Question

The linear mass density of a ladder of length $l$ increases uniformly from one end $A$ to the other end $B$,
(i) Form an expression for linear mass density as a function of distance $x$ from end $A$ where linear mass density ${\lambda }_0$. The density at one end being twice that of the other end.
(ii) Find the position of the centre of mass from end $A$.

Answer 1

Let $\lambda$ be the linear mass density

$\lambda =$ Ax+B,

It is given that at $x=0,\lambda ={\lambda }_o$

and at $x=l,\lambda =2{\lambda }_o$

using these boundary conditions we get,

$A=\frac{{\lambda }_o}{l},B={\lambda }_o$

$⟹\lambda =\frac{{\lambda }_o}{l}x+{\lambda }_o$

we know that,

$X_{cm}=\frac{{\int }_0^{l}xdm}{{\int }_0^{l}dm},$ where $dm=\lambda dx$

$X_{cm}=\frac{{\int }_0^{l}x(\frac{{\lambda }_{o}x}{l}+{\lambda }_o)dx}{{\int }_0^l(\frac{{\lambda }_o}{l}x+{\lambda }_o)dx}=\frac{5l}{9}$