Question

If linear density of rod of length 3m varies as l=2+x then position of centre of gravity of rod is

Answer 1

total mass = density * length

that is

dm = lamda * dx

dm = (x+2)dx

integrating both sides,

we get,

m = x^2/2 + 2x

  = l^2/2 + 2l

put l = 3 meter.

m = 9/2 + 6 = 21/2 m that is approx. 10.5 m

Answer 2

Answer:

Position of Centre of Gravity of Rod = 1.71 m

Explanation:

[Concept Used; Centre of Mass]

(Refer to the attached image for the systematic representation of the case)

Given;-

            Length, L = 3m

            Density, λ = 2 + x

Let us assume at some x distance, for very small mass as dm, & for very small distance, dx. Then, dm = dx λ ____(1)

Now, we know by formula;-

            $Xcm = \frac{\int\limits^L_0 {x\, dm} }{\int\limits^L_0 {dm} }$

[where, Xcm represents centre of mass/gravity in x-axis]

            $Xcm = \frac{\int\limits^3_0 {x(2+x)\, dx} }{\int\limits^3_0 {x\, dx} }$

            $Xcm = \frac{\int\limits^3_0 {(2x + x^{2}) } \, dx }{\int\limits^3_0 {(2+x)} \, dx }$

            $Xcm = [\frac{\frac{2x^{2} }{2}+\frac{x^{3} }{3} }{2x + \frac{x^{2} }{4} }]^{3} \, to \, ^{0} .$

            $Xcm =\frac{2}{3} [\frac{3x^{2} + x^{3} }{4x + x^{2} } ] ^{3} \, to \, ^{0}$

            $Xcm = \frac{2}{3} [\frac{3x + x^{2} }{4 + x} ] ^{3} \, to \, ^{0}$

            $Xcm = \frac{2}{3} [\frac{18}{7} ]$

            $Xcm = 1.71 \, m$

Hope it helps! ;-))