The centre of mass of a non-uniform rod of le | vedclass

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(A) The mass of a small element of length $dx$ at a distance $x$ from the end is $dm = \lambda dx = \frac{k x^3}{L^3} dx$.The position of the centre o

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$\frac{4}{5} L$

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(A) The mass of a small element of length $dx$ at a distance $x$ from the end is $dm = \lambda dx = \frac{k x^3}{L^3} dx$.The position of the centre of mass $X_{cm}$ is given by the formula:$X_{cm} = \frac{\int x dm}{\int dm}$Substituting the expression for $dm$:$X_{cm} = \frac{\int_0^L x (\frac{k x^3}{L^3} dx)}{\int_0^L (\frac{k x^3}{L^3} dx)}$$X_{cm} = \frac{\frac{k}{L^3} \int_0^L x^4 dx}{\frac{k}{L^3} \int_0^L x^3 dx}$$X_{cm} = \frac{[\frac{x^5}{5}]_0^L}{[\frac{x^4}{4}]_0^L} = \frac{L^5 / 5}{L^4 / 4} = \frac{4}{5} L$

The centre of mass of a non-uniform rod of length $L$ whose mass per unit length $\lambda$ varies as $\lambda = \frac{k x^3}{L^3}$ (where $k$ is a constant and $x$ is the distance of any point on the rod from one end) is at a distance from the same end equal to:

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