A bicycle is turned upside down while its owner repairs a flattire. A friend spi

Question

A bicycle is turned upside down while its owner repairs a flattire. A friend spins the other wheel, of radius 0.377 m, and observes that drops of water fly offtangentially. She measures the height reached by drops movingvertically. A drop that breaks loose from the tire on one turnrises h = 54.7 cm above thetangent point. A drop that breaks loose on the next turn rises 51.0cm above the tangent point. The height to which the drops risedecreases because the angular speed of the wheel decreases. Fromthis information, determine the magnitude of the average angularacceleration of the wheel.
1rad/s2

Explanation / Answer

   from the given problem we can see that the first drophas velocity leaving the wheel given by    (1 / 2) m v12 = m gh1    v1 = (2 g h1)        = (2 x 9.80 m/ s2 x 0.547 m)        = ....... m /s    in the similar manner the second drop has avelocity    v2 = (2 g h2)        = (2 x 9.80 m/ s2 x 0.510 m)        = ....... m /s    we know that    v = r    so that    1 = v1 / r          =v1 / (0.377 m)          =....... rad / s    2 = v2 / r          =v1 / (0.377 m)          =....... rad / s    so the magnitude of the average acceleration ofthe wheel will be    = (22 -12) / 2       =(22 - 12)/ 4       = ........ rad /s2

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