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 0.386 m, and observes that drops of water fly off tangentially in an upward direction when the drops are at the same level as the center 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 = 54.6 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

For the first revolution Let the angular speed be w1 so velocity of drops separating from wheel = w1 * r = 0.386 w1 now height travelled = (v^2) /2g = (0.386 w1)^2 ) / 2g = 0.546 m so, w1 = 8.475 rad / sec For the next revolution Let the angular speed be w2 so velocity of drops separating from wheel = w2 * r = 0.386 w1 now height travelled = (v^2) /2g = (0.386 w1)^2 ) / 2g = 0.51 m so, w1 = 8.19 rad / sec as it was after one complete rotation of the wheel so , angle travelled = 2*pi = 6.28 rad now , w2 ^2 = w1^2 - 2*(alpha)*angle travelled alpha = angular acceleration (retardation in this case ) so, 8.19^2 = 8.475^2 - 2 * alpha * 6.28 so, alpha = 0.378 rad/sec2

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