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Tyro
Asked: 2021-03-30T12:27:02+00:00 In: ,

Integral Calculus: Center of Mass of A Rod

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The total mass of a rod 5 meters long is 33 kg. If the linear density varies directly as the cube of the distance of the point from one end, find the linear density at the midpoint of the rod.

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  2. acedstud
    This answer was edited.

    Since the linear density, \lambda is directly proportional to the cube of the distance from one end, let the distance from one end be x, then \lambda\prop x^3\Rightarrow\lambda=kx^3, where k is a constant of proportionality.

    Now, the total mass of the rod is given by M=\int_0^l{\lambda}dx, where l is the length of the rod.

    Thus, \int_0^5{kx^3}dx=33\Rightarrow k\left[\dfrac{x^4}{4}\right]_0^5=33

    \Rightarrow\dfrac{5^4}{4}k=33\Rightarrow156.25k=33\Rightarrow k=\dfrac{33}{156.25}=0.2112.

    Therefore, at the midpoint, x=2.5, thus \lambda=kx^3=0.2112(2.5)^3=3.3

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