Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the pipe with a speed of 2.0 m/s. What is the diameter of the column of water as it hits the bucket?

Question

Diagram showing vegetable oil flowing from a container into a funnel and bottle, illustrating fluid mechanics.

Accepted Answer

Hey, everyone, here's a problem that relates to a stream of water that has a changing radius and a changing speed. So the first thing I'm going to do is read the problem and make a note of the givens. But as I do so note that I'm going to use the subscript one to refer to variables pertaining to the flow as it comes out of this little bowl. And I'm going to use the subscript two to refer to the variables pertaining to the flow just as it enters this bottle at the bottom here. OK. Right. So aiming to build its own vegetable oil bottling machine. A student designed following setup, the oil drips vertically downwards from a one centimeter diameter tube, so placed at the bottom of the main reservoir into the bottles. So that means that the bottom of this bowl here, the tube where the oil comes out, the diameter is equal to one centimeter. It says the height difference between the tube's nozzle and the bottle's top is 1.0 m. So I'm gonna use L to refer to one m and that meter is actually shown in the diagram here. If the oil leaves the tube nozzle at a speed of 1.5 m per second, determine the radius of the oil stream as it enters the bottle. So we're looking for our sub two. So the first thing I'm going to do is clean up our givens a bit because we're looking for a radius, but we're only given a diameter. So recall that radius is equal to a diameter divided by two. So that means that R one is equal to one centimeter divided by two. So it's equal to 20.5 centimeters or alternatively, if you want to convert that into meters 0. m. Now, because we're dealing with a flow with a changing radius, we're going to have to use the continuity equation which states that when you have a flow, its cross sectional area times its speed is always constant. So in other words, a one V one must be equal to a two V two. I also recall that the area of a circle is equal to pi multiplied by the square of the circle's radius. Let me make that two look a little neater. There we go. So we can expand this formula out a bit by writing pi R one squared V one equals pi R two squared V two. Then these pis can cancel out since they exist in both term. Now R two is what we're trying to find. So I'm going to algebraically solve this for R two by dividing both sides of the equation by V two. And then squaring both sides of the equation to get rid of the square root here. So R two squared is equal to R one squared times V one over V two. And then taking the square to both sides, R two is equal to R one times the square root of V one over V two. So here's an equation for R two. And it, and it almost works because we have R one given to us in the problem. And V one is also given to us in the problem, but we don't have V two. So we're going to have to come up with some way to get that. And we can do that using our old kinematics equations. Now because we're looking for the final speed and the problem gives us the initial speed and the distance the flow travels, we can use the kinematics equation that states that the square of the, of the final speed is equal to the square of the initial speed plus two times the acceleration times the distance traveled between them. Now, for the case of our problem, we can say that V F is V two, V I is V one. Since we're looking at a flow that's falling A is just G and delta X, the distance the flow travels is for us going to be L. So all we gotta do is plug this in for V two in the equation we found earlier for R two. So I'm going to do that. Now R two is equal to R one times the square root A V one all divided by the square root A V one squared plus two G L. And that should be the formula to solve for V two. We simply just take the square root of both sides of the equation. So that V two is equal to V one squared plus two G L. And now we just plug that in to V two in our R two equation. So I'm gonna do that now. And you find that R two is equal to R one times the square root of V one divided by then I'm gonna switch colors again the square root of V one squared plus two G L. And that should be it. That should be all we have to do because now this equation contains nothing but our givens. So let's try it. So R one is equal to 10.5 m times the square root of V one which is 1.5 m per second, 0.5 m per second divided by the square root of 1.5 m per second. And that term is square plus two times G which is just 9.8 m per second squared times L which is one m, gotta extend these lines out. And if you put that into your calculator then you should find a value for R two of about 2.8 times 10 to the negative five m. But if you look at the options we have in the multiple choice, you'll notice that every single option is written in millimeters. So if you want to get our answer, we're gonna have to take what we found and multiply it by the chain link conversion to convert it from meters into millimeters. So let's do that. I'm gonna multiply it by the chain link conversion that says that there are 1000 millimeters in one m and this gives us an answer of 2.8 millimeters. And now if we look at our options, we'll notice that we do indeed have that as an option as option B So that is the answer to this problem. That should be it. I hope this video helped you out if you'd like to learn more, please check out some of our other videos and I hope you all have a lovely day. Bye bye.

Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the pipe with a speed of 2.0 m/s. What is the diameter of the column of water as it hits the bucket?

Verified Solution

Key Concepts

Continuity Equation

Bernoulli's Principle

Kinematic Equations of Motion