The mass per unit length of a non-uniform rod of length L varies as $m = m_{0} \frac{X^{2}}{L}$ , where $m_{0}$ is a constant and X is the distance of any point on the rod measured from one end. The centre of mass of the rod is at $\frac{PL}{4}$ where P is

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Detailed Solution

$m = m_{0} \frac{X^{2}}{L} X_{CM} = \frac{\int_{0}^{L} (mdx) x}{\int_{0}^{L} mdx} = \frac{\int_{0}^{L} X (m_{0} \frac{X^{2}}{L}) dx}{\int_{0}^{L} (m_{0} \frac{X^{2}}{L}) dx} = {(\frac{\frac{X^{4}}{4}}{\frac{X^{3}}{3}})}_{0} = \frac{3 L}{4}$

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