A bicycle is turned upside down while nslowner repairs a nat tire on the rear wh

Question

A bicycle is turned upside down while nslowner repairs a nat tire on the rear wheel. A friend spins the front wheel of radius o 428 mw and observes that drops of water fly off tangentaty in an upward direction when the drops are at the same level as the center of the wheel, she measures the height reached by drops moving verticaly (see figure below) A drop that breaks loose from the tire on one turn rises h ss.1 cm above the tangent point. A drop that breaks loose on the next turn vises si o om above the tangent point. The height to which the drops rive decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel. Your response is within 10 of the correct value. This may be due to roundorf error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four digt accuracy to minimize roundoff error. rad/s? Need Help?

Explanation / Answer

Since at maximum heigh velocit is zero

So V1=sqrt (2*9.8*0.551)= 3.2863 m/s from ( 0=V1^2+2gh)

similarly

  V2=sqrt (0 -2*9.8*0.51)= 3.1616 m/s

Similarly V1=gt1 gives t1= 0.3353 seconds

V2=gt2 gives t2= 0.3226 seconds

time difference = 0.3353 - 0.3226 =0.0127 seconds during this time velocity changes from V1 to V2

we know that rate of change of velocity is accelaration

=(  3.2863 - 3.1616)/ 0.0127 = 9.8189 m/s2

we know accelaration= angular accelaration* radius

thus   angular accelaration=accelaration/ radius=9.8189/0.428= 22.9414 s-2

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