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 radius0.391m, 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 risesh=53.3cm 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.
rad/s2

Explanation / Answer

Relevant equations .5mvi^2=mgh1 w=v/r wf^2=wi^2+2(alpha)(thetafinal-thetainitial) 3. The attempt at a solution r=0.4 m h1= 0.498 m h2=0.468 m Change of theta=2pi (I'm a little confused on this. Would it be 2pi or 4pi?) I got v1=3.124227905 m/s v2=3.028663071 m/s w1 = 7.81059763 w2 = 7.571657678 w2^2=w1^2+2(alpha)(changetheta) 7.57165678^2=7.810569763^2+2(2pi)a a=-0.29244721 rad/s^2

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