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A machine part consists of a thin, uniform $4.00-\mathrm{kg}$ bar that is 1.50 $\mathrm{m}$ long, hinged perpendicular to a similar vertical bar of mass 3.00 $\mathrm{kg}$ and length 1.80 $\mathrm{m} .$ The longer bar has a small but dense 2.00 -kg ball at one end (Fig. 8.42$)$ By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through $90^{\circ}$ to make the entire part horizontal?

$(0.333,0.7)$

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

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problem. Elena 8.55. We have this assembly in some sort of machine. We have a bar that's horizontal, another bar that's vertical and at the end of this bar is a heavyweight heavy relative to its size. So what we want to do is figure out where the center of how much the center of mass moves when the bar hinged, whatever goes from this configuration. So this configuration with location like that. So let's set up some coordinates. It's easiest to center the Met the hinge, so that's what we will do. And it's also easiest tohave X and Y running along the along the initial positions of the bars. So the initial X component of the center of mass, it's going to be the sum of all of the, you know, the weighted sum of all of the ex components of of the things here now, because we've put X equals zero here, the vertical bar doesn't contribute anything to the ex center of mass of this position. So this is we can treat this as a point mass halfway along its length. That's actually right out. We call this one this to this three then it's em one x one plus m two x 20 and to our m three x three. It is also zero, but we have to add up all over the masses. Because although the exposition is here, that means that they actually have toe by us the center of mass towards them still. And then, if you put all of the numbers and we find that this is 1/3 of a meter three this and just to reiterate the X, is there going to be for what? For the for the ball? It's just the center of the ball. And for the bars, it's halfway along the length of the bar. So that's all good. Uh, let's use space up here, baby. So now the initial why we're going to be m one y one. Which zero? Because this is on the y equals zero lion and then em to y two plus 73 white three. Why? To us screen why three blighted by to us at three. And this is zero. Quite said. Now we turn our attention to hear recall that the this will be our explore x to y two on here, obviously is X three y three, and now we can see that the there's no why component anything because all of the masses are now entirely along the X axis. So that was easy enough. And now So this is the only thing here that's any different is that we need to recall that this is the positive direction and acts so that these now have negative X coordinates. And let's call the new ones primes. I guess this one hasn't moved, but the other two have so two x to crime three x three Crime invited by one plus plus and two plus m three and then putting the numbers in and recalling like I said, being very careful that ex to a next year Negative is this is our origin and we said that left words left of the origin is positive X and so right of it is negative x. We get that This is negative 0.366 meters. So just carve out some space here. The change in the X coordinate of the center of mass is the final minus the initial so X final minus ex initial and so 0.333 meters minus negative 0.366 meters is going to be 0.7 meters to three digits years and then felt why is the final victory said zero minus y initial? So just up a negative sign on to that. And so now we concede that it moved. Yeah. No, I'm sorry, I'm mistaken. This is negative. That's positive. Because this way, that's the center of mass moves. So that in the negative extraction Excuse me. All right, because this was up. His ex final is negative in that spot. More negative. And then Oh, no. This is also because down down the positive direction and it moved upward anyway, so it moved 7/10 of a meter, Dan y And then, if you like Thio, get the actual, like magnitude of the displacement, you could use the pipe, agree here. But we're not really asked for that in this problem, so we won't worry ourselves with it. It's also fairly simple thing to do in it. You want to know what that is? It's it's good practice for you

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