Question

A rod of length is 3 m 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

Answer 1

Answer: (B)

2 m

Answer 2

Here $\rho= kx$ where $k$ is a constant mass of small element of $dx$ length is
$dm = kx \cdot dx$
$x _{ cm }=\frac{\int x . dm }{\int dm }=\frac{\int_{0}^{3} x ( x dx )}{\int_{0}^{3} x \cdot d x }=\frac{\left[\frac{ x ^{3}}{3}\right]_{0}^{3}}{\left[\frac{ x ^{2}}{2}\right]_{0}^{3}}=\frac{\frac{27}{3}}{\frac{9}{2}}=2$