Moment Of Inertia Of A Solid Sphere

Moment of Inertia of a Solid Sphere

Let I = Moment of inertia of a solid sphere about an axis through its centre

To calculate I

Consider a sphere of mass M, radius R and centre O. Mass per unit volume of the sphere = $\rho=\frac{M}{\frac{4}{3} \pi R^3}$

Take an elementary disc of mass dm, whose centre is C and which lies between two planes perpendicular to the axis at a distance x and x + dx from its centre

Disc has Radius=AC and thickness =dx.

As shown in the figure

So, $A C=\sqrt{R^2-x^2}$
So, $d V=\pi\left(R^2-x^2\right) d x$
$
d m=\rho d V=\frac{3 M}{4 \pi R^3} * \pi\left(R^2-x^2\right) d x=\frac{3 M}{4 R^3} *\left(R^2-x^2\right) d x
$

The moment of inertia of the elementary disc about the axis:

$d I=\int \frac{(A C)^2 d m}{2} \quad\left(\because I_{\text {disc }}=\frac{m R^2}{2}\right)$

Now integrate this dI between the limits x=-R to x=+R

$\begin{aligned}
& I=\int d I=\int \frac{(A C)^2 d m}{2} \\
& =\int_{-R}^R\left(R^2-x^2\right) \frac{3 M}{8 R^3}\left(R^2-x^2\right) d x \\
& =\int_{-R}^R \frac{3 M}{8 R^3}\left(R^2-x^2\right)^2 d x \\
& =\frac{3 M}{8 R^3} \int_{-R}^R\left(R^4-2 R^2 x^2+x^4\right) d x \\
& \Rightarrow \mathbf{I}=\frac{2}{5} \mathrm{MR}^2
\end{aligned}$

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Solved Examples Based on Moment of Inertia of a Solid Sphere

Example 1: Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure) The moment of inertia (MR²) of the system about the axis passing perpendicularly through the centre of the rod is :

1) 10.13

2) 9.13

3) 13.93

4) 1.13

Solution:

Moment of inertia for a solid sphere

$
I=\frac{2}{5} M R^2
$

About a diameter.
$
\begin{aligned}
I_{\text {ball }} & =\frac{2}{5} M R^2+M(2 R)^2 \\
& =\frac{22}{5} M R^2
\end{aligned}
$
2 balls are there
$
\begin{aligned}
& \therefore \frac{44}{5} M R^2=I_{\text {balls }} \\
& I_{\text {rod }}=\frac{M(2 R)^2}{12}
\end{aligned}
$
$
\begin{aligned}
\mathrm{I}_{\text {system }} & =\mathrm{I}_{\text {balls }}+\mathrm{I}_{\text {rod }} \\
& =\frac{137}{15} M R^2
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: Three solid spheres each of mass $\mathrm{m}$ and diameter $\mathrm{d}$ are stuck together such that the lines connecting the centres form an equilateral triangle of the side of length d. The ratio $\frac{I_O}{I_A}$ of the moment of inertia $\mathrm{I}_0$ of the system about an axis passing the centroid and about the centre of any of spheres $\mathrm{I}_{\mathrm{A}}$ and perpendicular to the plane of the triangle is:

1) $\frac{15}{13}$
2) $\frac{13}{15}$
3) $\frac{13}{23}$
4) $\frac{23}{13}$

Solution:

As a moment of inertia of one sphere of radius R about its $centre= I=\frac{2}{5} m R^2$

From the figure given it is clear that $O C=\frac{d}{\sqrt{3}}$

$\text { M.I about O }=3\left[\left(\frac{2}{5} M\left(\frac{d}{2}\right)^2+M\left(\frac{d}{\sqrt{3}}\right)^2\right)\right]=\frac{13}{10} M d^2$

$\text { M.I about A }=2\left[\left(\frac{2}{5} M\left(\frac{d}{2}\right)^2+M(d)^2\right)\right]+\frac{2}{5} M\left(\frac{d}{2}\right)^2=\frac{23}{10} M d^2$

$\text { New Ratio }=\frac{13}{23}$

Hence, the answer is option (3).

