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, of radius 0.357 m, and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically (Fig. P10.63). A drop that breaks loose from the tire on one turn rises h = 53.1 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. _______ rad/s^2

Explanation / Answer

Relation between linar velocity and angualr velocity is

V = r W

where r is radius = 0.357 m

from Conservation of eenrgy

KE = PE

0.5 mv^2 = mgh

v^2 = 2 gh

V1^2 = 2* 9.81 * 0.531

V1 = 3.22 m/s

V2^2 = 2* 9.81 * 0.51

V2 = 3.163 m/s

timelag of 1 turn will make a diff of pth by 2pi radians

so

Wf = 3.163/0.357 = 8.85 rad/s

Wi = 3.22/0.357 = 9.01rad/s

using wf^2 - wi^2 = 2 Alpha * theta

Alpha = angualr accleration = (8.85^2 -9.01^2)/(2 * 2*pi)

Alpha = -0.227 rad/s^2

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