Chapter 8, Problem 86GP

(a)

To determine

To Find: The angular acceleration of each tire.

Answer to Problem 86GP

The angular acceleration of each tire $-1.605\mathrm{rad}/s^2.$

Explanation of Solution

Given:

The diameter of each tire, $d=0.90\mathrm{m}$

The revolution made by car, $N=85$

The initial speed of car, $v_i=90\text{km/h}$

The final speed of car, $v_f=60\text{km/h}$

Formula used:

The angular acceleration formula is:

  $\alpha =\frac{{\omega }_f{}^2-{\omega }_i{}^2}{2\theta }\mathrm{....}(1)$

Where, ${\omega }_f$ is final angular velocity.

  ${\omega }_i$ is initial angular velocity.

  $\theta$ is angular displacement.

The angular velocity is $\omega =\frac{v}{r}$

  $v$ is linear velocity.

  $r$ is the radius.

Angular displacement is $\theta =2\pi N$

  $N$ is the number of revolutions.

Calculation:

Substitute the given values in equation (1):

  $\begin{array}{l}{\omega }_f=\frac{v_f}{r}\\ \theta =2\pi N\end{array}$

  $\begin{array}{l}{v}_i=90\text{km/h}\\ =90\times \frac{1000}{3600}=25\text{m/s}\\ v_f=60\text{km/h}\\ =60\times \frac{1000}{3600}=16.67\text{m/s}\\ \theta =2\pi (85)=534.07\text{rad}\\ \alpha =\frac{{\left(\frac{16.67}{0.45}\right)}^2-{\left(\frac{25}{0.45}\right)}^2}{2(534.07)}\\ =\frac{1372.29-3086.42}{1068.17}\\ =\frac{-1714.13}{1068.17}=-1.605{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

The value of angular acceleration of each tire is $-1.605{\text{rad/s}}^{\text{2}}$

(b)

To determine

To Find: The time taken by the car to stop.

Answer to Problem 86GP

The time taken by the car to stop is $34.6\mathrm{s}.$

Explanation of Solution

Given:

The initial velocity of car, $v_i=90\text{km/h}=25\text{m/s}$

The angular acceleration, $\alpha =-1.605{\text{rad/s}}^{\text{2}}$

Formula used:

  $v_f=v_i+at$

Here,

The initial and final velocity are denoted as $v_i\&v_f$ respectively.

Calculation:

  $\begin{array}{l}{v}_f=0\text{m/s}\\ v_i=25\text{m/}s\\ \alpha =-1.605{\text{rad/s}}^{\text{2}}\\ r=.45\mathrm{m}\\ a=\alpha r=(-1.605)(0.45)=-0.72225\text{m/s}\\ 0=25+(-0.72225)t\\ t=\frac{25}{0.72225}=34.6\mathrm{s}\end{array}$

Conclusion:

The time required to stop the car is $34.6\mathrm{s}$ .