Question

The center of gravity of a rod \[\left( {of{\text{ }}length{\text{ }}L} \right),\]whose linear mass density varies as the square of the distance from one end is at:
(A) \[\;\;\dfrac{{3L}}{5}\]
(B) $\dfrac{{2L}}{5}$
(C) \[\;\;\dfrac{L}{3}\]
(D) $\dfrac{{3L}}{4}$

Answer 1

Hint:In this question we use the concept of center of gravity and centre of gravity \[\left( {CG} \right)\] of an object is the point at which weight of the object is evenly distributed and all sides are in balance. Center of gravity is calculated by using the formula \[CG = \dfrac{{\int\limits_0^L {\rho xdx} }}{{\int\limits_0^L {\rho dx} }}\]

Complete step by step answer:
Centre of gravity is given by the formula,
\[CG = \dfrac{{\int\limits_0^L {\rho xdx} }}{{\int\limits_0^L {\rho dx} }}\], where ρ is the linear mass density. It is the mass per unit length \[.{\text{ }}X\] is the distance from its one end. L is the total length of the rod.
According to the question, ρ varies as the square of the distance from one end,
\[ \Rightarrow \rho = k{x^2}\] , where k is the proportionality constant.
Putting in equation \[\left( 1 \right)\] and integrating over the full length of the road.
\[CG = \dfrac{{\int\limits_0^L {k{x^3}dx} }}{{\int\limits_0^L {k{x^2}dx} }} = \dfrac{{\mathop {3{x^4}|}\nolimits_0^L }}{{\mathop {4{x^3}|}\nolimits_0^L }} = \dfrac{{3L}}{4}\]
The center of gravity of the rod, who’s linear mass density varies as the square of the distance from one end is at \[\dfrac{{3L}}{4}\].
Hence the answer is option \[\left( D \right)\]

Note: Centre of gravity \[\left( {CG} \right)\] of an object is the point at which weight of the object is evenly distributed and Centre of mass \[\left( {CM} \right)\] is the point where the whole mass of the body is concentrated. Center of gravity and Centre of mass are the same only when there is a uniform gravitational field. Center of gravity of a human being can change as he takes different positions.