Chapter 12, Problem 35PQ

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.200 m from the center. He then mounts the tire in a vertical plane and notes that the radius vector to the spot is at an angle of 35.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 3.00 rad/s². a. What is the angular speed of the wheel after 4.00 s? b. What is the tangential speed of the spot after 4.00 s? c. What is the magnitude of the total accleration of the spot after 4.00 s?" d. What is the angular position of the spot after 4.00 s?

To determine

The angular speed of the wheel after $4.00\text{s}$.

Answer to Problem 35PQ

The angular speed of the wheel after $4.00\text{s}$ is $\underset{\_}{12.0\text{rad}/\text{s}}$.

Explanation of Solution

Write the expression for the angular velocity of the wheel.

    $\omega ={\omega }_0+\alpha t$                                                                                                               (I)

Here, $\omega$ is the angular velocity after a time $t$, ${\omega }_0$ is the initial angular velocity, $\alpha$ is the angular acceleration

Conclusion:

Substitute $0$ for ${\omega }_0$, $3.00\text{rad}/{\text{s}}^2$ for $\alpha$ and $4.00\text{s}$ for $t$ in equation (I) to find $\omega$.

    $\begin{array}{c}\omega =0+\left(3.00\text{rad}/{\text{s}}^2\right)\left(4.00\text{s}\right)\\ =12.0\text{rad}/\text{s}\end{array}$

Therefore, the angular speed of the wheel after $4.00\text{s}$ is $\underset{\_}{12.0\text{rad}/\text{s}}$.

To determine

The tangential speed of the spot after $4.00\text{s}$.

Answer to Problem 35PQ

The tangential speed of the spot after $4.00\text{s}$ is $\underset{\_}{2.40\text{m}/\text{s}}$.

Explanation of Solution

Write the relation connecting the tangential speed and the angular speed.

    $v=r\omega$                                                                                                                     (II)

Here, $v$ is the tangential velocity of the object, $r$ is the radius of the orbit.

Conclusion:

Substitute $0.200\text{m}$ for $r$ and $12.0\text{rad}/\text{s}$ for $\omega$ in equation (II), to find $v$.

    $\begin{array}{c}v=\left(0.200\text{m}\right)\left(12.0\text{rad}/\text{s}\right)\\ =2.40\text{m}/\text{s}\end{array}$

Therefore, the tangential speed of the spot after $4.00\text{s}$ is $\underset{\_}{2.40\text{m}/\text{s}}$.

To determine

The magnitude of the total acceleration of the spot after $4.00\text{s}$.

Answer to Problem 35PQ

The magnitude of the total acceleration of the spot after $4.00\text{s}$ is $\underset{\_}{28.8\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the magnitude of the total acceleration.

    $a=\sqrt{a_r^2+a_t^2}$                                                                                                          (III)

Here, $a_r$ is the radial acceleration, $a_c$ is the centripetal acceleration.

The magnitude of the inward radial acceleration equals the centripetal acceleration so that,

    $a_r=a_c$                                                                                                                    (IV)

Write the expression for the centripetal acceleration.

    $a_c=\frac{v^2}{r}$                                                                                                                    (V)

Use equation (II) in equation (IV),

    $\begin{array}{c}{a}_c=\frac{{\left(r\omega \right)}^2}{r}\\ =r{\omega }^2\end{array}$                                                                                                             (VI)

The tangential acceleration equals the angular acceleration times the radius of the orbit.

    $a_t=r\alpha$                                                                                                                (VII)

Here, $a_t$ is the tangential acceleration.

Conclusion:

Substitute $0.200\text{m}$ for $r$ and $3.00\text{rad}/{\text{s}}^2$ for $\alpha$ in equation (VII) to find $a_t$.

    $\begin{array}{c}{a}_t=\left(0.200\text{m}\right)\left(3.00\text{rad}/{\text{s}}^2\right)\\ =0.600\text{m}/{\text{s}}^2\end{array}$

Substitute $0.200\text{m}$ for $r$ and $12.0\text{rad}/{\text{s}}^2$ for $\omega$ in equation (VI) to find $a_c$.

    $\begin{array}{c}{a}_c=\left(0.200\text{m}\right)\left(12.0\text{rad}/{\text{s}}^2\right)\\ =28.8\text{m}/{\text{s}}^2\end{array}$

Substitute $0.600\text{m}/{\text{s}}^2$ for $a_t$ and $28.8\text{m}/{\text{s}}^2$ for $a_c$ in equation (III) to find $a$.

    $\begin{array}{c}a=\sqrt{{\left(28.8\text{m}/{\text{s}}^2\right)}^2+{\left(0.600\text{m}/{\text{s}}^2\right)}^2}\\ =28.8\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude of the total acceleration of the spot after $4.00\text{s}$ is $\underset{\_}{28.8\text{m}/{\text{s}}^2}$.

To determine

The angular position of the spot after $4.00\text{s}$.

Answer to Problem 35PQ

The angular position of the spot is $\underset{\_}{24.6\text{rad}}$.

Explanation of Solution

Write the expression for the angular position after a time $t$.

    ${\theta }_f={\theta }_i+{\omega }_{o}t+\frac{1}{2}\alpha t^2$                                                                                            (VIII)

Here, ${\theta }_f$ is the angular position after a time $t$, ${\theta }_i$ is the initial position.

Conclusion:

Substitute $35°$ for ${\theta }_i$, $3.00\text{rad}/{\text{s}}^2$ for $\alpha$, $0$ for ${\omega }_0$, $4.00\text{s}$ for $t$ in equation (VIII) to find ${\theta }_f$.

    $\begin{array}{c}{\theta }_f=\left(35°\right)+\left[0\times \left(4.00\text{s}\right)\right]+\left[\frac{1}{2}\left(3.00\text{rad}/{\text{s}}^2\right){\left(4.00\text{s}\right)}^2\right]\\ =\left(35°\times \frac{\pi \text{rad}}{180°}\right)+\left[\frac{1}{2}\left(3.00\text{rad}/{\text{s}}^2\right){\left(4.00\text{s}\right)}^2\right]\\ =24.6\text{rad}\end{array}$

Therefore, the angular position of the spot is $\underset{\_}{24.6\text{rad}}$.