Chapter 8, Problem 68A

To determine

The moment of inertia of the toy consisting of two balls at the end of a $0.46\text{m}$ lightweight rod, about the center of the rod.

Answer to Problem 68A

  $0.04761\text{kg}\cdot {\text{m}}^{\text{2}}$

Explanation of Solution

Given:

The mass of the each of the two ball is, $m=0.45\text{kg}$

The length of the rod, $l=0.46\text{m}$

Formula used:

The moment of inertia of a system about a point is given by,

  $I={\displaystyle \sum _{i=1}^n{m}_i{r}_i{}^2}$

Where $m_i$ are the mass of the objects in the system, and $r_i$ are their respective distance from the point about which moment of Inertia is considered.

Calculation:

Now, the distance of each ball from the center of the rod is,

  $r_1=r_2=\frac{l}{2}=0.23\text{}\text{m}$

Then, moment of Inertia of the toy is,

  $\begin{array}{c}I=m_1{r}_1{}^2+m_2{r}_2{}^2\\ =\left(0.45\right){\left(0.23\right)}^2+\left(0.45\right){\left(0.23\right)}^2\\ =0.04761\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the moment of inertia of the toy is $0.04761\text{kg}\cdot {\text{m}}^{\text{2}}$ .