Chris Cruz

2021-12-13

A standard table tennis ball has a diameter of 40 mm, weighs 2.7 g, and is made of a thin, spherical shell. What is its moment of inertia about an axis that passes through its center?

kalfswors0m

Beginner2021-12-14Added 24 answers

You may calculate the ball's moment of inertia using equality:

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

$I$ is the inertia of the tennis ball, $m$ is the mass of the ball, $r$ is a rafius of the ball.

$d=40mm$, $m=2.7g$

Calculating the radius:

$r=\frac{1}{2}d=\frac{1}{2}\times 40\times {10}^{-3}m=0.02m$

$m=2.7g\times {10}^{-3}kg=0.0027kg$

Substitute the values to the formula:

$I=\frac{2}{3}\times 0.0027kg\times {\left(0.02m\right)}^{2}=7.2\times {10}^{-7}kg\times {m}^{2}$

xandir307dc

Beginner2021-12-15Added 35 answers

Good job, thanks!

22+64

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