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 0.344 m.

(a) Why does the first drop rise higher than the second drop?

(b) 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

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). ____ rad/s2 Physics - drwls, Monday, November 30, 2009 at 6:41am 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 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 0.44 m above the tangent point. A drop that breaks loose on the next turn rises 0.58 m above the tangent point. The radius of the wheel is 0.40 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). Physics - MathMate, Friday, November 27, 2009 at 1:00pm The key is to find the tangential velocities at each of the turns, v1 and v2. The initial velocity v for an object thrown upwards will reach a height of h is related by the formula: h=v²/2g where g is acceleration due to gravity. Conversely, given h, we can find v from: v=v(2gh) So the tangential velocities v1 and v2 can be determined from h1 and h2 (work in metres). Since the tangential velocities v1 and v2 are known, the average tangential velocity is va=(v1+v2)/2 The time it took to make the turn was t=2pr/va From angular velocity = v/r radians/sec We can determine angular acceleration, a =(v2-v1)/r/t Solve for a. I get about 1.4 rad./s. Check my calculations

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