Four solid spheres each of diameter and mass 0.5 kg are placed with their centres at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square isN × 10

Engineering

Physics

Moment of Inertia

Question

Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centres at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is
N × 10^–4 kg-m², then N is

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$r = \frac{\sqrt{5}}{2} cm = \frac{\sqrt{5}}{2} \times 10^{- 2} m$

$m = \frac{1}{2} kg$

$l$ = 4 × 10^–2 m

Using parallel axis theorem

$I_{total} = [4 \times \frac{2}{5} \times \frac{1}{2} \times \frac{5}{4} \times 10^{- 4}] + [2 \times \frac{1}{2} \times 8 \times 10^{- 4}]$

$\Rightarrow$ 10^–4 + 8 × 10^–4 $\Rightarrow$ 9 × 10^–4 kg m²