Posted by Ed on .
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 from point A. She measures the heights reached by drops moving vertically (see figure). A drop that breaks loose from the tire on one turn rises vertically 0.54 m above the tangent point. A drop that breaks loose on the next turn rises 0.47 m above the tangent point. The radius of the wheel is 0.44 m. 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).
The wheel is decelerating in angular velocity. If the drops rise a distance H, they were released with a velocity V such that
V = sqrt(2gH)
For turn 1, V1 = 3.255 m/s
Fot turn 2, V2 = 3.037 m/s
The average wheel velocity between the release of the two drops is
Vav = 3.146 m/s.
The elapsed time between release of the two drops is
T = 2 pi R/ Vav = 0.88 s
The angular acceleration rate is
(1/R)(V2 - V1)/T = __ rad/s
Thank you for helping! got the answer correct (: