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| *Vultures Knob>>>Bike Repair |
Tough one to figure.....Bike turned upside down? |
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. 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. |
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