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## Homework Statement

A 15 kg bucket of water is suspended by a rope wrapped around a windlass, that is a solid cylinder .3 m in diameter with mass 12 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of the well and falls 10 m to the water. You can ignore the weight of the rope.

a) what is the tension in the rope while the bucket is falling?

b)with what speed the bucket strike the water?

c) what is the time of the fall?

d) while the bucket is falling what is force exerted on the cylinder by the axle

## The Attempt at a Solution

a)Tension is equal to the torque of the cylinder while is equal to its moment of inertia I times its acceleration a. The tension is also equal to the weight of bucket minus the downward force that accelerates the system:

T=aI ==>a=T/I

T=m(g-a)

I=mr^2/2=(12)(.15^2)/2=.135

Since a is unknown, substitute T/I. After simplifying:

T=(mgI)/(m+I)=(15)(9.8)(.135)/(15+.135)=1.31N

b)Initially all energy is gravitational potential so the final energy must equal the kinetic of the bucket and the cylinder.

mgd=.5mv^2+.5I(v/r)^2

v=sqrt(2mgd/(m+I/r^2)=sqrt(2(15)(9.8)(10)/(15+(.135/.15^2))=11.8 m/s

c)x-xi=.5(v-vi)t

t=2(x-xi)/(v-vi)=2(10)/(11.8)=1.69s

d)The upward force exerted by the axle would just be the force to fight gravity, so it would be equal to the weight of the cylinder but in the opposite direction. But if thats the case would that force serve to slow down the cylinder? and wouldn't that affect all of the above calculations which I'm not sure are correct so thanks for the look over.