Question

Linear mass density of rod of length l is directly proportional to x cube where x is the distance from one end of rod.center of mass of rod lies at a distance

Answer 1

Answer:

4/5 l

Explanation:

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Answer 2

centre of mass of continuous distributing object is given by, $\bf{C.M=\frac{\int{x}\,dm}{\int \,dm}}$

given, linear mass density of the rod, λ is directly proportional to x³.

i.e., λ = kx³

so, elementary mass , dm = λ dx

= kx³ dx

now, centre of mass , C.M = $\frac{\int\limits^l_0{x.kx^3}\,dx}{\int\limits^l_0{kx^3}\,dx}$

= $\frac{k}{k}\frac{\int\limits^l_0{x^4}\,dx}{\int\limits^l_0{x^3}\,dx}$

= $\frac{\left[\frac{x^5}{5}\right]^l_0}{\left[\frac{x^4}{4}\right]^l_0}$

= $\frac{4l^5}{5l^4}$

= 4l/5

hence, centre of mass of rod lies at a distance 4l/5.