Question

Four identical spheres of masses 2.0 kg each and radius 0.25 m are situated at the four corners of a square. One side of the square measures 3.00 m. Find the moment of inertia about an axis. passing through one of the masses and perpendicular to its plane

Answer 1

Explanations & Calculations

• The moment of inertia should be considered accordingly their density either hollow or solid.
• Let's take them to be solid.
• M.O.I about the centre of a sphere is $\small \frac{2}{5}mr^2$
• When the axis is placed outside the centre, then the inertia is calculated as $\small I_{cc}+mx^2$ where $\small x$ is the displacement from the centre.

• Then the moment inertia of the system will be

$\qquad\qquad \begin{aligned} \small I&=\small\Sigma mr^2\\ &=\small mr^2+(mr^2+mx^2)+(mr^2+mx^2)+(mr^2+m[\sqrt2 x]^2)\\ &=\small 4mr^2+3mx^2+2mx^2\\ &=\small m(4r^2+5x^2)\\ &=\small 2.0\,kg(4\times0.25^2+5\times3.0^2)\\ &=\small 90.5\,kgm^2 \end{aligned}$