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Point) A long thin bar of length cm is made of material, whose mass density changes along the bar according tofk) =x+4, 0<x<9Find:a) The total mass of the bar...

Question

Point) A long thin bar of length cm is made of material, whose mass density changes along the bar according tofk) =x+4, 0<x<9Find:a) The total mass of the bar:b) The average mass density along the bar:c) The center of mass of the bar:What point along the bar should you cut to get two equal pieces of mass?

point) A long thin bar of length cm is made of material, whose mass density changes along the bar according to fk) =x+4, 0<x<9 Find: a) The total mass of the bar: b) The average mass density along the bar: c) The center of mass of the bar: What point along the bar should you cut to get two equal pieces of mass?



Answers

The mass density of a metal bar of length 3 meters is given by $\rho(x)=1000+x-\sqrt{x}$ kilograms per cubic meter, where $x$ is the distance in meters from one end of the bar. What is the average mass density over the length of the entire bar?

In this question, we're asked to find a mass of a one meter cylindrical bar with given constant that constant densities. Let's just write that down First were given that role one. Is he going one grand per centimeter? This is would be the left side, Let's say and roll to would be two grams per centimeter. And we're also given that length IHS one meter which I will convert two centimeters because I see that dest entity mess is measured in grams for centimeter. So that's 100 centimeters. We are asked to find a mass. So when we every right, uh, that city function, um, as a piece wise function first to make it more clear. So we are saying that row is equal to one when your your value of X is between zero and halfway, which is 50 and your role is equal to all right when next is between 50 and 100. And we know that, um, mass formula is equal to integral from a to B of three, the X So in this case we have you need to grow from Cyril to 50 of one t x plus in to grow of 50 to 100 of two d x and so the first in the grow one DX The interval We just x and we're evaluating from their all to 50 plus and integral to D X would be two x and we are evaluating from 50 to 100 which means it's 50 minus zero plus to 100 minus 50. This gives us 50 plus two times 50 which would be 150 and that's it.

Okay, Figure out the length of the still bar that we need to cut that has a cross section, this equilateral triangle. First figure out the area of the equilateral triangle. You can look up the equation online. It's going to be squared. Three over for me. Bye bye. This one side of the equal lateral triangle square. Now this was given to US intentions. So they are. Answer will be 2.706 inches square because this was 2.5 inches. So now convert the interests where the centimeters squared. Recognizing for everyone inch there's 2.54 centimeters. But when you square it, so we get squared. Centimeters 17.46 centimeters squared. Density is equal to grams from no leaders. Density was given to us. The grams was given to us as one kilogram or 1000 grams. Well, then divide by the volume and milliliters or centimeters cube. To get that, we're going to need to take the centimeters squared, multiplied by some unknown centimeter value. Well, then rearrange the equation to solve for that unknown centimeter value, which will be the length 7.44 centimeters

Problem. 11.1. We have a bar that's 50 centimetres long and has a mass with a couple of other masses. Intuit, whose size is such that we don't need to worry about how big they are, we can treat them as his point masses. And so we want to know if we want to balance this rod with some sort of full crew where along it we would want to put the fulcrum, which is the same was asking, Where is its center of gravity? And we want. It is a length from the left. So when the variation in the force of gravity with you know elevation could be ignored, like in most cases the center of gravity in the center of Mass were the same and the make that look like a signal, the, um, center of mass is just the weighted average of all of the positions of things in the in the system. No. So you some of the massive, each object times it's positioned. That's all divided by the total mass. So in this case since we're measuring, we wanted from the left end. So let's go ahead and just, you know, measure things from from the left end. So here it would be zero, the center of mass of the bar, which we can use as its position is in the middle of it exits uniform. And then this is 50 centimeters away. So we have zero for this one. We don't need to really worry too much about that right now. Uh and then we have are what, 0.120 kilograms of the bar times the 25 centimeters it is away from the left end. And then our 0.11 kilograms of the red mass here times half a meter away. It iss. And then all of this will be divided by the sum of the masses. 12 kilograms. Scare of playing girl, five by kilograms, plus 0.11 kilograms. And so this works out to be 29.8 centimeters. Since this is originally given in centimeters, might might as well make sure answers and seven meters to it doesn't really matter for this. It's as long as you don't get confused. With what? You know how you're using the units? Yeah, you could. Didn't you report this in meters? Are do this calculation and centimeters. You can convince yourself that, you know, you'll get the same answer if all of these air in centimeters instead of meters, because you're just multiplying by 100 or dividing by 100 good. Anyway, So, yeah, So you would want to put it, you know, somewhere about here ish? Yeah. To the right of the center of the bar.

All right. For the first part of this question, we're trying to figure out which values of L is. Bar one heavier than bar to bar one has a density of four e to the negative X power and bar to has a density of six e to the negative two x power. And this is for all X between zero and a length. L Okay, So in order to do this, we have to use our relationship. Mass is equal to the integral of density. End from zero to else. Let's plug in or two values and eventually will set up in inequality. So for the 1st 1 we'll plug in for E to the negative x, and we could integrate that you pull out the four, we would have e to the negative X, and that's all divided by negative one. With respect for in regards to a range of zero to l. This negative one could be pulled out with the four and we could be right this as the following Take this a step further, just rearranging stuff at this point and each the zero equals one. So just rearranging, rearranging and that is your mess one, not a fine mass to will do the same exact thing. So in this case, it's six e to the negative two x, and I'll jump a little bit faster for this storm. The negative, too gets brought to the bottom because you're dividing it. That's gonna become negative. Three. You to the negative to L Same thing. We just rearranged this and you'll get whether this should be just to make it clear. Not absolute value bars. He should be brackets. They noticed. I'm doing that, the other brackets. So this is what you get from Mass to. And now let's set up our equality. So we're trying to find we're trying to set it up. That mass one is heavier than mass to. So in inequality of this and we plug in what MASS one is, we get four minus four over E to the L. That's just distributing this four, and we could do the same exact thing for Mass to distribute the three. And now this algebra may look a little bit intimidating, but let's walk through it so we could bring the let's move like terms to the same side. So do four minus three, we'll do. We'll keep the ease on the same side and this could be rewritten as eight minus three. We're finding a common denominator. And if we do that, we'll get eight. Four Ministries one. And now we just rearrange each the Tooele. Okay, we have five on this side. And to get rid of this E, we need to take the natural log. So let's do that. That will give us to L. Ellen five, and at the end you'll get that 0.5 or 1/2 times. The natural log of five is less than L, and that is your answer for port A meaning that that is the L value for which bar one is heavier than bar to when it comes to Part B. Were asked about the relationship between density, um, the length of a bar and mess so, typically for a one dimensional object, linear densities used. So it's linear density, and it has Kunitz of mass per length. So those are the units of linear deck city, and this relationship is a directly proportional relationship that is, mass is equal to density multiplied by the length when and if we increase length, then mass would also increase directly proportional


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