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.

c. Find an expression in terms of L and M for the moment of inertia of the rod for rotation about an axis through the center.

Question

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.

c. Find an expression in terms of L and M for the moment of inertia of the rod for rotation about an axis through the center.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with moment of inertia problem. We have a non homogeneous bar of length lb and a mass MB and it's placed along the Y direction. The question was just to write an expression for the moment of inertia of the bar about a horizontal axis rotating through the center of the bar. The mass per unit length bar extending along the Y direction is given by the equation lambda equals 3 kg per meter cubed multiplied by Y squared or Y as in meters and is computed from the center of the bar. We're given four possible answers. A is MB multiplied by lb squared. B is three divided by five multiplied by MB multiplied by lb squared. We C is three divided by 20 multiplied by MB multiplied by lb squared. And finally OD we have five divided by 13 multiplied by MB multiplied by lb squared. So the first thing we're gonna do is draw a diagram of what we were given the problem. So I'm gonna draw my bar as a rectangle and we'll draw that in the um vertical direction since it's along the Y direction and we have an axis of rotation going through the center of the bar and that's a horizontal axis of rotation. And we'll label this as Y equal to zero. So that would be our coordinate um origin. That means that the upper, um the top of the bar is located at Y equals lb divided by two. And the lower end of the bar is located at negative lb divided by two. So we need to find the moment of inertia. Let's recall our formula for moment of inertia. So our moment of inertia, I can be equal to the integral of R squared DM. So here is R is the perpendicular, perpendicular distance from the axis of rotation NDM. Here is a small element of the mass. So what we need to do first is identify what that DM is. And so on our figure here, I'm going to take a little slice of our bar here and that's gonna be BM. So it's gonna be a small unit of mass. But we actually want to write that in terms of a distance. That way, we can actually integrate it. Since our um formula for the density here, the linear mass density has a value of Y in it. So we're gonna write our DM in terms of the density and ad Y so we'll have DM is equal to the linear mass density. And if I wanted to get a mass with that, I'd have to multiply that by distance, well, that DM covers a distance Dy in our diagram. So I'll multiply our LAMBDA by DY. Now, when I um plug that into our formula for the moment of inertia, I'm actually going to be getting, I'm going to be integrating over the Y variable. So I have I is equal to the integral going from negative lb divided by 22, I'll be over two and in place of R we're going to be using the variable. Why? Because the DM there is located distance Y from the um center axis rotation. So I have Y squared multiplied by lambda multiplied by bye. We were given a formula for our linear mass density lampa. So I'll plug that into our equation. So I have I is equal to the integral of negative lb over two, two lb divided by two have Y squared multiplied by 3 kg per meter cubed and that's multiplied by Y squared and dy. So if I notice here, um I'm gonna have Y squared multiplied by Y squared Y to the fourth. That's going to be an even function. So I can simplify the uh integration a little bit by multiplying it by two and integrating from zero to lb over two. I have I is equal to you. I'm gonna go ahead and factor out um pull out the constant term outside the interval. I'll have two multiplied by 3 kg per meter cubed and that's gonna be multiplying the integral from zero to lb over two of white to the fourth. Ey. So at this point, I can actually do the integration I'll have I is equal to two multiplied by 3 kg per meter cubed, multiplying the quantity. So when I integrate um Y to the fourth, I get Y to the fifth divided by five. So I'll have one divided by five multiplying for the Y value I'll be plugging in the lb, divided by two. So I'll have lb divided by two, all raised to the fifth power. Then I'll have minus lower limit which is just going to be zero. So I can simplify some things a little bit in this expression. So I'm gonna have 3 kg per meter cubed and I look at the constant term here. Um The two that I have here will cancel out with one of the twos in the denominator, the three to the fifth power. And so I'll have one divided by and I'll have five multiplied by two, the fourth which is 16. So that'll give me 80 on the denominator and then I'll just have that multiplying lb to the fifth at this point. I've calculated the moment of inertia. However, if I look at my um answers that I'm given, they're in terms of the total mass of the bar which is MB and not this um 3 kg per meter cubed. What I need to do now is figure out what the mass is of the bar. And for that, we just need to add up the total mass of the system. So our mass envy is going to be equal to the integral of DM. So if I add up all the little tiny masses, which is what I'm doing with the interval sign here, then I'll get the total mass of the system. And since I've found an expression for DM, so that means that MB is going to be equal to the integral going from negative lb over two, two lb divided by two of the lambda. Dy. Again, I can plug in our formula for the lambda. And so I get MB is equal to the integral going from negative lb divided by two to lb divided by two of the 3 kg per meter cubed multiplied by Y squared Dy. Again, I have an even function here. So I can change the limits of integration from zero to lb divided by two. If I multiply this in a gram by two, so I'll have MB is equal to two multiplied by the 3 kg per meter cubed double constant outside the integral. That's gonna be multiplied by the integral of 02 lb over two of Y squared. Dy did not perform the integration. They have MB equal to two multiplied by the 3 kg per meter cubed multiplying the quantity. So when I integrate um Y squared, I get Y cubed divided by three. So I'll have one third mo pine lb five by two cubed, the upper limit and I'll have minus zero for the lower limit. Just simplifying things a little bit. I have MB is equal to the 3 kg per meter cubed. And with constant term, I'm gonna have um the factor of two canceling out with one of the twos in the denominator that's been cubed. I'm left with two squared and that's gonna be multiplying by three. So I'll have 12. So I'm on the denominator. So I have the 3 kg per meter cubed multiplied by 1/12. And that's gonna be multiplied by lb at this point. Um I need to figure out how to input this mass into my moment of inertia equation. So I noticed that I have a 3 kg per meter cubed in my moment of Nursia equation and a 3 kg per meter cubed in my mass equation. So I wanna solve the mass equation for the 3 kg per meter cubed. So I'll have 3 kg per meter cubed, gonna be equal to 12 MB divided by LBH. So I can now take that and plug that into my moment of inertia equation. So here I'll have the moment inertia I equal to and in the place of the 3 kg per meter cubed, I'm gonna have the 12 MB divided by I'll be cut and that's is gets multiplied by the 1/80 and that is multiplied by lb to the fifth. So I can just simplify some things here in this expression. So 12 divided by 80 is going to give me um three divided by 20. And then that's going to be multiplied by the MB and I have lb to the fifth divided by lb to the third. So that just leaves me with LB. Where, so that's gonna be my final answer. And if I look at what answer choices that corresponds to, that corresponds to answer C so just to recap what we did, we found that moment of inertia by um integrating, using our definition moment of inertia. And at the end, we also needed to write it in terms of mass. So we basically used um the integral of our DM, our small unit of mass to figure out what it was in terms of the total mass. And then substituted that back in to find the total uh the moment of inertia in terms of the total mass and the length of the object. So when you're done with these types of questions, the trick is to figure out what your DM, what your small unit of mass is in terms of something that is typically in terms of AD YDX or DZ. So some sort of small change in the distance because that's typically what you're going to be integrating over. So I hope that this has been helpful and I'll see you in the next video

Verified Solution

Key Concepts

Moment of Inertia

Linear Mass Density

Integration in Physics