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). A drop that breaks loose from the tire on one turn rises vertically 54.0 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The radius of the wheel is 0.363 m.

(a) Why does the first drop rise higher than the second drop?

(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).

rad/s2

Explanation / Answer

v1 = 2gh1 = (2*9.8*0.54) = 3.2533 m/s

v2 = 2gh2 = (2*9.8*0.51) = 3.16165 m/s

w1 = v1/R = 3.2533/0.363 = 8.96226 rad/s

w2 = v2/R = 3.16165/0.363 = 8.7098 rad/s

w2^2 - w1^2 = 2

8.7098*8.7098 - 8.96226*8.96226 = 2 * * (-2*3.1416)

==> = 0.355 rad/s2

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