A rod of length L has non-uniform linear mass density given by (x)=a+b (x/L)2, where a and b are constants and 0 x L. The value of x for the centre of mass of the rod is at

(A) \frac{4}{3}\left(\frac{a+b}{2 a+3 b}\right) L

(B) \frac{3}{2}\left(\frac{a+b}{2 a+b}\right) L

(C) \frac{3}{2}\left(\frac{2 a+b}{3 a+b}\right) L

JEE Mains · Physics · STD 11 - System of particles and rotational motion

A rod of length \(L\) has non-uniform linear mass density given by \(\rho(\mathrm{x})=\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{L}}\right)^{2},\) where \(a\) and \(\mathrm{b}\) are constants and \(0 \leq \mathrm{x} \leq \mathrm{L}\). The value of \(\mathrm{x}\) for the centre of mass of the rod is at

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Correct Answer

(D) \(\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) L\)

Detailed explanation

\(\mathrm{x}_{\mathrm{cm}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}=\frac{\int(\lambda \mathrm{dx}) \mathrm{x}}{\int \mathrm{dm}}\)

\(=\frac{\int_{0}^{L}\left(a+\frac{b x^{2}}{L^{2}}\right) x d x}{\int_{0}^{L}\left(a+\frac{b x^{2}}{L^{2}}\right) d x}\)

\(=\frac{\frac{a L^{2}}{2}+\frac{b}{L^{2}} \cdot \frac{L^{4}}{4}}{a L+\frac{b}{L^{2}} \cdot \frac{L^{3}}{3}}\)

\(=\frac{\left(\frac{4 a+2 b}{8}\right) L}{\frac{(3 a+b)}{3}}=\frac{3}{4} \frac{(2 a+b) L}{(3 a+b)}\)

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