The density (mass per unit length) of a thin rod of length ℓ i...

Question

The density (mass per unit length) of a thin rod of length ℓ increases uniformly from λ₀ at one end to 3λ₀ at the other end. Determine the moment of inertia about an axis perpendicular to the rod through its geometric center.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the moment of inertia. So in this problem, we need to determine the moment of inertia of a slender rod of length L which has a linear mass density that increases uniformly from lambda at one end to two lambda at the other about an axis that's perpendicular to the rod and passing through its center. Rum. Four possible choices as are answers. Choice A is Lambda L cube divided by 24. Choice B is lambda L cube divided by 12. Choice C is lambda L cube divided by eight and choice D is lambda L cube divided by six. The first thing we're gonna do is draw a diagram of our problem. So we'll draw a rectangle or the slender rod and we're gonna draw the axis rotation and we're told that it's perpendicular to the rod through its center. So we'll draw that as a dash line, I wanna set this as our origin of our coordinate system. So this will be at position X equal to zero. So that means that on the right hand side of the rod, I'll have X is equal to L divided by two. And that's because the entire length of the rod is L. And on the left hand side, we'll have X is equal to minus L divided by two. Now, we're asked to calculate the moment of inertia. So for that, we need to recall our formula for the moment of inertia. And that is I is equal to the integral of R squared DM. So here I is our moment of inertia, R is the perpendicular distance from the axis of rotation to some small piece of mass of the rod. And DM is a small piece of the mass of the rod. So DM is gonna be represented as a small sliver of a rod. And we need to write that DM in terms of the quantities that we're given in the problem the L and the lambda as well as our coordinate position, uh coordinate X for our position. So we're gonna write DM in terms of those quantities. So we'll have DM is equal to. And one way I can get a small piece of mass is to take our linear mass density, which will label as lambda. But this is gonna be a function of X since it's not constant. And if we take our linear mass density and multiply it by a small length, that'll give me some small piece mass. So our small length will label as DX. Now I need to write um our linear mass density as a function of X. So we can use the two limbs that we're given the lambda and two lambda about the two ends to come up with an equation. So we'll have DM is going to be equal to. And here I'm just gonna have an equation for a line since it says that my linear mass density is increasing uniformly. So since it's increasing uniformly, that means it basically has a slope. So the slope of our linear mass density function is just gonna be our change in our linear mass, which is the two lambda minus lambda. And we'll divide that by the change in the X cord. It which is just gonna be the entire length of the rod, which is just L and now the slope gets multiplied by the coordinate X. And then we'll have to add to this the value of the linear mass density when X is equal to zero. Now, since it's increasing uniformly from lambda to two lambda, when X is equal to zero, which is the center point, it's just gonna be the midpoint between lambda and two lambda. And that happens to be three lambda divided by two. And now that we have the equation, we take that equation and multiply the whole thing by DX, I can simplify my slope there a little bit. So I'll get DM is going to be equal to the quantity. So I have two Lambda minus Lambda that just gives me lambda. Divided by L, that fraction is multiplied by X. And then we'll have plus three lambda divided by two. That whole quantity is multiplied by DX. Now, I have an expression for the DM in terms of the lambda, L and X. So I can take that and plug it into our formula for the moment of inertia. So we'll have I is equal to the integral going from minus L divided by two L divided by two. And for R here, the perpendicular distance from the axis of rotation to some small DM is just gonna be X. So I have X squared. So basically replacing R with X and then we multiply this by the expression that we found for DM, it gets multiplied by the quantity of lambda, L lambda divided by L multiplied by X plus three lambda divided by two. And this quantity is multiplied by DX. Now I can distribute the X squared inside the parentheses. So I'll get I is equal to the integral from minus L divided by two to L divided by two of quantity of lambda divided by L multiplied by X cubed plus three lambda divided by two multiplied by X squared. And that quantity is multiplied by DX. So at this point, I can actually do the integration and I noticed that I have basically two integrations to do one for each of the two terms and uh inside the parentheses. And the first term is an odd function X cubed. And anytime you integrate a odd function uh about limits that are symmetric about X equal to zero, that whole integral is actually going to be zero. So since I'm integrating from minus L divided by two to L divided by two of X cubed, that whole integral is going to be zero. Let's just use the fact that I have that executed as an odd function. Now the other integral is gonna be X squared. That's an even function. So I can simplify the integration a little bit by changing my limits of integration to go from zero to L divided by two. If I multiply the entire integral by two, so we'll do that. So we'll have I is equal to two multiplying the integral going from zero to L divided by two of three lambda divided by two multiplied by X squared DX. So here I can now go ahead and do the integration of X squared. I first noticed that the factor of two and one half will cancel out. And so I'll have I is equal to three lambda. And when I integrate X squared that gives me X cubed divided by three. And this is evaluated at zero and L divided by two. So we can go ahead and plug in our limits of integration. So I have I is equal to the factor of three and one third will cancel out. And so I'll have lambda multiplying the quantity then we'll just have X cubed. When I plug in that limit, we'll just have L divided by two quantity cubed minus zero for the lower limit. This means that my moment of inertia, I is gonna be equal to lambda L cubed divided by eight. So that's the answer we were looking for. And I wanna look at my answer choices, this corresponds to answer choice C. So while this problem was a direct application of our definition for the moment of inertia, we did have to be careful when looking at our DM element. And we have to write the linear mass density as a function of position. And so that meant writing it in terms of the lambda, the L and our position X. But once we did that, we could plug that into our formula for the moment of inertia and it was a straightforward integration to find our answer. So I hope that this has been useful and I'll see you in the next video.

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The density (mass per unit length) of a thin rod of length ℓ increases uniformly from λ₀ at one end to 3λ₀ at the other end. Determine the moment of inertia about an axis perpendicular to the rod through its geometric center.