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Use the methods of Challenge Problem 8.113 to calculate the $x-$ and $y$ -coordinates of the center of mass of a semicircular metal plate with uniform density $\rho$ and thickness $t$ . Let the radius of the plate be $a$ . The mass of the plate is thus $M=\frac{1}{2} \rho \pi a^{2} t$ . Use the coordinate system indicated in Fig. $8.51 .$

$x_{c m}=0 \quad-\quad y_{c m}=\frac{4 a}{3 \pi}$

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Numerade Educator

University of Washington

University of Sheffield

McMaster University

{'transcript': "problem 8.105 So here we're going to do a bit of calculus to find the center of mass of this semi circular plate. Got a radius a thickness t told x and y go like this were helpfully told what the total mass of the disc is, so we don't have to go to the bother figuring that out ourselves. So that's right down a couple of preliminary things we're going to want. We know that the center of mass is the weighted average of the position of, you know, an entire system waited by the mass so in a continuous case and averages one over sort of interval that we're looking at times X, the M and then similarly, the white coordinate of the center of mass is one over m I'm gonna go. Why, Dia, where this is the infinite decibel mass of, you know, whatever elements were looking Yeah, and then we're also going to want to know the relationship between X and Y at the surface of the plate. Here is a straight line. So that's, you know, find Phyllis. But since this is a circular art than the equation for a circle lives where basically the distance of the origin as to be equal to the radius or swearing both sides. The sum of the squares of the components of the position vector have to be equal to the radius squared. So now let's start thinking about the export in it. First off, since there's an equal amount of mass on either side of the Y axis, we're hoping to see the ex center of Mass. Uh, ex coordinated center of Mass is going to be zero. It should be in the middle here because if you flip this around so that exposed to negative acts, it looked exactly the same. So the only place the center mass should be is an X equals zero or along the line X equals zero. Why will be more interesting? Because it's not a complete circle. So it it actually does change when you flip it around. And so the, uh, have actually been a little bit more work. So for eggs, we want to think about dividing this into little vertical strips that have a height. Why with the X, and obviously their position is acts. So just taking this equation here, why is always going to be a function of X, the square root of a squared minus expired. So plugging that in here, okay? And also noting that d ever this is the massive. This little strip is going to be equal to its volume times its density. So now what's its volume? Uh, it's going to be Oh, yeah, that's what we need to fight for. But, uh so sorry. Um it's gonna be the density times, the thickness times the height we know in terms of X is a squared minus X squared times its wits, which is the ex. And so now this is going to be in a girl from X equals. Negative, Eh? Eh? Uh, well, that's put the constants back on the ground. So road he are Constance. So roti over and from X equals negative. A positive A of x times the square root of B squared minus next week. Now, this is an odd function. We could see that because this part is even if you take extra negative X, this doesn't change. But this is odd, because ext two negative X, you know, changes the time. Negative X and so on. Odd function times and even function is odd. And we know that if we integrate anod function across the symmetric and interval, it's metric about the origin. Then you know, this works out to be zero. And like we said in the beginning, this is what we're hoping to get. So all is well with the world so far for the why center of mass. We want to integrate these little B Y strips. They have the wits of to X and then the thicknesses. Of course, t always and then be why is their height? So here are the m gonna be road times the this times two x the wits which just solve this one for ex now a squared minus y squared The root of that and then the height is the Why just change colors here is getting a little bit crowded. So the white, the white co ordinate of the center of mass don't look quite similar, But we just did. Except there's a to here now, why only goes from from zero to a why times square root eighths minus y squared. Why? So you can use a, uh, a U substitution to solve this integral. We're not gonna walk all the way through that. But just to tell you what the substitution is, you want few equals a squared minus y squared. We need to get through to this problem here. Then the U is going to be negative too wide E y. So this is a substitution you want to use to evaluate this and working through all that, you'll find that this is for a divided by three pies are white co ordinate for the center."}