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

Answer:

$4.3\cdot 10^{-7} kg m^2$

Explanation:

The moment of inertia of a sphere about an axis passing through its center is given by:

$I=\frac{2}{5}MR^2$

where

M is the mass of the sphere

R is the radius of the sphere

For the tennis ball in this problem, we have the following data:

Mass: M = 2.7 g = 0.0027 kg

Radius: R = d/2 = 40 mm / 2 = 20 mm = 0.020 m

Therefore, the moment of inertia is

$I=\frac{2}{5}(0.0027 kg)(0.020 m)^2=4.3\cdot 10^{-7} kg m^2$

Answer 2

Answer:7.2 x 10^-7 kg•m^2

Explanation:

2/3 MR^2= 2/3 x 2.7 x10^-3 x (0.02)^2