The density of a linear rod of length L varies as
$ρ$ = A + Bx where x is the distance from the left end.
Locate the centre of mass.

$\frac{2 A L + 3 B L^{2}}{2 (3 A + B L)}$

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$\frac{3 A L + 4 B L^{2}}{2 (3 A + 2 B L)}$

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$\frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

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$\frac{3 A L + B L^{2}}{(2 A + B L)}$

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Solution

The correct option is C
$\frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

Let the cross-sectional area be $α$ . the mass of an element of length dx located at a distance x away from the left ends is (A + Bx) dx. The x-coordinate of the centre of mass is given by
$X_{c m} = \frac{\int x d m}{\int d m} = \frac{\int_{O}^{L} x (A + B x) a d x}{\int_{O}^{L} (A + B x) a d x}$
$= \frac{A \frac{L^{2}}{2} + B \frac{L^{3}}{3}}{A L + B \frac{L^{2}}{2}} = \frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

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Similar questions

The density of a linear rod of length L varies as
$ρ$ = A + Bx where x is the distance from the left end.
Locate the centre of mass.

L

ρ = A + B x

x

ρ

[X_{C M} = \frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}]

L

ρ = A + B x

x

L = 1 m

ρ = A + B x

A = 10^{4} {kg/m}^{3} & B = 10^{3} {kg/m}^{2}

x

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