Question 597: Centre of Mass and Rotational Motion

Question

Three points masses, each of mass $m$ are placed at the corners of an equilateral triangle of side $\ell$. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining other two vertices, will be

Accepted Answer

The given situation is as shown below
We have to find the moment of inertia about axis $X X^{\prime}$.
The distance of line $X X^{\prime}$ from base of triangle is
$$d=\sqrt{I^2-\frac{l^2}{4}}=\frac{\sqrt{3} I}{2}$$
$	herefore$ Moment of inertia, due to each mass,
$$\begin{aligned}
I & =m 	imes 0+m\left(\frac{\sqrt{3}}{2} Iight)^2+m\left(\frac{\sqrt{3}}{2} Iight)^2 \
& =\frac{3 m l^2}{4}+\frac{3 m l^2}{4} \
& =\frac{6 m l^2}{4}=\frac{3 m l^2}{2}
\end{aligned}$$

Answer

$\frac{3}{4} m \ell^2$

Answer

$\frac{1}{4} m \ell^2$

Answer

$\frac{3}{2} m \ell^2$

Answer

$\frac{1}{2} m \ell^2$

Centre of Mass and Rotational Motion

Three points masses, each of mass m are placed at the corners of an equilateral triangle of side \ell. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining other two vertices, will be

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