v1=√(2gh)=√(2*9.8*0.53)=3.223 m/s
v1=√(2gh)=√(2*9.8*0.53)=3.162 m/s
Let ω=angular velocity
r=0.363 m
ω1=v1/r=8.879 rad/s
ω2=v2/r=8.710 rad/s
Time for one turn between t1 & t2
Δt=2π/((8.879+8.710)/2)
= 0.7145s
Angular acceleration
=(ω2-ω1)/Δt
= -0.237 rad/s²
A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel, of radius 0.363 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 = 53.0 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.
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