A moped tire was observed to rotate at 34.0 rot/min at t = 1.0 s, and 75.0 rot/m

Question

A moped tire was observed to rotate at 34.0 rot/min at t = 1.0 s, and 75.0 rot/min at t = 6.0 s. Assume that the moped is moving on a horizontal flat road and that the tires have a 33.0 cm radius. What is the angular velocity in rad/s of the moped tire at t=6.0s? CHECK What is in rad/s^2 the angular acceleration of the moped tire during this time span from 1.0 s to 6.0s? CHECK What angle in radians does the tire rotate through during the same time span? CHECK How far did the moped move in meters during the same time span? CHECK What is the magnitude of the centripetal acceleration of at point on the edge of the tire at t=6.0s? CHECK

Explanation / Answer

a. Angular speed

= 75 * 2 * pi/60

= 7.85 rad/s.

b. Angular acceleration

=(7.85 - (34* 2 *pi/60))/ t2-t1

=0.858 rad/ s^2

c.Angle of rotation

= (7.85^2 - 3.56^2)/(2*0.858)

= 28.53 rad

= (28.53 / 2 * pi)

= 4.54 rotation

d. distance it moved.

= 28.53 * r =28.53 *.33 = 9.41 m.

E. a = ^2 r

= 7.85^2 * .33

=20.33 m/s^2

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