Definitions of basic quantities used in this analysis

To answer the questions posed here you need to know the following:

• The average rate of change of A with respect to B is  

average rate = (change in A) / (change in B).

• The volume of any region with a uniform cross-sectional area is the product of the cross-sectional area and the altitude or length of the region as measured perpendicular to the cross section.

It is assumed you know how to find the area of a circle.

It is assumed you know that an inch is about 2.54 centimeters.

Information about the pictures

You are given the following information about the pictures shown below:

• The small tube has an inner diameter of 1/4 inch.
• The region inside the container consists of a cylinder whose inner diameter is about 1 11/16 inch.
• The first 14 frames occur at intervals of 1/30 second. These frames are numbered 1-14.
• The remaining frames occur at intervals of 10/30 second, or 1/3 second.  They are numbered in intervals of ten (i.e., 24, 34, 44, ...).  These numbers represent the number 1/30-second intervals that have elapsed since frame 0 (frame 0 is not pictured).
• The person standing behind the system is wearing a blue shirt and khaki pants.  The pants have unoccupied belt loops, of this two are visible in the picture.  The distance between the belt loops is about 18 centimeters.

In this series of pictures you see water from the cylindrical container filling the small tube, then flowing out of the end of the tube and arcing towards the ground.

Questions

Using the given information:

• Investigate, with the best accuracy you can achieve, whether the liquid flowing through the tube is speeding up, slowing down, or maintaining a constant speed between frame 2 and the last frame before it first reaches the end of the tube.
• After reaching the end of the tube the liquid remains in a fairly coherent stream as it begins arcing toward the ground.  There are two frames in which you can clearly see the entire stream of the exiting liquid.  Determine as best you can whether during these two frames the liquid is moving faster, slower or at the same speed as it was while filling the tube.
• Figure out as nearly as possible the volume of the liquid required to fill the tube. 
• Use this result to figure out by how much the level of the liquid in the container must have decreased during the time required to fill the tube.

• Starting with the second frame, at what average rate with respect to clock time was water volume flowing into the tube?  At what average rate with respect to clock time was water flowing out of the container?
• While the tube filled, at what average rate was the level of the liquid in the container therefore changing, with respect to clock time?

• From the first frame to the last, it is clear that the level of the liquid in the cylinder fell (see the two pictures at the end).  Did the level fall faster and faster, did it fall more and more slowly, or did it fall at a constant rate?  What is your specific evidence for this and how did you use this evidence to support your conclusion?
• At what average rate with respect to clock time did the level of the liquid in the container fall between the first frame and the last? 
• Is this result consistent with your calculation of the average rate of change of water level during the filling of the tube.
• What evidence is there that the level of the liquid in the cylinder was changing with respect to clock time at a rate that was changing with respect to clock time? 
• Does the above question even make sense?  If it does, what in the world does it mean, and can you answer it with the information in the pictures?

1

3

side-by-side comparison of first and last frames

More good stuff

In the absence of dissipative forces the velocity of water flowing from a hole at depth y relative to the water surface is

• v = sqrt(1960 y),

where v is the velocity in cm/s when y is given in centimeters.

(Note:  In more general terms v = sqrt( 2 g y), where g is the acceleration of gravity and the units of v depend on the units used for g and y.  The given equation v = sqrt(1960 y) is based on g = 980 cm/s^2.).

How does the velocity of water in the tube compare to the velocity predicted by this equation?  What is the nature of the dissipative forces that hinder the motion of the water?

Under ideal conditions the arc of the water stream that exits the tube at constant velocity is parabolic in shape.  If the velocity of the outflowing water is changing this won't quite be the case; but if the velocity is changing slowly (on the time scale relative to the time required for the water to fall to the ground) the shape at any instant will be very nearly parabolic.  The equation of the parabola can be used to determine the speed of the exiting water.

By locating three points on a parabola it is possible to determine its equation.  If x and y are the horizontal and vertical coordinates of the parabola, then the equation of the parabola has form y = a x^2 + b x + c.  If the coordinates of a point are substituted for x and y we obtain an equation with unknowns a, b and c.  If we do this with three points on the parabola we will have three equations in a, b and c.  These three equations are solved simultaneously to obtain the values of a, b and c.  It is the value of a that determines the speed of the water.  The equation of the parabola is obtained by substituting these values of a, b and c into the form y = a x^2 + b x + c.

We get a different parabola for each of the frames.  There is significant uncertainty in measuring the coordinates of any given parabola from the picture; apart from issues of parallax, it was a breezy day, which affected the arc differently in each picture.  Despite this there is a clear trend to the values of a.

The question is, what does the value of a appear to be changing at a constant, and increasing or a decreasing rate with respect to clock time?