Question

If the linear density of a rod of length L varies as lambda = A + Bx, determine the position of its centre of mass.
(where x is the distance from one of its ends)

Answer 1

centre of mass is (3AL + 2BL²)/(6A + 3BL)

let us consider small mass of the rod of small length dxdx , therefore the small mass will be given in differential form as $dm=\lambda dx$

so, $\int\limits^M_0\,{dm}=\int\limits^L_0{A+B x}\,dx$

or, $M=A L+\frac{1}{2}B$

now, centre of mass of rod is given by,

$R_{cm}=\frac{1}{M}\int\limits^L_0{\lambda x}\,dxR$

=$\frac{1}{M}\int\limits^L_0{A x+B x^2}\,dx$

=$\frac{1}{M}\left(\frac{1}{2}AL^2+\frac{1}{3}B L^3\right)$

=$\frac{\left(\frac{1}{2}A L^2+\frac{1}{3}B L^3\right)}{\left(A L+\frac{1}{2}B L^2\right)}$

hence center of mass of rod is $R_{cm}$= $\frac{\left(3A L+ 2B L^2\right)}{(6A +3B L)}$

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Answer 2

$\huge\bold\purple{Answer:-}$

let us consider small mass of the rod of small length dxdx , therefore the small mass will be given in differential form as dm=\lambda dx

so, \int\limits^M_0\,{dm}=\int\limits^L_0{A+B x}\,dx

or, M=A L+\frac{1}{2}B

now, centre of mass of rod is given by,

R_{cm}=\frac{1}{M}\int\limits^L_0{\lambda x}\,dxR

=\frac{1}{M}\int\limits^L_0{A x+B x^2}\,dx

=\frac{1}{M}\left(\frac{1}{2}AL^2+\frac{1}{3}B L^3\right)

=\frac{\left(\frac{1}{2}A L^2+\frac{1}{3}B L^3\right)}{\left(A L+\frac{1}{2}B L^2\right)}

hence center of mass of rod is R_{cm}= \frac{\left(3A L+ 2B L^2\right)}{(6A +3B L)}