A bicycle is turned upside down while its owner repairs a flat tire on the rear

Question

A bicycle is turned upside down while its owner repairs a flat tire on the rear wheel. A friend spins the front wheel, of radius 0.432 m, and observes that drops of water fly oft tangentially in an upward direction when the drops are at the same level as the canter of the wheel. She measures the height reached by drops moving vertically (see figure below). A drop that breaks loose from the tire on one turn rises h = 55.8 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel.

Explanation / Answer

Here, tangential V on 1st. measured turn = sqrt.(2gh) = sqrt.(2x9.81x 0.558), = 3.31 m/sec.
Tangential V on next turn = sqrt.(2x9.81x 0.51) = 3.16 m/sec.
Perimeter of wheel = (pi 2r) = 3.14x2x0.432 = 2.713 m
Rotational distance turn 1 = (3.31/2.713) = 1.22 turns.
Rotational distance turn 2 = (3.16/2.713) = 1.165 turns.
Difference = (1.22 - 1.165) = 0.055 turn.
Angular difference = (360 x 0.055) = 19.8 degrees.
Angular acceleration = [(2*pi)/360]xAngular difference = (6.284/360) x 19.8 = 0.3456 rads/s^2

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