A bicycle is turned upside down while its owner repairs a flat tire. A friend sp

Question

A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tangentially. She measures the heights reached by drops moving vertically (see figure below). A drop that breaks loose from the tire on one turn rises vertically 60.0 cm above the tangent point. A drop that breaks loose on the next turn rises 40.0 cm above the tangent point. The radius of the wheel is 0.395 m. (a) Why does the first drop rise higher than the second drop? This answer has not been graded yet. (b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant). (Indicate the direction with the sign of your answer. Take the clockwise direction to be positive.) rad/s2

Explanation / Answer

a) because the wheel is slowing down.

b) v1 = sqrt(2*g*h1) = sqrt(2*9.8*0.6) = 3.43 m

v2 = sqrt(2*g*h2) = sqrt(2*9.8*0.4) = 2.8 m

w1 = v1/R = 3.43/0.395 = 8.68 rad/s

w2 = v2/R = 2.8/0.395 = 7.1 rad/s

average angular speed, w = (w1+w2)/2

= (8.68 + 7.1)/2

= 7.89 rad/s

Time taken for one revolution, T = 2*pi/w

= 2*pi/7.89

= 0.796 s

angular acceleration of the wheel, alfa = (w2 - w1)/T

= (7.1 - 8.68)/0.796

= -1.98 rad/s^2

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