Question

The centre of gravity (centre of mass) of a rod (of length $L$) whose linear mass density varies as the square of the distance from one end is at

• A. $\frac{L}{3}$
• B. $\frac{3L}{4}$
• C. $\frac{3L}{5}$
• D. $\frac{2L}{5}$

Answer 1

Answer: (B)

Let $A$ be the area of cross section of the rod.
Then, $\bar{x}=\frac{{\int }_0^{L}x\cdot k{x}^{2}adx}{{\int }_0^{L}k{x}^2\cdot adx}=\frac{{\int }_0^L{x}^{3}dx}{{\int }_0^L{x}^{2}dx}=\frac{\frac{L^4}{4}}{\frac{L^3}{3}}=\frac{3}{4}L$