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

i think it's 35 meters

Answer from: Quest

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

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