A thin bar of length \(L\) has a mass per unit length \(\lambda \), that increases linearly with distance from one end. If its total mass is \(M\) and its mass per unit length at the lighter end is \(\lambda_0\), then the distance of the centre of mass from the lighter end is

\(\mathrm{M}=\int_{0}^{\mathrm{L}}\left(\lambda_{0}+\mathrm{kx}\right) \mathrm{dx}\)

\(M=\lambda_{0} L+\frac{K \times L^{2}}{2}\)

\(\frac{2 \mathrm{M}-\lambda_{0} \mathrm{L}}{\mathrm{L}^{2}}=\mathrm{K}\)

\(\frac{2 \mathrm{M}}{\mathrm{L}^{2}}-\frac{\lambda_{0}}{\mathrm{L}}=\mathrm{K}\)

\(\frac{\int \mathrm{d} \mathrm{m}(\mathrm{r})}{\int \mathrm{d} \mathrm{m}}=\frac{\int(\lambda \mathrm{d} n) \mathrm{x}}{\mathrm{M}}=\frac{\int_{0}^{\mathrm{L}}\left(\lambda_{0} \mathrm{x}+\mathrm{k} \mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}}{\mathrm{M}}\)

\(r_{c m}=\frac{\lambda_{0} L+\frac{k L^{2}}{2}}{M}\)

substitute \('k"\)

\(r_{c m}=\frac{2 L}{3}-\frac{\lambda_{0} \ell^{2}}{6 M}\)