Question

A regulation table tennis ball has a mass of 2.7g and is 40mm in diameter. What is its moment of inertia about an axis that passes through its center?

Answer 1

Concepts and reason

The concepts required to solve this problem are moment of inertia of the hollow sphere and conversion of units.

First solve for the radius of the sphere from diameter and convert mass to kilogram. Then, substitute the values in the expression of moment of inertia of hollow sphere to calculate the moment of inertia of the tennis ball about an axis that passes through its center.

Fundamentals

The radius of a sphere is half of the diameter. That is,

$r = \frac{d}{2}$

Here, $r$ is the radius and $d$ is the diameter.

The moment of inertia of hollow sphere about an axis that passes through its center is,

$I = \frac{2}{3}m{r^2}$

Here, $I$ is the moment of inertia, $m$ is the mass, and $r$ is the radius.

The relations used to coverts units are as follows:

$\begin{array}{c}\\1{\rm{ mm}} = {10^{ - 3}}{\rm{ m}}\\\\1{\rm{ g}} = {10^{ - 3}}{\rm{ kg}}\\\end{array}$

The relation of diameter and radius is.

$r = \frac{d}{2}$

Substitute $40{\rm{ mm}}$ for $d$ in the equation $r = \frac{d}{2}$ .

$\begin{array}{c}\\r = \frac{{40}}{2}{\rm{ mm}}\\\\ = 20{\rm{ mm}}\\\end{array}$

Convert mm to m by multiplying with $\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)$ .

$\begin{array}{c}\\r = 20{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)\\\\ = 2.0 \times {10^{ - 2}}{\rm{ m}}\\\end{array}$

Convert mass $m = 2.7{\rm{ g}}$ to kg by multiplying with $\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)$ .

$\begin{array}{r}\\m = 2.7{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\\\\ = 2.7 \times {10^{ - 3}}{\rm{ kg}}\\\end{array}$

The expression of moment of inertia of the tennis ball is,

$I = \frac{2}{3}m{r^2}$

Substitute $2.7 \times {10^{ - 3}}{\rm{ kg}}$ for $m$ , and $2.0 \times {10^{ - 2}}{\rm{ m}}$ in the equation $I = \frac{2}{3}m{r^2}$ and calculate the moment of inertia of the tennis ball about an axis that passes through its center.

$\begin{array}{c}\\I = \frac{2}{3}\left( {2.7 \times {{10}^{ - 3}}{\rm{ kg}}} \right){\left( {2.0 \times {{10}^{ - 2}}{\rm{ m}}} \right)^2}\\\\ = 7.2 \times {10^{ - 7}}{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

Ans:

The moment of inertia of the tennis ball about an axis that passes through its center is $7.2 \times {10^{ - 7}}{\rm{ kg}} \cdot {{\rm{m}}^2}$ .