a bucket of water of mass 14.7 is suspended by a rope wrapped around a windlass,

Question

a bucket of water of mass 14.7 is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.280m with mass 11.6 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 10.3m 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/s squared

Part B
with what speed does the bucket strike the water?
-Take the free fall acceleration to be g=9.80 m/s^2

Part C
What is the time of fall
-Take the free fall acceleration to be g=9.80m/s^2

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/s^2

Explanation / Answer

Torque t = I a =F*r where F is the tension in the rope. and a is the angular acceleration and a = a /r where a is the linear acceleration. (Mr²/2) (a/r) = F*r F = (M/2) a F is also = m (g-a) (M/2) a= mg - ma a( [M/2m] +1) = g a = g / (1+ [M/2m]} a = 9.8/ (1+ (11.6/29.4)) = 7.05 m/s^2 1. Tension F = 14.7*(9.8-7.05) = 40.41 or F = (M/2) a = 5.8*7.05 = 40.89N v = v (2ah) = v (2*7.05*10.3) = 12.05 m/s 3. t = v/a = 12.05/7.05= 1.70m/s² 4 Force on the axle = Mg + F = 11.6*9.8 + 40.89 = 154.48 N

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