course Phy 201
The position vs. clock time graph
We extensively discussed the position vs. clock time graph in class.
In class I saw a number of carefully constructed graphs, many with sufficient detail to give good results. Some of the graphs did not have particular smooth curves, and/or were too small to yield good results.
If necessary you should sketch another graph, of sufficient size and accuracy to give you reasonably good results. Refer to the instructions given for the preceding class.
Using your graph estimate the following. You may use cycles of your pendulum, or half-cycles, as your time unit, or you can convert these to seconds.
What was the length of your pendulum (you can give this in centimeters, inches, miles, textbook widths, lines on your notebook paper or whatever units are convenient, as long as these units can later be measured in centimeters)? &$&$ 7cm.
What is the change in position corresponding to the first half of the time interval corresponding to motion down the incline (we will use `dt_total to refer to this time interval)? &$&$ In class we sketched the graph and used a horizontal arrow to represent the time down the ramp. Then we used horizontal arrows to represent the first half of the time, and the second half of the time down the ramp. If you were to do the same thing with your graph, you would be able to estimate how much change in clock time corresponds to each arrow. This would not necessarily help you answer this question, but would serve to help orient you to the graph.
We then used vertical arrows to estimate the corresponding positions on the graph. Based on estimates of the distance corresponding to each vertical arrow, you can estimate the position of the ramp at each clock time. Based on your estimates you can estimate the distances requested in this question and the next.
Answer the same for the second half of the interval. &$&$
Darken the part of the graph which corresponds to motion down the fourth ramp. For this interval estimate the change in position and the change in clock time. &$&$
Mark the point of the graph that corresponds to the ball's first contact with the seventh ramp. Give the coordinates of that point. &$&$ 7.5, 6
Do the same for the ball's last contact with the seventh ramp. &$&$ 7.8,7
What is q_rise between these points (recall that q_rise stands for 'the quantity represented by the rise')? &$&$ 7.5/7.8=.961
You get the rise by subtracting, not by dividing the two 'graph altitudes'.
W$hat is q_run between these points? &$&$ 6/7=.857
You get the run by subtracting, not by dividing the two horizontal coordinates.
Note also that the 6 ft and 7 ft positions are on the 'vertical' axis, not the horizontal. So the rise is determined by these coordinates. Similarly the run is determined by your clock times 7.5 seconds and 7.8 seconds.
Your estimates 7.5 sec and 7.8 sec are perfectly reasonable, so if you use subtraction and correct the order of your rise and run, you will get a very good result.
$What therefore is q_slope between these points? .961/.857=1.121
Mark on your graph the points corresponding to the transitions from one ramp to the next (i.e., the ball leaves one ramp and first encounters the other at the same instant; mark each on the graph at which this occurs).
Sketch a series of short straight line segments connecting these points.
Find q_rise, q_run and q_slope for each of these line segments. Report q_rise, q_run and q_slope, in that order and separated by commas, starting in the line below. Report three numbers in each line, so that each line represents the quantities represented by the rise, run and slope of one of your segments.&$&$
The v0, vf, `dt trapezoid
The altitudes of a certain graph trapezoid are symbolically represented by v0 and vf, indicating initial and final velocity.
The base is represented by `dt, the change in clock time t. The base therefore represents the time interval `dt.
Sketch a graph trapezoid. Label its altitudes v0 and vf and its base `dt.
Now answer the following questions:
If v0 = 5 meters / second and vf = 13 meters / second, with `dt = 4 seconds, then
What is the rise of the trapezoid and what does it represent? &$&$ The rise is Vf-Vo=13-5=8 Good, but that would be 13 meter/second - 8 meters/second = 5 meters/second. The units are essential.
What is the run of the trapezoid and what does it represent? &$&$ 4 Good, but the run would be 4 seconds, not just the number 4.
What is the slope of the trapezoid and what does it represent? &$&$ 8/4=2 Very good, but this would be (8 meters/second) / (4 seconds) = 2 meters / second^2. Many students will still be a little 'fuzzy' with the units; students who are should do a good review of division of fractions.
What are the dimensions of the equal-area rectangle and what do they represent? &$&$ 6.5*4 The altitude will have the same units as the 'graph altitudes' of the trapezoid. The 'graph altitudes' are expressed in meters/second, so the altitude of the equal-area rectangle will be meters/second.
The altitude of the equal-area rectangle would be halfway between the altitudes 5 meters/second and 13 meters/second. Halfway between 5 and 13 is 9 (most easily calculated by adding 5 and 13 and dividing by 2). The altitude of the equal-area rectangle is therefore 9 meters/second.
The width is not 4 but 4 seconds.
So the area is 9 meters/second * 4 seconds = 36 meters.
Be sure to review multiplication of fractions so you will fully understand the units calculation.
What therefore is the area of the trapezoid and what does it represent? &$&$ 6.5*4=26 area of trapezoid The vertical coordinate of this graph represents velocity in meters / second. The horizontal coordinate represents clock time in seconds.
So the average 'graph altitude' represents the average velocity in meters / second, and the base of the trapezoid represents the change in clock time in seconds.
