A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windla
ID: 1792597 • Letter: A
Question
A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.340 m with mass 11.0 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.6 m to the water. You can ignore the weight of the rope. Part A What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80 m/s2 . 32.89 N SubmitMy AnswersGive Up Incorrect; Try Again; 11 attempts remaining Part B With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80 m/s2 . m/s SubmitMy AnswersGive Up Part C What is the time of fall? Take the free fall acceleration to be g = 9.80 m/s2 . s SubmitMy AnswersGive Up Part D While the bucket is falling, what is the force exerted on the cylinder by the axle? Take the free fall acceleration to be g = 9.80 m/s2 .
Explanation / Answer
(a)
Aplly torque momnet of inertia relation
T = I alpha
F r= ( Mr^2/2) ( a/r)
F = ( M/2) a
(M/2) a= mg - ma
a( [M/2m] +1) = g
a = g / (1+ [M/2m]}
=9.8/ 1 + 11/28)
=7.03 m/s^2
T = m( g-a) = 14 ( 9.8-7.03) = 38.69 N
(b)
v= sqrt 2gh = sqrt 2 ( 9.8) 10.6 = 14.41 m/s
(c)
t= v/a = 14.41 m/s/7.03 = 2.05 s
(d)
he force exerted on the cylinder by the axle is
F _ axle = Md+ F = 11(9.8) + (5.5) ( 7.03) = 146.46 N
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