2002 AIPMT System of Particles and Rotational MotionCentre of Mass of a Non-uniform Rod Medium Question 97 of 100 in System of Particles and Rotational Motion

A rod of length is 3m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at : -

A 1.5 m
B 2 m
C 2.5 m
D 3.0 m

The answer is B.

Solution

Mass per unit length: $\lambda = kx$, where k is a constant and x is measured from the given end O.
Mass of a small element of length dx: $dm = kx\,dx$
$x_{cm} = \dfrac{\int x\,dm}{\int dm} = \dfrac{\int_0^3 x(kx)\,dx}{\int_0^3 kx\,dx} = \dfrac{\left[\dfrac{x^3}{3}\right]_0^3}{\left[\dfrac{x^2}{2}\right]_0^3}$
$= \dfrac{27/3}{9/2} = \dfrac{9}{4.5} = 2$ m
So the centre of gravity is 2 m from that end (closer to the heavier end).

Concept Centre of mass by integration

System of Particles and Rotational Motion in past papers

100 questions from this chapter have appeared across 27 exam years.