Average velocity is average rate of change of position with respect to clock time, which is (change in position) / (change in clock time). When average velocity is multiplied by change in clock time we get
average velocity * change in clock time = (change in position) / (change in clock time) * (change in clock time) = change in position.
In terms of just the symbols v0, vf and `dt:
What expression represents the rise? &$&$ Vf-Vo
What expression represents the run? &$&$ ‘dt
What expression therefore represents the slope? &$&$ Vf+Vo/2 The slope is rise / run. You correctly identified the rise as vf - v0, and the run by `dt. However your result (vf + v0) / 2 is not what you get when you divide your rise by your run.
What expression represents the width of the equal-area rectangle? &$&$ ‘dt
What expression represents the altitude of the equal-area rectangle? &$&$ Vo vo is one of the two 'graph altitudes' of the trapezoid. The equal-area rectangle has an altitude which is halfway between the two 'graph altitudes'.
The two widths are v0 and vf.
What quantity is halfway between v0 and vf? (review how the number halfway between 5 and 13 was obtained in the preceding example).
What expression therefore represents the area of the trapezoid? &$&$ Vo*’dt The 'graph width' is `dt. If the altitude of the equal-area rectangle was v0, then the area would indeed by v0 * `dt. Howevr the altitude of that rectangle is not v0.
What is the altitude of the equal-area rectangle, and what therefore is its area?
What is the meaning of the slope? &$&$ How step the trapezoid is
The meaning is interpreted in terms of the situation.
What is the meaning of the rise? Is it the average velocity, the average clock time, the change in velocity, the change in clock time, the initial velocity, the initial clock time, the final velocity, the final clock time, etc.?
What is the meaning of the run? Is it the average velocity, the average clock time, the change in velocity, the change in clock time, the initial velocity, the initial clock time, the final velocity, the final clock time, etc.?
What therefore is the meaning of the slope, which is obtained by dividing the rise by the run?
What is the meaning of the area? &$&$ how much space is in the trapezoid
The area is the product of the base and the altitude of the equal-area rectangle.
How was the altitude of this rectangle obtained, and what does it therefore represent in terms of the motion?
What is the base of the rectangle and what does it therefore represent in terms of the motion?
If the ball on the ramp changes its velocity from v0 to vf during time interval `dt, then
If you have numbers for v0, vf and `dt how would you use them to find the following:
the change in velocity on this interval &$&$ Vf-Vo
the change in clock time on this interval &$&$ ‘dt
the average velocity on this interval, assuming a straight-line v vs. t graph &$&$ your answers to the two questions before this and the question after this are correct, but you didn't answer this question
the average acceleration on this interval &$&$ Vf-Vo/’dt
the change in position on this interval &$&$ vf-vo/2
vf and v0 are velocities.
When you subtract two velocities you get a velocity.
When you divide a velocity by 2 you get a velocity which is half as great.
So when you subtract two velocities and divide by 2 you get a velocity.
A velocity is found by calculating (change in position) / (change in clock time). So a velocity can't be a position or a change in position.
To get change in position, you could multiply (change in position) / (change in clock time) by (change in clock time), which is the same as multiplying an average velocity by an change in clock time.
What are the average velocity and the change in clock time, and what therefore do we get when we multiply the average velocity by the change in clock time?
In terms of the symbols for v0, vf and `dt, what are the symbolic expressions for each of the following:
the change in velocity on this interval &$&$ Vf-Vo
the change in clock time on this interval &$&$ Vf
the average velocity on this interval, assuming a straight-line v vs. t graph &$&$ v0-vf This is the change in velocity, not the average velocity.
the average acceleration on this interval &$&$ V0/Vf Average acceleration is (change in velocity) / (change in clock time). You can't get an acceleration by dividing one velocity by another.
What is the expression for change in velocity?
What is the expression for change in clock time?
What therefore is the expression for average acceleration?
the change in position on this interval &$&$ Graph Area, vf+vo/2
The graph area requires that you multiply average 'graph altitude' by the base. (vf + v0) / 2 is part of the correct answer. What is the missing part, and what therefore is the result?
How are your answers to the above questions related to the v0, vf, `dt trapezoid? &$&$ V0, initial velocity, VF final velocity, ‘dt, time it takes
If v0 = 50 cm / sec and vf = 20 cm / sec, and the area of the trapezoid is 140 cm, then
What is the rise of the trapezoid and what does it represent? &$&$ 50/2=2.5
What are the two 'graph altitudes'?
How do you use the 'graph altitudes' to get the rise?
What therefore is the rise of this trapezoid?
What is the altitude of the equal-area rectangle? &$&$ Vf-Vo
The altitude of the equal-area rectangle is halfway between the two altitudes.
The two altitudes are v0 = 50 cm/s and vf = 20 cm/s.
What is halfway between these two quantities?
Can you use one of your answers, with the given area, to determine the base of the trapezoid? &$&$ Yes.
'Yes' and 'no' are both wrong answers to any question, if they are unsupported. You need to say why the answer is 'yes'.
Can you now find the slope of the trapezoid? &$&$ Yes. Vf/vo
vf is not the rise and v0 is not the run, so vf / v0 is not the slope.
