Tough one to figure.....Bike turned upside down?

Question

and the owner is repairing a flat tire. A friend spins the other wheel (radius r=0.4m) and observes drops of water fly off tangentially. They measure the height reached by the drops moving vertically. A drop that breaks loose from the tire on one turn rises h1 = 78 cm above the tangent point. A drop that breaks loose on the next turn rises h2 = 75 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. Acceleration of gravity is 9.8 m/s^2 and angular decelaration is constant. Find the magnitude of the angular deceleration fo the wheel in units of rad/s^2.

Answer 1

Height reached by water drop = h
then since at top v =0
use conservation of energy to find initial speed of drop (when it left the tire)
mgh = 1/2mv^2 or v = sqrt(2gh)

then angular velocity of wheel is w = v/r (since the drops were attached to the wheel before flying off, and had the same angular velocity as the wheel).

calculate w for both heights, lets call that w1 and w2. In one turn the wheel travels 2 pi radians.
Therefore
2 * alpha * 2*pi = (w2)^2 -(w1)^2

where alpha is the angular deceleration. The above is one of the three kinematic equations for rotation.