Question

Seven identical circular planar disks, each of mass $M$ and radius $R$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:

• A. $\begin{array}{l}\frac{73}{2}M{R}^2\end{array}$
• B. $\begin{array}{l}\frac{181}{2}M{R}^2\end{array}$
• C. $\begin{array}{l}\frac{19}{2}M{R}^2\end{array}$
• D. $\begin{array}{l}\frac{55}{2}M{R}^2\end{array}$

Answer 1

The correct option is B

$\begin{array}{l}\frac{181}{2}M{R}^2\end{array}$

Step 1: Given Data:

Number of disks$=7$

Mass of each disk$=M$

Radius of each disk$=R$

Step 2: Formula used:

Moment of inertia of a uniform circular plate about its axis-

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

According to parallel axis theorem-

$I_o=I_{cm}+M{d}^2$

$d$ is the distance between two axis

Step 3: Calculation of the moment of inertia of the arrangement about the axis normal to the plane and passing through the point P:

Using parallel axis theorem-

The moment of inertia of the arrangement about the axis normal to the plane-

$$\begin{gathered} \begin{array}{l}{I}_0=\frac{7M{R}^2}{2}+6\left[M\right(2R{)}^2]\end{array} \\ \begin{array}{l}{I}_0=\frac{7M{R}^2}{2}+24M{R}^2\end{array} \\ \begin{array}{l}{I}_0=\frac{55}{2}M{R}^2\end{array} \end{gathered}$$

The moment of inertia about an axis passing through the point P

$$\begin{gathered} \begin{array}{l}{I}_p=I_0+7M(3R{)}^2\end{array} \\ \begin{array}{l}{I}_p=\frac{55}{2}M{R}^2+63M{R}^2\end{array} \\ \begin{array}{l}{I}_p=\frac{181}{2}M{R}^2\end{array} \end{gathered}$$

Hence option B is the correct answer.