Looking for classes? Ksquare Career Institute, Bengaluru →
Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis XX' which is touching two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX' axis is
A
$\frac{11}{5}mr^2$
B
$3mr^2$
C
$\frac{16}{5}mr^2$
D
$4mr^2$
Detailed Solution
About a diameter of a thin spherical shell: $I = \frac{2}{3}mr^2$, so $I_1 = \frac{2}{3}mr^2$.
About a tangent (parallel axis theorem): $I = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2$, so $I_2 = I_3 = \frac{5}{3}mr^2$.
$I_{XX'} = \frac{2}{3}mr^2 + \frac{5}{3}mr^2 + \frac{5}{3}mr^2 = 4mr^2$
About a tangent (parallel axis theorem): $I = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2$, so $I_2 = I_3 = \frac{5}{3}mr^2$.
$I_{XX'} = \frac{2}{3}mr^2 + \frac{5}{3}mr^2 + \frac{5}{3}mr^2 = 4mr^2$