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.
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).

Explanation / Answer

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. now,plug the numericals.you will answers to your question.

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