Chapter 6.1, Problem 86E

Bicycle Wheel The sprockets and chain of a bicycle are shown in the figure. The pedal sprocket has a radius of 4 in., the wheel sprocket a radius of 2 in., and the wheel a radius of 13 in. The cyclist pedals at 40 rpm.

1. (a) Find the angular speed of the wheel sprocket.
2. (b) Find the speed of the bicycle. (Assume that the wheel turns at the same rate as the wheel sprocket.)

To determine

The angular speed of the wheel sprocket.

Answer to Problem 86E

The angular speed of the wheel sprocket is $\underset{¯}{160\pi \mathrm{rad}/\min }$.

Explanation of Solution

Formulas used:

Relation between linear and angular speed is $v=r\omega$, where $v$ is the linear speed, $r$ is the radius of the wheel and$\omega$ is the angular speed.

Calculation:

Radius of the pedal sprocket is 4 in. That is, $R_1=4\mathrm{in}$.

Radius of the wheel sprocket is 2 in. That is, $R_2=2\mathrm{in}$.

Radius of the wheel is 13 in. That is, $R_3=13\mathrm{in}$.

Cyclist pedals at 40 rpm. That is, $N_1=40\mathrm{rpm}$.

For $1$ complete rotation a circle is covered. That is, angle for $1$ rotation is $2\pi$.

Both sprockets move the same distance then, they have the same rotational speed.

To find the angular speed, equate the speeds of both sprockets. That is,

$\begin{array}{l}v=\omega R\\ v=2\pi NR\\ {\omega }_2{R}_2={\omega }_1{R}_1\end{array}$

Simplify further as follows:

$\begin{array}{l}{\omega }_1{R}_1={\omega }_2{R}_2\\ 2\pi N_1{R}_1={\omega }_2{R}_2\\ {\omega }_2=2\pi N_1\frac{R_1}{R_2}\end{array}$

Substitute $N_1=40\mathrm{rpm}$, $R_1=4\mathrm{in}$ and $R_2=2\mathrm{in}$ in ${\omega }_2=2\pi N_1\frac{R_1}{R_2}$:

$\begin{array}{c}{\omega }_2=2\pi N_1\frac{R_1}{R_2}\\ =2\pi \left(40\right)\frac{4}{2}\\ =160\pi \end{array}$

Thus, the angular speed of the blade is $\underset{¯}{160\pi \mathrm{rad}/\min }$.

To determine

The speed of the bicycle.

Answer to Problem 86E

The speed of the bicycle is $\underset{¯}{2080\pi \mathrm{in}/\min }$.

Explanation of Solution

Formula used:

Linear speed is $v=\frac{s}{t}=\frac{r\theta }{t}=r\left(\frac{\theta }{t}\right)=r\omega$, where $\theta$ is the angle change in a rotation, $s$is the distance travelled in time $t$ and $\omega$ is the angular speed.

Calculation:

From part (a), the angular speed of the blade is $160\pi \mathrm{rad}/\min$.

Substitute $\omega =160\pi$ and $R_3=13\mathrm{in}$ in $v={\omega }_2{R}_3$:

$\begin{array}{c}v={\omega }_2{R}_3\\ =160\pi \left(13\right)\\ =2080\pi \end{array}$

Thus, the speed of the bicycle is $\underset{¯}{2080\pi \mathrm{in}/\min }$.