Answer:
[tex]4.3\cdot 10^{-7} kg m^2[/tex]
Explanation:
The moment of inertia of a sphere about an axis passing through its center is given by:
[tex]I=\frac{2}{5}MR^2[/tex]
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
[tex]I=\frac{2}{5}(0.0027 kg)(0.020 m)^2=4.3\cdot 10^{-7} kg m^2[/tex]
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
