. Advertisement .

..3..

. Advertisement .

..4..

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?

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Concepts and Reason

This problem can be solved using two concepts: moment of inertia for the hollow sphere, and conversion of units.

Calculate the mass in kilograms by first dividing the radius of the sphere into its diameter. Substitute the values in the expression for moment of inertia hollow sphere to calculate moment of inertia for the tennis ball around an axis passing through its center.

Fundamentals

This problem can be solved using two concepts: moment of inertia for the hollow sphere, and conversion of units.

Calculate the mass in kilograms by first dividing the radius of the sphere into its diameter. Substitute the values in the expression for moment of inertia hollow sphere to calculate moment of inertia for the tennis ball around an axis passing through its center.

A sphere’s radius is equal to half its diameter. This is:

[katex]r = \frac{d}{2}[/katex]

Here, [katex]r[/katex]

The moment of inertia for hollow spheres around an axis which passes through its center is

[katex]I = \frac{2}{3}m{r^2}[/katex]

Here, [katex]I[/katex]

These are the relations that covert units use:

[katex]\begin{array}{c}\\1{\rm{ mm}} = {10^{ – 3}}{\rm{ m}}\\\\1{\rm{ g}} = {10^{ – 3}}{\rm{ kg}}\\\end{array}[/katex]

The relationship between radius and diameter is.

[katex]r = \frac{d}{2}[/katex]

Substitute [katex]40{\rm{ mm}}[/katex]

[katex]\begin{array}{c}\\r = \frac{{40}}{2}{\rm{ mm}}\\\\ = 20{\rm{ mm}}\\\end{array}[/katex]

Multiply [katex]\left( {\frac{{{{10}^{ – 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)[/katex] to convert mm to m

[katex]\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}[/katex]

Convert mass [katex]r[/katex]0

[katex]r[/katex]1

The tennis ball’s moment of inertia is described as,

[katex]I = \frac{2}{3}m{r^2}[/katex]

Substitute [katex]r[/katex]3

[katex]r[/katex]4Ans:

The moment of inertia for the tennis ball around an axis passing through its center is [katex]r[/katex]5