CALC A slender rod with length L has a mass per unit length that ...

Question

CALC A slender rod with length L has a mass per unit length that varies with distance from the left end, where x = 0, according to dm/dx = gx, where g has units of kg/m^2. (b) Use Eq. (9.20) to calculate the moment of inertia of the rod for an axis at the left end, perpendicular to the rod. Use the expression you derived in part (a) to express I in terms of M and L. How does your result compare to that for a uniform rod? Explain.

Accepted Answer

in this problem a non uniform then spoke has a total length of its mass per unit length depends on the position X. Measured from the left end and it is described by the function DM over the X equal to a X. Or a is units kilograms per meter script. The first part we wanted to determine its moment of inertia about a perpendicular access through its left hand, X equals zero using this equation. And second we want to know if the moment of inertia differs from the moment of inertia of the uniform spoke and how it differs. So we can first start off by writing the moment of inertia equation that is given to us. I is equal to the integral of R squared D. M. You're also given position packs measured from the left hand is described by the function G. M. Over D X is equal to X. So D. M. The ads to some constant K. Times. And we want to relate our two packs. R is simply just a shorthand for distance. So we know that our is simply just X. In this case. So we can make that substitution and we have this D. M. But M is a function of X. As we can see from this differential so we can sulfur dM from this differential equation. D. M is equal to a times X. D X. Now we can substitute D. M into this equation. We get that the inertia is simply just the integral of X squared times A X. D X. Further solve this, taking out a as a constant, integrating X cubes respect to x. And the The limits of integration for this problem are from Mexico 0 to the total length l. So we can do this simple integration and we get that The moment of Inertia is simply eight times X. The fourth over four evaluated from zero to L. So we get it that it is a times out the fourth over. We want the moment of inertia in terms of M and L. But we have this constant pain. We know that M is simply equal to the integral of all the small D. M elements. And we also have the so we can evaluate this total mass simply A X D X constant. So we get that M is equal to a times the integral of X D X. Just X squared over two. And we evaluate this likewise from zero to now you get that the total mass is a times L squared over two. We could rearrange this by getting that A is equal to M over L squared by two x or two. M. over all spring. You make the substitution into the equation for a to get it into terms of hell When we seven a we get the inertia is to em over sprayed Times out the 4th over four. This 4 to catch us out with this and this becomes a two in terms of that with this term we left With I is equal to 1/2 ml spread for a non uniform spoke. Now. In the second part, we recalled after a uniform spoke, The moment of inertia is one third ml squared, so we can see that the non uniform spoke is just one half M. L squared, and the uniform spoke is one third ml squared, so you can see that the non uniform one is going to be greater than the, you know, yes than the uniform. So the answer choice for this problem would be a one a half mile squared and that it is greater than a moment of inertia of a uniform spoke. Hope to help have a great day.

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