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System of Particles and Rotational Motion
Centre of Mass of a Non-uniform Rod
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Concepts tested
- Centre of mass by integration
All Questions
2002 AIPMT 1 question
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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 : -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).