Chapter 5, Problem 11P

To determine

To rank the angular speed from largest to smallest.

Answer to Problem 11P

The ranking of angular speed from largest to smallest is $\underset{\_}{\left(c\right)=\left(d\right)>\left(a\right)=\left(b\right)=\left(e\right)}$.

Explanation of Solution

Write the expression for angular speed of the wheel 1.

    ${\omega }_1=\frac{2\pi }{T_1}\text{rad}/\text{s}$        (I)

Here, ${\omega }_1$ is the angular speed of wheel 1, and $T_1$ is the time period of wheel 1.

Given that time period and radius of the wheel 1 is $4.0\text{ms}$ and $8.0\text{cm}$ respectively.

Apply the above condition in equation (I).

    $\begin{array}{c}{\omega }_1=\frac{2\left(3.14\right)}{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{rad}\\ \text{=1}\text{.57}\times {\text{10}}^3\text{rad}/\text{s}\end{array}$

Write the expression for angular speed of the wheel 2.

    ${\omega }_2=\frac{2\pi }{T_2}\text{rad}/\text{s}$        (II)

Here, ${\omega }_2$ is the angular speed of wheel 2, and $T_2$ is the time period of wheel 2.

Given that time period of the wheel 2 is $4.0\text{ms}$ and the radius of the wheel 2 is $8.0\text{cm}$.

Apply the above condition in equation (II).

    $\begin{array}{c}{\omega }_2=\frac{2\left(3.14\right)}{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{rad}\\ \text{=1}\text{.57}\times {\text{10}}^3\text{rad}/\text{s}\end{array}$

Write the expression for angular speed of the wheel 3.

    ${\omega }_3=\frac{2\pi }{T_3}\text{rad}/\text{s}$        (III)

Here, ${\omega }_3$ is the angular speed of wheel 3, and $T_3$ is the time period of wheel 3.

Given that time period of the wheel 3 is $1.0\text{ms}$ and the radius of the wheel 3 is $8.0\text{cm}$.

Apply the above condition in equation (III).

    $\begin{array}{c}{\omega }_3=\frac{2\left(3.14\right)}{\left(1.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{rad}\\ \text{=6}\text{.28}\times {\text{10}}^3\text{rad}/\text{s}\end{array}$

Write the expression for angular speed of the wheel 4.

    ${\omega }_4=\frac{2\pi }{T_4}\text{rad}/\text{s}$        (IV)

Here, ${\omega }_4$ is the angular speed of wheel 4, and $T_4$ is the time period of wheel 4.

Given that time period of the wheel 4 is $1.0\text{ms}$ and the radius of the wheel 4 is $8.0\text{cm}$.

Apply the above condition in equation (IV).

    $\begin{array}{c}{\omega }_4=\frac{2\left(3.14\right)}{\left(1.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{rad}\\ \text{=6}\text{.28}\times {\text{10}}^3\text{rad}/\text{s}\end{array}$

Write the expression for angular speed of the wheel 5.

    ${\omega }_4=\frac{2\pi }{T_4}\text{rad}/\text{s}$        (V)

Here, ${\omega }_4$ is the angular speed of wheel 5, and $T_4$ is the time period of wheel 5.

Given that time period of the wheel 5 is $1.0\text{ms}$ and the radius of the wheel 5 is $8.0\text{cm}$.

Apply the above condition in equation (V).

    $\begin{array}{c}{\omega }_5=\frac{2\left(3.14\right)}{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{rad}\\ \text{=1}\text{.57}\times {\text{10}}^3\text{rad}/\text{s}\end{array}$

The ranking of angular speed from largest to smallest is ${\omega }_3={\omega }_4>{\omega }_1={\omega }_2={\omega }_5$.

Conclusion:

Therefore, the ranking of angular speed from largest to smallest is $\underset{\_}{\left(c\right)=\left(d\right)>\left(a\right)=\left(b\right)=\left(e\right)}$.