Chapter 8, Problem 27P

To determine

To find: The moment of inertia of 10.8kg sphere.

Answer to Problem 27P

  $\text{1}.{\text{81 kg·m}}^{\text{2}}$

Explanation of Solution

Given information:

The mass of sphere, $\left(M\right)=10.8\text{ kg}$

The radius of sphere, $\left(R\right)=0.648\text{ m}$ .

Formula used:

The moment of inertia of a solid sphere about an axis through its centre of mass is

  $I_{CM}=\frac{2}{5}M{R}^2$

Where

M = the total distributed mass

R = the radius of the sphere.

Calculation:

The moment of inertial of sphere when the axis of rotation is through its center

  $I=\frac{2}{5}M{R}^2$

Substituting the values as shown below,

  $\begin{array}{l}I=\frac{2}{5}M{R}^2\\ I=\frac{2}{5}\left(10.8\text{kg}\right){\left(0.648\text{m}\right)}^2\\ I=1.81{\text{kgm}}^{\text{2}}\end{array}$

Conclusion:

The value of Moment of Inertia $\text{1}.{\text{81 kg·m}}^{\text{2}}$ .