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Find a mass density function so the center of mass is at L/4

by RoRi

Find a mass density function so that \overline{x}, the center of mass of a rod of length L, is at a distance \frac{L}{4} from one end.

Let

    \[ \rho(x) = x^2 \quad \text{for} \quad 0 \leq x \leq L. \]

Then we compute the center of mass of the rod,

    \begin{align*} \overline{x} &= \frac{\int_0^L x \rho(x) \, dx}{\int_0^L \rho(x) \, dx} \\ &= \frac{\int_0^L x^3 \, dx}{\int_0^L x^2 \, dx } \\ &= \frac{L^4/4}{L^3/3} \\ &= \frac{3L}{4}. \end{align*}

Thus, the center of mass is a distance \frac{L}{4} from one end.

4 comments

  2. Waleed says:

    How did you know that the density should be x squared.. I mean is there a method for that or you have guessed it?

    • Anonymous says:

      L/4 from the end which mean 3L/4 from the start. then. 3L/4 = 3x^4/4x^3 from 0 to L=> int(x^3)/int(x^2)=int(x*p(x))/int(p(x)). =>p(x)=x^2.

      hope it helps

    • richardsull says:

      Probably the easiest way to guess that the density function should be x^2 is to look back to Exercise 20. In that exercise, the density function was given as x^2, and you calculate the center of mass as 3L/4, which is L/4 from one end of the rod. So, Apostol already gave you the solution, just two exercises earlier.

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