Question

The density of a rod $AB$ increases linearly from $A$ to $B$. Its midpoint is $O$ and its centre of mass is at $C$. Four axes pass through $A,B,O$ and $C$, all perpendicular to the length of the rod. The moments of inertia of the rod about these axes are $I_A$ , $I_B$, $I_O$, and $I_C$ respectively.

• A. $I_A>I_B$
• B. $I_A<I_B$
• C. $I_O>I_C$
• D. $I_O<I_C$

Answer 1

Answer: (A), (C)

The correct options are
A $I_A>I_B$
C $I_O>I_C$

$M={\int }_0^L\rho xdx=\frac{\rho L^2}{2}{I}_A={\int }_0^L{x}^2\rho xdx=\frac{\rho L^4}{4}{I}_B={\int }_0^L{x}^2\rho (L-x)dx=\frac{\rho L^4}{12}{I}_O={\int }_{-\frac{L}{2}}^{\frac{L}{2}}{x}^2\rho (\frac{L}{2}+x)dx=\frac{\rho L^4}{24}COM=\frac{{\int }_0^L\rho x.xdx}{{\int }_0^L\rho xdx}=\frac{2L}{3}{I}_C={\int }_{-\frac{2L}{3}}^{\frac{L}{3}}{x}^2\rho (2L/3+x)dx=\frac{\rho L^4}{36}$
Therefore, $I_A>I_B\&I_O>I_C$