A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ = cx^2 , where x is measured from the center of the rod and c is a constant. b. Find an expression for c in terms of L and M.

Video transcript

Hey everyone. Let's go through this practice problem. The linear density of a thin bar with length L sub B and mass M sub B varies with position in accordance with the function lambda of Y equals a multiplied by Y cubed where Y is the distance in meters from one end of the bar and A is a constant, determine the relationship between the constant A and the bar's mass and length. Option AM sub B divided by L sub B. Option B two multiplied by M sub B divided by L sub B squared. Option C three multiplied by M sub B divided by L sub B cubed. And option D four multiplied by M sub B divided by L sub B raised to the power of four because A is a constant that's only defined in the problem as being related to the linear density formula. We want to find an expression for the mass that relates to the linear density and then solve for a to complete the problem. Recall that in order to find a mass of a more complex shape, you have to integrate the density along the area or the volume or the length of the material. In this case, we're looking at a thin bar, which if we assume is a one dimensional shape, then we'll just want to integrate the linear density across the length of the bar. So we'll say that M sub B, the mass of the bar is the integral of the linear density of the bar. The formula given to us and the problem with the variable of integration being D Y because Y is the variable distance in meters from one end of the bar to the points being integrated, we're integrating this along the full length of the bar. So the bounds of integration are going to be from zero to L A B because that'll be an integral across the bar's full length. So to make this integral a little more clear, I'm gonna actually write the formula itself in there. So instead of lambda of Y, it's a multiplied by Y cubed and that's what we're integrating with respect to Y. So the first step of course is to simplify the integral slightly by removing the constant. So that's a multiplied by the integral of L sub B, the integral of Y cubed with respect to Y. Fortunately, for us, this is a fairly simple integral uh integrating Y cubed is just a matter of doing a power rule integration. And recall that power rule integrating is simply done by taking the exponent and raising it by one. So three plus one is four and then dividing the whole thing by that new integral. So that's Y to the power of four divided by four. And then of course, we have to apply the boundaries by using the fundamental theorem of calculus. So the integral becomes a multiplied by. And then we take the upper bound substitute that into the formula, then subtract from that the formula with the lower bound substituted in, that's a multiplied by L sub B to the power of four instead of Y divided by four minus. And then we'll put in the lower bound zero, raise the power of four divided by four. And then of course, zero race. The power of anything is zero and zero divided by anything is zero. So this second term goes away. And the simplest way to write the integral we found is a multiplied by L sub B race, the power of four all divided by four and this is equal to M sub B. So the final and easiest step is to solve this equation for A by doing some simple algebra. So we multiply both sides of the equation by four and divide both sides of the equation by L sub B raised to the power of four. And we find that A is equal to four, multiplied by M sub B divided by Elsa B raised to the power of four. And so that is the answer to this problem. And if we look at our multiple choice options. We can see that this agrees with option D. So option D is the correct answer to this problem and that's it for the problem. I hope this video helped you out if it did and you need more practice. Please consider checking out some of our other videos which will give you more experience with these types of problems, but that's all for now. I hope you all have a lovely day. Bye bye.