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(B) The linear mass density is given by $\lambda(x) = ax$.Consider a small element of length $dx$ at a distance $x$ from the origin.The mass of this e

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$2L/3$

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(B) The linear mass density is given by $\lambda(x) = ax$.Consider a small element of length $dx$ at a distance $x$ from the origin.The mass of this element is $dm = \lambda(x) dx = ax dx$.The total mass $M$ of the rod is the integral of $dm$ from $0$ to $L$:$M = \int_{0}^{L} ax dx = a [x^2/2]_{0}^{L} = aL^2/2$.The position of the centre of mass $X_{cm}$ is given by:$X_{cm} = \frac{1}{M} \int_{0}^{L} x dm = \frac{1}{M} \int_{0}^{L} x (ax dx) = \frac{a}{M} \int_{0}^{L} x^2 dx$.$X_{cm} = \frac{a}{aL^2/2} [x^3/3]_{0}^{L} = \frac{2}{L^2} \cdot \frac{L^3}{3} = \frac{2L}{3}$.Thus,the centre of mass is at $2L/3$.

$A$ uniform rod of length $L$ is placed along the $x$-axis with one end at the origin. If the linear mass density of the rod varies as $\lambda(x) = ax$,where $a$ is a constant,find the position of the centre of mass of the rod.

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