In testing an automobile tire for proper alignment, a technician marks a spot on

Question

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.160 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 32.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.70 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?
rad/s

(b) What is the tangential speed of the spot after 1.30 s?
m/s

(c) What is the magnitude of the total acceleration of the spot after 1.30 s?
m/s2

(d) What is the angular position of the spot after 1.30 s?
rad

Explanation / Answer

Angular acceleration = 1.70 rad/s^2

a)

What is the angular speed of the wheel after 1.30 s?

angular speed = alpha*t = 1.70*1.30 = 2.21 rad/s

b)

What is the tangential speed of the spot after 1.30 s?

The tangential speed = omega R = 2.21*0.160 = 0.354 m/s

c)

What is the magnitude of the total acceleration of the spot after 1.30 s?

The tangential acceleration = alpha*R = 1.70*0.160 = 0.272 m/s^2

The radial acceleration = v^2/R = 0.354^2/0.160 = 0.783 m/s^2

Net acceleration = sqrt(0.783^2+0.272^2)=0.829 m/s^2

d)

What is the angular position of the spot after 1.30 s?

angular position = 32.0 degree + 0.5*1.70*1.30^2*(180/pi) degree = 114 degree. = 1.99 rad.

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