Chapter 8, Problem 18P

To determine

To Find: The distance travelled by the point on the edge of the wheel.

Answer to Problem 18P

  $33.67\text{m}$

Explanation of Solution

Given information:

A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s.

Formula used:

Conversion factor:

  $1\text{rpm}=\frac{2\pi }{60}\text{rad/s}$

First equation of rotational motion:

  $\alpha =\frac{\omega -{\omega }_o}{t}$

Third equation of rotational motion:

  $\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2$

Here,

  $\alpha$ is the angular acceleration.

  $\omega$ is the final angular velocity.

  ${\omega }_o$ is the initial angular velocity.

  $\theta$ is the angular displacement.

Relation between linear displacement and angular displacement:

  $d=r\theta$

Here, r is the radius.

Calculation:

Convert the units of initial angular velocity:

  $\begin{array}{l}{\omega }_o=240\text{rpm}\times \frac{2\pi }{60}\frac{\text{rad/s}}{\text{rpm}}\\ {\omega }_o=25.12\text{rad/s}\end{array}$

Convert the units of final angular velocity:

  $\begin{array}{l}\omega =360\text{rpm}\times \frac{2\pi }{60}\frac{\text{rad/s}}{\text{rpm}}\\ \omega =37.68\text{rad/s}\end{array}$

Angular acceleration:

  $\begin{array}{c}\alpha =\frac{37.68-25.12}{6.5}\\ =\frac{12.56}{6.5}\\ =1.93{\text{rad/s}}^{\text{2}}\end{array}$

  $\begin{array}{l}\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2\\ \theta =\left(25.12\right)\left(6.5\right)+\frac{1}{2}\left(1.93\right){\left(6.5\right)}^2\\ \theta =163.28+40.77\\ \theta =204.05\text{rad}\end{array}$

  $d=r\theta$

  $\begin{array}{l}d=204.05\times \left(\frac{0.33}{2}\right)\\ d=33.67\text{m}\end{array}$

Conclusion:

The point on the edge of the wheel travelled $33.67\text{m}$ .