If you open a faucet, often it produce a a clear, circular stream of water that

Question

If you open a faucet, often it produce a a clear, circular stream of water that gets narrower as it gets further from the faucet. We can use continuity to find the shape of this stream. Assume the stream initially has a radius of 1 cm and is moving at 0.5 m/s downward. What will the stream's velocity be 30 cm below the tap? (Use energy or kinematics - both work.) Using continuity, what is the radius of the stream at that point? In this way, we can leave the change in height as a variable and find that the radius of the stream changes as 4 square root |Delta y|. (You don't have to do this, but it is kinda neat to know.)

Explanation / Answer

a)

Conserving the envergy

Energy of water at the faucet = mgh + 1/2*m*v^2

= m*g*0.3 + 1/2*m*0.5^2

Energy of water 30 cm down :

= m*g*0 + 1/2*m*v^2

Conserving Energy :

m*g*0.3 + 1/2*m*0.5^2 = m*g*0 + 1/2*m*v^2

Cancelling m

g*0.3 + 1/2*0.5^2 = 1/2*v^2

v^2 = 6.13

v = 2.476 m/s

b) Not very sure about this part.

R = k*(y)^1/4

At the faucet :

R = k*(0.3)^1/4

R = 0.74 cm

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