Example 3: One solid sphere A and another hollow sphere B are of the same mass and the same outer radii. Their moment of inertia about their diameters are respectively $I_A$ and $I_B$ such that:

$\text { Where } d_A \text { and } d_B \text { are their densities }$

1) $I_A=I_B$
2) $I_A>I_B$
3) $I_A<I_B$
4) $I_A / I_B=d_A / d_B$

Solution:

Solid sphere $I_s=\frac{2}{5} M R^2$
Hollow sphere $I_H=\frac{2}{3} M R^2$
$
\frac{I_s}{I_H}=\frac{3}{5}
$
i.e. $I_s<I_H$
$
\begin{aligned}
& I_s=I_A \\
& I_H=I_B
\end{aligned}
$
i.e. $I_A<I_B$

Hence, the answer is option (3).

Example 4: The moment of inertia (M.I.) of four bodies, having the same radius is reported as;
$I_1=$ M.l. of the thin circular ring about its diameter,
$I_2=$ M.I. of the circular disc about an axis perpendicular to the disc and going through the centre,
$I_3=$ M.i. of the solid cylinder about its axis and
$I_4=$ M.I. of the solid sphere about its diameter.

Then:

1) $I_1=I_2=I_3>I_4$
2) $I_1=I_2=I_3<I_4$
3) $I_1+I_3<I_2+I_4$
4) $I_1+I_2=I_3+\frac{5}{2} I_4$

Solution

Ring $\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}$ about diameter

Disc $I_2=\frac{M R^2}{2}$
Solid cylinder $I_3=\frac{M R^2}{2}$
Solid sphere $\mathrm{I}_4=\frac{2}{5} \mathrm{MR}^2$
$
\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_3>\mathrm{I}_4
$

Hence, the answer is the option (1).

Example 5: The moment of Inertia (M.I) of four bodies having the same mass ${ }^{\prime} \mathrm{M}^{\prime}$ and radius ${ }^{\prime} 2 \mathrm{R}^{\prime}$ are as follows :
$\mathrm{I}_1=$ M.I of the solid sphere about its diameter.
$\mathrm{I}_2=$ M.I of the solid cylinder about its axis.
$\mathrm{I}_3=$ M.I of the solid circular disc about the diameter.
$\mathrm{I}_4=$ M.I of the thin circular ring about its diameter.
If $2\left(\mathrm{I}_2+\mathrm{I}_3\right)+\mathrm{I}_4=\mathrm{x} \cdot \mathrm{I}_1$, then the value of $\mathrm{x}$ will be $\qquad$

1) 5

2) 6

3) 7

4) 8

Solution:

$\begin{aligned}
& \mathrm{I}_1=\frac{2}{5} \mathrm{M}(2 \mathrm{R})^2=\frac{8 \mathrm{MR}^2}{5} \\
& \mathrm{I}_2=\frac{\mathrm{M}(2 \mathrm{R})^2}{2}=2 \mathrm{MR}^2 \\
& \mathrm{I}_3=\frac{\mathrm{M}(2 \mathrm{R})^2}{4}=\mathrm{MR}^2 \\
& \mathrm{I}_4=\frac{\mathrm{M}(2 \mathrm{R})^2}{2}=\frac{4 \mathrm{MR}^2}{2}=2 \mathrm{MR}^2 \\
& 2\left(\mathrm{I}_2+\mathrm{I}_3\right)+\mathrm{I}_4=\mathrm{xI}_1 \\
& 6 \mathrm{MR}^2+2 \mathrm{MR}^2=\mathrm{x} \frac{\left(8 \mathrm{MR}^2\right)}{5} \\
& 8 \mathrm{MR}^2=\mathrm{x} \frac{\left(8 \mathrm{MR}^2\right)}{5} \\
& \mathrm{x}=5
\end{aligned}$

Hence, the answer is option (1).

Summary

The moment of inertia of a solid sphere quantifies its resistance to rotational motion about an axis. This value depends on the sphere's mass and radius and is crucial for predicting the sphere's behaviour in rotational dynamics. Understanding this concept is essential for solving the example based on rotational motion.