Question

Three identical solid spheres each have mass $‘m’$ and diameter $‘d’$ are touching each other as shown in the figure. Calculate ratio of moment of inertia about the axis perpendicular to plane of paper and passing through point $P$ and $B$ as shown in the figure. Given $P$ is centroid of the triangle.

• A. $\frac{13}{23}$
• B. $\frac{13}{15}$
• C. $\frac{15}{13}$
• D. $\frac{23}{13}$

Answer 1

The correct option is A

$\frac{13}{23}$

Step 1. Given Data,

$r=\frac{d}{2}$, $d$ is the diameter touching each other.

Step 2. Formula used,

Moment of inertia (MI) of one sphere about $P$,

$I=I_{CM}+{\left(PB\right)}^2\times m$

$PB$ is the distance between $P$ and $B$
The moment of inertia of the sphere with respect to an axis passing through its center is .$\frac{2}{5}m{r}^2$

.$I_{CM}=\frac{2}{5}m{r}^2$

$m$ is the mass of the solid sphere, $r$ is the distance of the particle from the rotational axis.

Step 3. Calculating the ratio of the moment of inertia,

The moment of inertia of the sphere with respect to an axis passing through its center is$\frac{2}{5}m{r}^2$.

$I_{CM}=\frac{2}{5}m{r}^2$ , $r=\frac{d}{2}$, $d$ is the diameter touching each other

$I_{CM}=\frac{2}{5}m{\left(\frac{d}{2}\right)}^2$

$I_{CM}=\frac{m{d}^2}{10}$

$$\begin{gathered} PB=\frac{2}{3}\left(BC\times cos30\right)=\frac{2}{3}\times d\times \frac{\sqrt{3}}{2} \\ PB=\frac{d}{\sqrt{3}} \end{gathered}$$,$BC$ is the distance between $B$ and $C$

Now, the Moment of inertia (MI) of one sphere about $P$,

$I=I_{CM}+{\left(PB\right)}^2\times m$

Putting the values we get,

$$\begin{gathered} I=\frac{m{d}^2}{10}+m{\left(\frac{d}{\sqrt{3}}\right)}^2 \\ I=m{d}^2\left(\frac{1}{10}+\frac{1}{3}\right) \\ SoTotalMIaboutP, \\ {I}_P=3m{d}^2\left(\frac{1}{10}+\frac{1}{3}\right) \\ {I}_P=3m{d}^2\left(\frac{3+10}{30}\right) \\ {I}_P=3m{d}^2\left(\frac{13}{30}\right) \\ {I}_P=\left(\frac{13}{10}\right)m{d}^2 \end{gathered}$$

Moment of Inertia about $B$,

$$\begin{gathered} I_B=\frac{2}{5}m{\left(\frac{d}{2}\right)}^2+2\left[\frac{2}{5}m{\left(\frac{d}{2}\right)}^2+m{d}^2\right] \\ {I}_B=\frac{6}{5}m{\left(\frac{d}{2}\right)}^2+2m{d}^2 \\ {I}_B=\frac{3}{10}m{d}^2+2m{d}^2=\frac{23}{10}m{d}^2 \end{gathered}$$

Ratio,$\frac{I_P}{I_B}=\frac{{\displaystyle \frac{13}{10}}m{d}^2}{{\displaystyle \frac{23}{10}}m{d}^2}=\frac{13}{23}$

Hence the moment of inertia about the axis perpendicular to the plane of the paper and passing points point $P$ and $B$ is $\frac{13}{23}$.

Hence option A is the correct answer.