Question

Determine the moment of inertia of a 10.8-kg sphere of radius 0.648 m when the axis of rotation is through its center.

Short answer

The moment of inertia of the solid sphere rotating about an axis through its centre is \(1.81\;{\rm{kg}}\;{{\rm{m}}^2}\).

Step-by-step solution

Identification of the given data

The mass of the sphere is \(m = 10.8\;{\rm{kg}}\).

The radius of the sphere is \(r = 0.648\;{\rm{m}}\).

Definition of the moment of inertia

A moment of inertia is a quantity that expresses a body’s tendency to resist angular acceleration about an axis of rotation.

It is given as the product of the mass of the rigid body and the square of the distance from the axis of rotation.

\(I = m{r^2}\)

Determination of the moment of inertia of the solid sphere

As the sphere rotates about the axis of rotation passing through its center, the moment of inertia is given as follows:

\(\begin{align}I &= \frac{2}{5}m{r^2}\\ &= \frac{2}{5}\left( {10.8\;{\rm{kg}}} \right){\left( {0.648\;{\rm{m}}} \right)^2}\\ &= 1.81\;{\rm{kg}}\;{{\rm{m}}^2}\end{align}\)

Thus, the moment of inertia of the solid sphere is \(1.81\;{\rm{kg}}\;{{\rm{m}}^2}\).