Question

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

Answer 1

linear mass density of the rod is given by

$\lambda = \alpha + \beta x$

now we have

$\frac{dm}{dx} = \alpha + \beta x$

integrate both sides

$\int dm = \int \alpha + \beta x dx$

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

so above is the total mass of rod

now by the formula of center of mass

$r_{cm} = \frac{1}{M}\int xdm$

$r_{cm} = \frac{1}{M} \int x*(\alpha + \beta x) dx$

$r_{cm} = \frac{1}{M} (\alpha*\frac{L^2}{2} + \beta*\frac{L^3}{3})$

now plug in value of mass of rod

$r_{cm} = \frac{(\alpha*\frac{L^2}{2} + \beta*\frac{L^3}{3})}{( \alpha L+\frac{ \beta L^2}{2})}$

so above is the position of center of mass of rod

Answer 2

Answer:

al^5/5+bl^3/3