Question

Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. The moment of inertia of this system about an axis passing through one of the particle and perpendicular to plane of disc is

• A. $16mR^2$
• B. $\left(3M+16m\right) \frac{R^{2}}{2}$
• C. $\left(3m+12M\right) \frac{R^{2}}{2}$
• D. zero

Answer 1

Answer: (B)

According to the theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc is
$=\frac{1}{2}MR^{2}+MR^{2}=\frac{3}{2}MR^{2}$
Total moment of inertia of the system
$=\frac{3}{2}MR^{2}+m\left(2R\right)^{2}+m\left(\sqrt{2}R\right)^{2}+m\left(\sqrt{2}R\right)^{2}$
$=\left(3M+16m\right) \frac{R^{2}}{2}$