Discussion Forum

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 $N \times 10^{-4}$ kg-m² then N is

Answer:

Explanation:

 $r= \frac{d}{2}=\frac{\sqrt{5}}{2}$ cm

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

 $m=0.5 kg$

 a=4cm

 =$4 \times 10^{-2}$ cm

 $I_{XX}=I_{1}+I_{2}+I_{3}+I_{4}$

$=\left[\frac{2}{5}mr^{2}+m\left(\frac{a}{\sqrt{2}}\right)^{2}\right]+\frac{2}{5}mr^{2}$

                   +$\left[\frac{2}{5}mr^{2}+m\left(\frac{a}{\sqrt{2}}\right)^{2}\right]+\frac{2}{5}mr^{2}$

 Substituting the values , we get

 $I_{XX}=9 \times 10^{-4} Kgm^{-2}$

 $\therefore$  N=9

 answer is 9