Question

The linear mass density( ) of a rod of length l kept along x-axis varies as = + x; where and are positive constants. The centre of mass of the rod is atchoose answer:

Answer 1

your question is -----> The linear mass density lambda of a rod of length L kept along x- axis varies as $\lambda=\alpha+\beta x$ ; where alpha and beta are positive constants. The centre of mass of the rod is at.

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

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

or, $M=\alpha L+\frac{1}{2}\beta L^2$

now, centre of mass of rod is given by,

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

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

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

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

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