Physics, 24.06.2020 02:01 atlantahead7011

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.220 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 30.0Â° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.90 rad/s2. (Assume the spot's position is initially positive, and assume the angular acceleration is in the positive direction).A) What is the angular speed of the wheel after 1.30 s? B) What is the tangential speed of the spot after 1.30 s?
C) What is the magnitude of the total acceleration of the spot after 1.30 s?
D) What is the angular position of the spot after 1.30 s?

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Answers

Answer from: kira3039

a) The angular speed of the wheel after 1.30 seconds is $2.47\,\frac{rad}{s}$ , b) The tangential speed of the spot after 1.30 seconds is $0.543\,\frac{m}{s}$ , c) The magnitude of the total acceleration of the spot after 1.30 seconds is $1.406\,\frac{m}{s^{2}}$ , d) The angular position of the spot is 2.130 radians (122.011°).

Explanation:

a) Given that tire accelerates at constant rate, final angular speed can be predicted by using the following formula:

$\omega = \omega_{o} + \alpha \cdot \Delta t$

Where:

$\omega$ - Final angular speed, measured in radians per second.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\Delta t$ - Time, measured in seconds.

Given that $\omega_{o} = 0\,\frac{rad}{s}$ (starts at rest), $\alpha = 1.90\,\frac{rad}{s^{2}}$ and $\Delta t = 1.30\,s$ , the final angular speed is:

$\omega = 0\,\frac{rad}{s} + \left(1.90\,\frac{rad}{s^{2}} \right) \cdot (1.30\,s)$

$\omega = 2.47\,\frac{rad}{s}$

The angular speed of the wheel after 1.30 seconds is $2.47\,\frac{rad}{s}$ .

b) The tangential speed of the spot is the product of the distance between the center of the wheel and spot. That is:

$v = r \cdot \omega$

Where r is the distance between the center of the wheel and spot. The tangential speed of the spot after 1.30 seconds is:

$v = (0.220\,m)\cdot \left(2.47\,\frac{rad}{s} \right)$

$v = 0.543\,\frac{m}{s}$

The tangential speed of the spot after 1.30 seconds is $0.543\,\frac{m}{s}$ .

c) The magnitude of the total acceleration of the spot is the magnitude of the vectorial sum of radial and tangential accelerations (both components are perpendicular to each other), which is determined by the Pythagorean theorem, that is:

$a = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where $a_{r}$ and $a_{t}$ are the radial and tangential accelerations.

$a = r\cdot \sqrt{\omega^{4} + \alpha^{2}}$

If $r = 0.220\,m$ , $\omega = 2.47\,\frac{rad}{s}$ and $\alpha = 1.90\,\frac{rad}{s^{2}}$ , then, the resultant acceleration is:

$a = (0.220\,m)\cdot \sqrt{\left(2.47\,\frac{rad}{s} \right)^{4}+\left(1.90\,\frac{rad}{s^{2}} \right)^{2}}$

$a \approx 1.406\,\frac{m}{s^{2}}$

The magnitude of the total acceleration of the spot after 1.30 seconds is $1.406\,\frac{m}{s^{2}}$ .

d) Let be 30° (0.524 radians) the initial angular position of the spot with respect to center. The final angular position is determined by the following equation of motion:

$\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta - \theta_{o})$

Final angular position is therefore cleared:

$\theta - \theta_{o} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}$

$\theta = \theta_{o} + \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}$

Given that $\theta_{o} = 0.524\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 2.47\,\frac{rad}{s}$ and $\alpha = 1.90\,\frac{rad}{s^{2}}$ , the angular position of the spot after 1.30 seconds is: