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(B) For a continuous body,the centre of mass $X_{cm}$ is given by the integral formula: $X_{cm} = \frac{1}{M} \int x \, dm$ Since the rod is uniform,i

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$L/2$

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(B) For a continuous body,the centre of mass $X_{cm}$ is given by the integral formula: $X_{cm} = \frac{1}{M} \int x \, dm$ Since the rod is uniform,its linear mass density $\lambda$ is constant,where $\lambda = \frac{M}{L}$. For a small element of length $dx$ at a distance $x$ from the origin,the mass is $dm = \lambda \, dx = \frac{M}{L} dx$. Substituting this into the integral: $X_{cm} = \frac{1}{M} \int_{0}^{L} x \left( \frac{M}{L} \right) dx$ $X_{cm} = \frac{1}{L} \int_{0}^{L} x \, dx$ $X_{cm} = \frac{1}{L} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{1}{L} \left( \frac{L^2}{2} - 0 \right) = \frac{L}{2}$. Thus,the centre of mass is at $L/2$.

$A$ uniform thin rod of length $L$ and mass $M$ is placed along the $x$-axis with one end at the origin. Find the position of the centre of mass of the rod.

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