Question

Out of the following bodies of same mass, which one will have maximum moment of inertia about an axis passing through its center of gravity and perpendicular to its place?

• A. ring of radius $r$
• B. disc of radius $r$
• C. square frame of sides $2r$
• D. square lamina of side $2r$

Answer 1

Answer: (C)

square frame of sides $2r$

$I_{ring}=m{r}^2$.
$I_{disc}=\frac{1}{2}m{r}^2$
$I_{squarelamina}=\frac{1}{12}\left(\right(2r{)}^2+\left(2r{)}^2\right)=\frac{2}{3}m{r}^2$
Square frame:
We consider each side of the frame as a thin rod of mass $\frac{m}{4}$. Thus moment of inertia about the center using parallel axis theorem $\frac{1}{12}(\frac{m}{4}\left(2r{)}^2\right)+\frac{m}{4}{r}^2=\frac{m{r}^2}{3}$. Multiplying this value by 4 for 4 rods we get the highest value of moment of inertia for square frame as $\frac{4}{3}m{r}^2$
$I_{squareframe}=\frac{4}{3}m{r}^2$
Thus, square frame of side $2r$ has the highest moment of inertia.