A bicycle is turned upside down while in the shop being repaired. See Figure. A

Question

A bicycle is turned upside down while in the shop being repaired. See Figure. A friend spins the wheel, radius R, and notices that water drops fly off the wheel tangentially to the wheel, and in the vertical direction. The owner measures the maximum height h that the drops attain.
In a first trial, a drop breaks away from the tire and reaches a maximum height h1. On the next revolution, another identical drop attains a height h2, where h2 < h1.
a) Calculate the tangential velocities in each case. v1 and v2
b) Calculate the average velocity V during one revolution.
c) Calculate the time interval T for one revolution, using the average velocity V and tangential distance traveled .
d)Calculate ?v
e)From this information, determine the magnitude of the angular acceleration of the wheel.(Use the variables ?v, R, and T.)

Explanation / Answer

Tangential V on 1st. measured turn = sqrt.(2gh) = sqrt.(19.6 x 0.556), = 3.3m/sec.
Tangential V on next turn = sqrt.(19.6 x 0.51) = 3.162m/sec.
Perimeter of wheel = (pi 2r) = 2.4382m.
Rotational distance turn 1 = (3.3/2.4382) = 1.35346 turns.
Rotational distance turn 2 = (3.162/2.4382) = 1.29686 turns.
Difference = (1.35346 - 1.29686) = 0.0566 turn.
Angular difference = (360 x 0.0566) = 20.376 degrees.
Angular acceleration = (6.284/360) x 20.376 = -0.3567rads/s^2. approx.

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