Chapter 8, Problem 29P

(a)

To determine

To find: The moment of inertia of the ball.

Answer to Problem 29P

The value is $0.936\text{kg}\cdot {\text{m}}^{\text{2}}$ .

Explanation of Solution

Given information:

A small $650\text{g}$ ball on the end of a thin, light rod is rotated in a horizontal circle of radius $1.2\text{m}$ .

Formula used:

  $I=M{R}^2$

Here,

  $I$ is the Moment of inertia of ball,

  $M$ is the Mass of ball,

  $R$ is the radius.

Calculation:

Mass of ball $\left(M\right)=0.650\text{kg}$

Radius of circle $\left(R\right)=1.2\text{m}$

Moment of inertia of ball about centre of circle $I=M{R}^2$

  $\begin{array}{l}=\left[0.650\times {\left(1.2\right)}^2\right]\text{kg}\cdot {\text{m}}^2\\ =0.936\text{kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The value is $0.936\text{kg}\cdot {\text{m}}^{\text{2}}$ .

(b)

To determine

To find: The torque needed for rotating the ball.

Answer to Problem 29P

The torque needed to keep the ball rotating is $0.024\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

A small $650\text{g}$ ball on the end of a thin, light rod is rotated in a horizontal circle of radius $1.2\text{m}$ .

Formula used:

  $\tau =F\times \text{Radius of circle}$

Here,

  $\tau$ is the torque,

  $F$ is the Force .

Calculation:

Force exerted by air $F=0.020\text{N}$

The torque needed to keep the ball rotating

  $\tau =F\times \text{Radius of circle}$

Substituting the values as shown below,

  $\begin{array}{l}\tau =F\times \text{Radius of circle}\\ \text{=}\left(0.020\times 1.2\right)\text{N}\cdot \text{m}\\ \tau =0.024\text{N}\cdot \text{m}\end{array}$

Hence, the torque needed to keep the ball rotating is $0.024\text{N}\cdot \text{m}$ .

Conclusion:

The torque needed to keep the ball rotating is $0.024\text{N}\cdot \text{m}$ .