Question

Four particles each of mass m are lying symmentrically on rim of disc of mass M and radius R. Moment of inertia of this system about an axis passsing through one of the particles and $\perp$ to plane of disc is

• A. $\frac{M{R}^2}{2}+4M{R}^2$
  
• B. $\frac{M{R}^4}{2}+4M{R}^2$
• C. $\frac{M{R}^3}{2}+4M{R}^2$
• D. $\frac{M{R}^5}{2}+4M{R}^2$

Answer 1

The correct option is A $\frac{M{R}^2}{2}+4M{R}^2$

The axis passes through centre $C$ perpendicular to plane of the disk moment of inertia ($I_1$) of disc about axis is:

$I_1=\frac{M{R}^2}{2}$

Moment of inertia of mass '$m$' lying at distance $R$ from axis is given by:

$I_2=m{R}^2$ (for each m)

So, net moment of inertia about axis would be:

$I=I_1+4I_2$

$I=\frac{m{R}^2}{2}+4m{R}^2$