Question

Find the MOI of four identical solid spheres each of mass $M$ and radius $R$ placed in a horizontal plane as shown in figure. Line $X{X}^{\prime }$ touches two spheres and passes through the diameter of the other two spheres.

• A. $\frac{9}{10}M{R}^2$
• B. $\frac{5}{18}M{R}^2$
• C. $\frac{18}{5}M{R}^2$
• D. $\frac{9}{5}M{R}^2$

Answer 1

The correct option is C $\frac{18}{5}M{R}^2$

MOI of solid sphere about its COM is
$I=\frac{2}{5}M{R}^2$
So MOI for 1 and 2 sphere will be
$I_1=I_2=\frac{2}{5}M{R}^2$
MOI for sphere 3 and 4 along axis $X{X}^{\prime }$ by parallel axis theorem:
$I_4=I_3=\frac{2}{5}M{R}^2+M{R}^2$
$I_3=I_4=\frac{7}{5}M{R}^2$
Hence, the moment of inertia of the system will be
$I=I_1+I_2+I_3+I_4$
$I=2(I_1+I_3)$
$I=2(\frac{2}{5}M{R}^2+\frac{7}{5}M{R}^2)$
$I=\frac{18}{5}M{R}^2$