Question

If moment of inertia of a point particle at a distance r from an axis would have defined as $L=mr$ instead of L=${mr}^2$, then moment of inertia of a uniform rod of mass M and length L about an axis passing through centre of mass and perpendicular to the rod will be (moment of inertia is still a scalar )

• A. $\frac{ML}{12}$
• B. $\frac{{ML}^2}{12}$
• C. $\frac{ML}{4}$
• D. Zero

Answer 1

Answer: (C)

$\frac{ML}{4}$

$\begin{array}{c}\text{Let a small element on the rod of Length}\left(dx\right)\text{at}\\ \text{a distance}x\text{from the centre-}\\ \text{mass of the element will be}=\frac{m\cdot dx=dm}{L}dx=dm\\ \text{Moment of Inertia of mall element will be}\\ dI=dm\cdot r=(\frac{m}{L}\cdot dx)\cdot \left(x\right)\end{array}$

$\begin{array}{c}\Rightarrow dI=\frac{m}{L}\cdot x\cdot dx\\ \text{Integrating both Sides with limits -}\\ \begin{array}{c}\text{If}\\ {\int }_0^{L}dI={\int }_{-L/2}^{L/2}\frac{m}{L}\cdot x\cdot dx=2{\int }_0^{L/2}\frac{m}{L}\cdot x\cdot dx\text{(Scalar)}\\ I={2\frac{m}{L}\frac{x^2}{2}|}_0^{L/2}=\frac{m}{L}\cdot \left(\frac{L^2}{4}\right)=\frac{mL}{4}\end{array}\end{array}$