Question

As shown in the figure discs of mass M and radius R are joined to the ends of a weightless rod of length 2R. The M.I. of the system about the perpendicular axis through its middle of the rod will be:

• A. $\frac{3}{2}{MR}^2$
• B. ${2MR}^2$
• C. $\frac{5}{2}{MR}^2$
• D. ${3MR}^2$

Answer 1

The correct option is C $\frac{5}{2}{MR}^2$

We know,

Moment of Inertia of a disk about its diameter$=\frac{M{R}^2}{4}$

$MoI$ of a disk $1$ about its diameter $=\frac{M{R}^2}{4}$

shifting it parallel by distance of R, $Mo{I}_1=\frac{5M{R}^2}{4}$

$MoI$ of a disk $2$ about its diameter $=\frac{M{R}^2}{4}$

shifting it parallel by distance of R, $Mo{I}_2=\frac{5M{R}^2}{4}$

Net moment of Inertia$=Mo{I}_1+Mo{I}_2=\frac{5M{R}^2}{4}+\frac{5M{R}^2}{4}=\frac{5M{R}^2}{2}$

Option $\text{C}$ is the correct answer