Chapter 8, Problem 21P

(a)

To determine

To Find: The value of the angular acceleration of the tires.

Answer to Problem 21P

  $\alpha =-4.1{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given information:

The tires of a car make $65$ revolutions as the car reduces its speed uniformly from $95\text{km/h}$ to $45\text{km/h}$ . The tires have a diameter of $0.80\text{m}$ .

Formula used:

  $\omega =v/r$

  ${\omega }^2={\omega }_0^2+2\alpha \theta$

Here, $\omega$ is the final angular velocity.

  ${\omega }_0$ is the initial angular velocity.

  $\alpha$ is the angular acceleration.

  $\theta$ is the angular displacement.

  $v$ is the linear velocity.

  $r$ is the radius.

Calculation:

The angular acceleration can be determined from angular kinematic equations, with the angular velocities being from $\omega =v/r$

  $\begin{array}{l}{\omega }^2={\omega }_0^2+2\alpha \theta \\ \alpha =\frac{{\omega }^2-{\omega }_0^2}{2\theta }\\ =\frac{v^2/r^2-v_0^2/r^2}{2\theta }\\ =\left(\frac{v^2-v_0^2}{2\theta r^2}\right)\\ =\left(\frac{{\left(45\text{km/h}\right)}^2-{\left(95\text{km/h}\right)}^2}{2\times {\left(0.40\text{m}\right)}^2\times \left(65\text{rev}\right)\times \left(2\pi \text{rad/rev}\right)}\right){\left(\frac{\text{1m/s}}{\text{3}\text{.6}\text{km/h}}\right)}^2\\ \alpha =-4.133{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

Angular acceleration is $\alpha =-4.1{\text{rad/s}}^{\text{2}}$ .

(b)

To determine

To Find: The time that is required to stop the car.

Answer to Problem 21P

  $t=7.6\text{s}$

Explanation of Solution

Given information:

The tires of a car make 65 revolutions as the car reduces its speed uniformly from $95\text{km/h}$ to $45\text{km/h}$ . The tires have a diameter of $0.80\text{m}$ .

Formula used:

  $t=-\frac{v}{r\alpha }$

  $\omega ={\omega }_0+\alpha t$

Here, $\omega$ is the final angular velocity,

  ${\omega }_0$ is the initial angular velocity.

  $\alpha$ is the angular acceleration.

  $t$ is the time.

  $v$ is the linear velocity,

  $r$ is the radius.

Calculation:

The time taken to stop can be finding by using equation of motion

  $\begin{array}{l}\omega ={\omega }_0+\alpha t\\ 0=\left(\frac{v}{r}\right)+\alpha t\\ t=-\frac{v}{r\alpha }\\ =-\frac{\left(45\text{km/h}\right)\left(\frac{1\text{m/s}}{3.6\text{km/h}}\right)}{\left(0.40\text{m}\right)\left(-4.133{\text{rad/s}}^2\right)}\\ t=7.6\text{s}\end{array}$

Conclusion:

Time needed to stop $t=7.6\text{s}$ .