Question

An automobile traveling at $80.0\text{km/h}$has tires of$75.0\text{\hspace{0.17em}cm}$ diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

Short answer

a) Angular Speed of tires$=59.2\text{\hspace{0.17em}rad/s}$

b) Angular Acceleration of wheel$=9.31{\text{\hspace{0.17em}rad/s}}^{\text{2}}$

c) Distance traveled during braking$=\text{70.5m}$

Step-by-step solution

Given

Diameter of the wheels, $d=75.0\text{cm}$.

The velocity of the automobile,$v=80\text{km}/\text{hr}$ .

Number of rotations of the tire before car stops is$30$ .

To understand the concept.

Using the linear velocity and radius, we can find the angular speed of the tire.

Using angular speed and kinematic equations for rotational motion we can find the rest of the unknowns. Using the formula for the length of the curve in terms of angle and radius, we can find the distance traveled.

The relation between angular velocity$\left(\mathbf{\omega }\right)$ and linear velocity$\left(\mathbf{v}\right)$ is given as-

$\mathbf{\omega }\mathbf{=}\frac{v}{r}$

The variation of angular velocity with angular displacement $\left(\mathbf{\theta }\right)$is represented as-

${\mathbf{\omega }}^2\mathbf{=}{\mathbf{\omega }}_o^2\mathbf{+}\mathbf{2}\mathbf{\alpha \theta }$

The relation between displacement$\left(\mathbf{s}\right)$ and angular displacement $\left(\mathbf{\theta }\right)$is given as-

$\mathbf{s}\mathbf{=}\mathbf{r\theta }$

Here$\mathbf{r}$is the radius of the sphere,$\mathbf{\alpha }$is the angular acceleration and${\mathbf{\omega }}_o$is the initial angular velocity.

Convert v and r into SI units

Before we go for calculation part, we will convert the $\text{vandr}$into SI units.

$\begin{array}{c}v=80\frac{\text{km}}{\text{hr}}\\ =80\frac{\text{km}}{\text{hr}}\times \frac{1000\text{m}}{1\text{\hspace{0.17em}km}}\times \frac{1\text{\hspace{0.17em}hr}}{3600\text{\hspace{0.17em}s}}\\ =22.2\text{\hspace{0.17em}m/s}\end{array}$

$\begin{array}{c}d=75.0\text{cm}\\ =75.0\text{cm}\times \frac{0.01\text{m}}{1\text{cm}}\\ =0.75\text{\hspace{0.17em}m}\end{array}$

$\begin{array}{c}r=\frac{d}{2}\\ =\frac{0.75\text{\hspace{0.17em}m}}{2}\\ =0.375\text{\hspace{0.17em}m}\end{array}$

(a) Calculate the angular speed of the tires about their axles

Angular velocity can be written in terms of linear velocity and the radius as follows:

$\begin{array}{c}\text{ω}=\frac{\text{v}}{\text{r}}\\ =\frac{22.2\text{\hspace{0.17em}m}/\text{s}}{0.375\text{\hspace{0.17em}m}}\\ =59.2\text{rad/s}\end{array}$

Angular speed of the tires is $59.2\text{rad/s}$

Step 5:(b) Calculate the magnitude of the angular acceleration of the wheels if the car is brought to a stop uniformly in 30.0 complete turns of the tires

$\begin{array}{c}\text{θ}=30.0\times 2\text{π\hspace{0.17em}rad}\\ =188.5\text{\hspace{0.17em}rad}\end{array}$

As the automobile is brought to a stop, the final angular velocity must be zero.

So, using the third equation of motion for rotational motion-

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

For the given values, the angular acceleration can be calculated as-

$\left|\text{α}\right|=\left|\frac{{(59.2\text{\hspace{0.17em}rad}/\text{s})}^2}{2\times 188.5\text{\hspace{0.17em}rad}}\right|$

$\text{α}=9.31{\text{rad/s}}^{\text{2}}$

Angular acceleration of tire is .${\text{9.31rad/s}}^{\text{2}}$

(c) Calculate how far the car moves during the braking

The displacement of car after applying the brakes-

$s=r\theta$

$s=0.375\text{\hspace{0.17em}m}\times 188.5\text{\hspace{0.17em}rad}$

$s=70.7\text{\hspace{0.17em}m}$

After applying the brakes, the automobile travels a distance$s=70.7\text{\hspace{0.17em}m}$ .