The 'run' is the base. You have to figure out the base in order to figure out the slope.
This is a challenging question, and most students won't get it on the first try. See if you can first correct as many of your original answers as necessary. Then spend a reasonable amount of time trying to answer these last two questions, but don't let yourself get bogged down on this one.
Lego toy car:
As shown in class on 090831, a toy car which moves through displacement 30 cm in 1.2 seconds, ending up at rest at the end of this time interval, has an average rate of change of position with respect to clock time of 25 cm / s, and by the definition of average velocity, this is its average velocity. If its v vs. t graph is a straight line, we conclude that its velocity changes from 50 cm/s to 0 cm/s during the 1.2 seconds, and the average rate of change of its velocity with respect to clock time is therefore about -41.7 cm/s.
The same toy car, given an initial push in the opposite direction, moves through displacement -60 cm in 1.5 seconds as it comes to rest. If you previously submitted the correct solution to this situation you found that the acceleration of this car was + 53.3 cm/s^2, approx.. If you didn't get this result, then you should answer the following questions (if you got -53.3 cm/s^2 and know what you did wrong to get the negative sign, you can just explain that): &$&$
Using the definitions of average velocity and average rate of change, determine the average velocity of the car during this interval. Explain completely how you got your results. &$&$ To get average rate of change you divide 30/1.2 which equals 1.2seconds. And the average rate of change of its velocity with respect to clock time is -41.7 50/1.2
'This interval' refers to the interval on which the displacement is -60 cm and the change in clock time is 1.5 seconds.
Your answer to this and the following two questions applies to the interval on which the car moves 30 cm in 1.2 seconds. You were given the reasoning used to analyze this interval, and can analyze the present interval by imitating that reasoning.
Describe your graph of velocity vs. clock time for this interval, give the altitudes of the corresponding v vs. t trapezoid and verify that the average altitude of this trapezoid is equal to the average velocity you obtained in the preceding step. &$&$ my graph of velocity vs. clock time is that its is going in a negative direction with the like going \. Because altitude represents velocity Your description should include the two altitudes and the width of the trapezoid, as well as its slope and area. You should specificy the two altitudes and the width, then explain how you use them to get the slope and area.
What is the car's initial velocity, its final velocity, and the change in its velocity on this interval? &$&$ 30, 0, 1.2
What therefore is its acceleration on this interval? &$&$ -41.7
Vertical rotating strap, ball on incline with magnets:
You are asked here to speculate on and think about the behavior of a couple of fairly complicated systems. These systems are complex enough that you could easily get carried away and spend weeks on your answers. Unless you just can't help yourself, limit yourself to 1/2 hour, or 1 hour at the most. You might spread that out over a few days to let you brain subconsciously sort out these ideas:
The rotating-strap system with the magnets is attracted to the straps on the table. At some points of its rotation the magnetic force exerted by the straps on the magnets tends to speed the system up, at other points it tends to slow the system. Obviously you aren't yet expected to know how to analyze this system (and a complete analysis is beyond the scope of first-year physics), but there are things about this system we will be able to reason out with the ideas we will be developing over the next few weeks. Just to get the process started, give me your best answers on to the following questions:
Describe in words how the system is oriented when the magnetic force acting on it is speeding it up. &$&$ Well the force is trying to connect to magnetic so it is speeding up.
Describe in words how the system is oriented when the magnetic force acting on it is slowing it down. &$&$ the slow down has a greater effect because the closer to the magnetic force it will slow it down.
At what position do you think the magnetic force is speeding it up the most? How could we experimentally test whether this is the case or not? &$&$ Well the closer you have it the slow down has a greater effect so it would only make since the further away it would speed it up.
At some points the magnetic interaction speeds the system up, and at some points it slows the system down. Which do you think has the greater effect? That is, do you think net effect of the presence of the magnetic force is to speed the system up or to slow it down? &$&$ The slow down has a greater effect because when observing you can actually see the slow down happen and I had to have someone point up the speed up part.
Do you think the net effect of the magnetic force is to increase or decrease the frequency of the oscillation? &$&$ I would say they both cancel each other out because you have and equal negative and positive effect. If you had the slowdown end up higher it would increase the number of oscillations, or if you had the faster part higher then it would decrease oscillations, it just depends on how you have the magnet.
Is it possible that the magnetic force slows the system down but increases its frequency of oscillation? &$&$ Yes. Because the slow down process may be something you cant even point out.
Does the system act like a pendulum in that the time required for a cycle is pretty much constant? How would we test this? What might we expect to find? &$&$
Comment also on what you think happens as the ball on the incline interacts with the magnet, and how we might test some of your ideas. &$&$ yes it does act like a pendulum. Test it with a pendulum, that the time required is either constant or not. If the ball interacts with the magnet on a incline it looks like it would slow it down because you are going against gravity, you could test it using a pendulum and see how many oscillations its takes.
Good job with your speculations.
You have a number of correct answers and procedures, and some are incorrect.
I've inserted a number of notes commenting on some of your answers and expanding on a number of the questions, breaking them down a bit more. If you have time to submit your questions and/or revisions into a copy of this document, marking your insertions with ####, I'll be glad to give you additional feedback.