Question

Centre of mass of a uniform rod of length l whose mass per unit length room varies as row equals to k x square upon l where k is and constant and x is a distance of any point from one is

Answer 1

Centre of mass is given by , x = $\frac{\int{x}\,dm}{\int dm}$

here it is given that,

mass per unit length , $\rho=\frac{k}{l}x^2$

so, dm = $\rho dx$

= $\frac{k}{l}x^2dx$

So, centre of mass , x = $\frac{\frac{k}{l}\int\limits^l_0{x^3}\,dx}{\frac{k}{l}\int\limits^l_0{x^2}\,dx}$

= $\frac{\left[\frac{x^4}{4}\right]^l_0}{\left[\frac{x^3}{3}\right]^l_0}$

= $\frac{\frac{l^4}{4}}{\frac{l^3}{3}}$

= $\frac{3l}{4}$

hence, answer is 3l/4

Answer 2

Answer:

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