A bucket of water of mass 15.4kg is suspended by a rope wrapped around a windlas

Question

A bucket of water of mass 15.4kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.340m with mass 12.0kg . 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 11.0m 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 = 9.80m/s2 .

N

Part B

With what speed does the bucket strike the water?

Take the free fall acceleration to be g = 9.80m/s2 .

Part C

What is the time of fall?

Take the free fall acceleration to be g = 9.80m/s2 .

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 = 9.80m/s2 .

Explanation / Answer

mass= 15.4kg

mg-T=ma
Tr=.5 Mr^2 * a/R
T=.5Ma
=>a=mg/(m+.5M)=15.4*9.8/(15.4+.5*12)=7.0523m/s^2
=>a)T=.5*12*7.0523=42.3138N
b)v^2=2as=2*7.0523*11
=>V=12.4559m/sec
c)v=at
=>12.4559=7.0523*t
=>t=1.766sec
d)N=T+mg=42.3138+12*9.8=159.91N

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