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 0.349 m, and observes thet 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 thet breaks loose from the tire on one turn rises h = 53.1 cm above the tangent point. A drop thet 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/s^2

Explanation / Answer

r1 =0.349 m , h1 =53.1 cm , h2 =51 cm

From kinematic equation for first drop

v^2 -u^2 = 2as

v1^2 =2*9.8*0.531

v1 = 3.23 m/s

v2^2 = 2*9.8*0.51

v2 =3.16 m/s

angular velocity w = v/r

w1 = v1/r = 3.23/0.349 = 9.255 rad/s

w2 = 3.16/0.349 = 9.054 rad/s

from rotational kinematic equation

w2^2 - w1^2 = 2(aplha)(theta)

(9.054)^2 -(9.255)^2 = 2(alpha)(2*3.14)

aplha = -0.293 rad/s^2

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