by e .i, where ei is the unit vectors like x had, y had, and x, i, means x, y, or z, depending on i.

01:41 Well, so when i is equal to 1, i have xi equals x, and when i is equals 2, xi equals y, and when i equals 3, x, i equals z.

01:53 But anyway, that's just notation.

01:57 So now let's first, first of all, let's go ahead and start looking for m, and we're done looking for m.

02:01 We're going to do three integrals.

02:04 So we have i equals 1 and 2 and 3.

02:06 But anyway, let's go ahead and get started.

02:10 Mass of the wire is just delta ds.

02:16 And we're given a function for delta, and delta is parameterized by the new variable t.

02:24 So we have, before plugging in delta, let's just go ahead and rewrite the s as x .2 plus y .2.

02:33 Plus, z dot squared, multiplied by dt.

02:38 Where i used the notation where dot on top of a letter means the derivative of that letter with respect to t.

02:46 All right.

02:46 Anyway, we have between 0 and 2, 3 times root 5 plus t, that's delta, multiplied by square root of 1 plus 4 plus t, dt.

03:03 And this is really trivial now because, um, um, okay, let's see, 5 plus t further.

03:14 D t 02 and when you crank this out you get 36 so that's the that's the mass and so we're done for for this part the denominator and now we're going to be we're going to be doing three integrals all of them look like this um the first one um is when i is one the second one is when i is two and the third one is when i is three okay so now i equals one we have x multiplied by dm integrated along the wire.

03:49 Okay, so x is just x.

03:54 Well, let's not, let's not start with 1.

03:59 Let's start with y, which is when i equals 2.

04:02 So we have y -dm.

04:06 That's equal to 2 times x -dm.

04:11 So if we have this integral, we also have this.

04:14 Anyway, we have 2 times t times 3, 5 plus.

04:24 T d t because that's um this is x x is um as a function of t is just t because that's given in the problem and dm is dm is just delta d s and if you look at this you you get 5 plus t d t um so i'm just you know i'm just reusing the results that we have previously obtained this thing is just delta ds and delta ds is equal to dm and so this is because this is because this is this is 3, 5 plus t d t, i took it and i plugged it here.

05:08 That's all i'm doing.

05:09 Anyway, we have, let's see, we have 6, 0 to 2, 5t plus t squared dt, and when you crank those numbers out, you get 76.

05:26 Okay, so as for xdm, that's simply 76 over 2, which is 30, let's see, 38.

05:37 Yeah, because this integral is twice this integral.

05:41 So to get this one, you just divide this by two.

05:45 Anyway...

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