course phy 201
about 5:15 am 9/9/09, vvvv
ps yes i know im a procrastinator" "Class 090902
Delta notation for 'change in'
`d is the symbol for the Greek capital letter Delta, the triangle thing you saw me write on the board. It has other meanings, but in any context involving rate of change, Delta pretty much universally stands for 'change in'.
• If we were to write `dA, it would be read 'delta A' of 'change in A'.
• If we were to write `dA / `dB, it would read 'change in A divided by change in B'. Thus the average rate of change of A with respect to B is
ave rate = `dA / `dB.
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)?
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• &$&$ my pendulum was the size of one of the tiles in the classroom floor
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• 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)?
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• &$&$ based on my graph it appears as if over the first 2.75 half cycles (the total was 5.5) the ball rolled approx. 3.25 feet.
• Answer the same for the second half of the interval.
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• &$&$ Over the final 2.75 ticks (half cycles) it traveled the remaining 6.75 feet (10 - 3.25).
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• 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.
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• &$&$ Well considering each ramp is 1 foot long, I assume the change in position is simply 1 foot. It also appears that it took about .5 ticks (between 3.1 and 3.6 overall)
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• Mark the point of the graph that corresponds to the ball's first contact with the seventh ramp. Give the coordinates of that point.
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• &$&$ (4.75,7)
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• Do the same for the ball's last contact with the seventh ramp.
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• &$&$ (5.05, 7.99)
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• What is q_rise between these points (recall that q_rise stands for 'the quantity represented by the rise')?
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• &$&$ .99
`dg11
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• What is q_run between these points?
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• &$&$ .3
`dg12
• What therefore is q_slope between these points?
&$&$ q slope = .99/.3
`dg13
• 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.
•
• &$&$
• 1,1, 1/1
• 1,.85, 1/.85
• 1,.65, 1/.65
• 1,.6, 1/.6
• 1,,55, 1/.55
• 1,.5, 1/.5
• 1,.45, 1/.45
• 1,.4, 1/.4
• 1,.3, 1/.3
• 1, .2, 1/.2
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?
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• &$&$ 13-5 = 8. This represents the change from the initial velocity to the final V, aka how much faster something got over a given time.
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• What is the run of the trapezoid and what does it represent?
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• &$&$ 4, the amount of clock time elapsed
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• What is the slope of the trapezoid and what does it represent?
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• &$&$ 8/4 = 2 this means that for every second passed the velocity increases 2 meters/second
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• What are the dimensions of the equal-area rectangle and what do they represent? &$&$ 13 + 5 = 18, 18/2 = 9…… 4 (base) x 9 (altitude)
&#You appear to understand, though there wasn't quite enough detail in your work and self-critiques for me to be sure. Be sure to let me know if you have specific questions or need me to clarify anything. &#
• What therefore is the area of the trapezoid and what does it represent?
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• &$&$ 36 meters/ second^2
The units of your calculation do not come out meters / second^2. You need to include the details of your calculation, which should include units.
In terms of just the symbols v0, vf and `dt:
• What expression represents the rise?
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• &$&$ vf - v0
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• What expression represents the run?
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• &$&$ dt
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• What expression therefore represents the slope?
• &$&$ (vf-v0)/dt
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• What expression represents the width of the equal-area rectangle?
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• &$&$ dt
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• What expression represents the altitude of the equal-area rectangle?
• &$&$ (v0 + vf)/2
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• What expression therefore represents the area of the trapezoid?
• &$&$ ((v0 + vf)/2) * dt
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• What is the meaning of the slope?
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• &$&$ the amount of speed gained or lost during the interval given
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• What is the meaning of the area?
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• &$&$ Total distance traveled during the given interval
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 &$&$ 10 feet/ 5.5 ticks = 1foot/ .55 ticks
• the change in clock time on this interval &$&$ 5.5 ticks - 0 = 5.5
• the average velocity on this interval, assuming a straight-line v vs. t graph &$&$ 1ft/.55 ticks
• the average acceleration on this interval &$&$ -1ft/ .55 ticks
• the change in position on this interval &$&$ 10 ft
• 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-v0
• the change in clock time on this interval &$&$ dt
• the average velocity on this interval, assuming a straight-line v vs. t graph &$&$ (vf-v0)/dt
• the average acceleration on this interval &$&$ (v0-vf)/dt
• the change in position on this interval &$&$ (vf-v0)/2) x dt
(vf-v0)/2) is half the change in velocity; it's not the average velocity. Common error, which I beleive you understand.
• How are your answers to the above questions related to the v0, vf, `dt trapezoid? &$&$ the altitudes are equal to feet traveled, the base is equal to amount of pendulum ticks.
Altitudes are velocities, which aren't measured in feet traveled but in feet/second.
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? &$&$ -30 cm/s the speed decreases 30 cm/ sec over the given time period
• What is the altitude of the equal-area rectangle? &$&$ 35 cm/s
• Can you use one of your answers, with the given area, to determine the base of the trapezoid? &$&$ 4 sec
• Can you now find the slope of the trapezoid? &$&$ -30/4 = -15/2
Introductory Problem Sets
Work through Introductory Problem Set 1 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 1). You should find these problems to be pretty easy, but be sure you understand everything in the given solutions.
You should also preview Introductory Problem Set 2 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 2). These problems are a bit more challenging, and at this point you might or might not understand everything you see. If you don't understand everything, you should submit at least one question related to something you're not sure you understand.
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):
&$&$ I got -53.33, I forgot the car was pushed in the opposite direction meaning that unlike in the other equation where you have -50/1.2, in this one there is no need for the - sign as the directions on the imaginary x axis have changed.
• 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. &$&$
• 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. &$&$
• What is the car's initial velocity, its final velocity, and the change in its velocity on this interval? &$&$
• What therefore is its acceleration on this interval? &$&$
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. &$&$
• Describe in words how the system is oriented when the magnetic force acting on it is slowing 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? &$&$
• 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? &$&$
• Do you think the net effect of the magnetic force is to increase or decrease the frequency of the oscillation? &$&$
• Is it possible that the magnetic force slows the system down but increases its frequency of oscillation? &$&$
• 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.
&$&$ Sorry this magnet experiment must have been on the day I missed (9/2) therefore since I have not seen the experiment I do not know how to go about answering it.
We'll see the magnet system again soon.
&#Your work looks good. See my notes. Let me know if you have any questions. &#