Consider a sphere of uniformly distributed mass of 1 kg/m³ and radius 1 m. Its moment of intertia about one diameter is:- 

Concept:

• Moment of inertia (I) is analogous to mass acts in linear motion.
• The Moment of inertia is rotational inertia for which torque (turning force)  is required for angular acceleration.
• Moment of inertia is also known as the second moment of mass, rotational inertia, angular mass, or mass moment of inertia.
• The moment of inertia is an extensive property by which its magnitude and values depend on the extent of the object which means how mass is concentrated at a different distance from the axis of rotation.
• The moment of inertia of a body about any axis is equal to the sum of the moment of the inertia of the body about a parallel axis passing through its center of mass and the product of Its mass and the square of the distance between the two parallel axes.
• As per definition the unit and dimension of the moment of inertia is a product of mass and square of the distance so, mass × square of distance hence its unit is kg m^-2 and dimensional quantity is [M¹L^-2T⁰].
   

​Calculation:

Here given a sphere is a uniformly distributed mass throughout its axis for as 1 kg m^-3 and the radius is about r = 1 m, because it's continuously distributed mass we can say that its density is ρ = 1 kg m^-3.

For finding the mass(m) we know that mass is the product of density (ρ) and Volume (V).

∴ m = ρ × V

∴ m = 1 × \(\frac {4}{3} \pi r^3 \)

∴ m = \(\frac {4}{3} \times 3.14 \times (1)^3\) = 4.186 kg _______(1)

→ For uniformly distributed sphere now finding a moment of inertia we know that

∴ I = \(\frac{2}{5}\)mr² = \(\frac {2}{5}\ \times \ 4.186\ \times (1)^3\)  [∵m = 4.186 kg from eq.(1)]

∴ I = 1.67 kg m²

So, the moment of inertia is about 1.67 kg m² for the solid sphere.

Additional InformationComparison of Translational and Rotational motion