Question

Moment of inertia of a uniform hexagonal plate about an axis $L L$' is ' $I^{\prime}$ as shown in the figure-$5.117$. The moment of inertia (about axis $X X$ ) of an equilateral uniform triangular plate of thickness double that of the hexagonal plate is (Ratio of specific gravity $\frac{\rho_{t}}{\rho_{h}}=3$ ):

• A. $\frac{I}{5}$
• B. $\frac{I}{10}$
• C. $I$
• D. Zero

Answer 1

Answer: (A)

We divide hexgon in six equilateral triangles by dotted lines as shown in figure and added two more triangles to it to form a diamond. If moment of inertia of triangle about $X X^{\prime}$ is $I_{1}$ then we use
$I+2 I_{1}=2\left[I_{1}(m \leftarrow 4 m\, \& \,l \leftarrow 2 l)\right]$
$I+2 I_{1}=32 I_{1}$
$\Rightarrow I_{1}=\frac{I}{30}$
Now as density of triangular plate is 3 times and thickness is double its moment of inertia will be
$I_{t}=I_{1} \times 3 \times 2=\frac{I}{5}$