Physics, 19.06.2020 03:57 jdkrisdaimcc11

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.190 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 33.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.50 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 1.20 s? rad/s

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Answer from: datzmypupppup

Complete Question

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.190 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 33.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.50 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 1.20 s? rad/s

b) What is the tangential speed of the spot after 1.20 s?

(d) What is the angular position of the spot after 1.20 s?

w = 1.8 rad/sec

$v =0.342 \ m/s$

$a_r = 0.6156\ m/s^2$

$\theta = 1.656 \ rad$

Explanation:

From the question we are told that

The distance of the spot from the center is $r = 0.190 \ m$

The angle of the radius vector with the horizontal is $\theta_o = 33^o = \frac{33\pi}{180} rad$

The acceleration is $\alpha = 1.50 \ rad/s^2$

The angular speed of the wheel after 1.20 s is mathematically represented as

$w =w_o + \alpha t$

Where w_o is the initial angular speed of the wheel which is zero

Substituting values

w = 0 + 1.5(1.20)

w = 1.8 rad/sec

the tangential speed of the spot after 1.20 is mathematically represented as

v = r w

substituting values

v = 0.190 * 1.8

$v =0.342 \ m/s$

The magnitude of the total acceleration of the spot after 1.30 s is mathematically represented as

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

The radial acceleration is

$a_r = \frac{v^2}{r}$

substituting values

$a_r = \frac{(0.342)^2}{0.190}$

$a_r = 0.6156\ m/s^2$

The tangential acceleration is

$a_t = r \alpha$

substituting values

a_t = 0.19 * 1.5

$a_t = 0.285 \ m/s^2$

=> $a = \sqrt{0.6156^2 + 0.285 ^2}$

$a = 0.6784 \ m/s^2$

the angular position of the spot after 1.20 s is mathematically represented as

$\theta = \theta _o + w_ot + \frac{1}{2} \alpha t^2$

substituting values

$\theta = \frac{33 \pi}{180} + 0 + 0.5 * 1.5 * 1.2^2$

$\theta = 1.656 \ rad$

Answer from: Quest

The answer is d) by requesting advice on science internet sites

Answer from: Quest

thermal energy is the answer

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