course phy 201
I really got confused with the car problem, it may have just been because I have looked at this for so long. I do not understand what data to use.
9/5 3pm" "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)? &$&$
12 inches
• 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)? &$&$
4’
• Answer the same for the second half of the interval. &$&$
10’ – 4’ = 6’
• 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. &$&$
2 1/2 feet and about 2 1/2 oscillations
• 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, 4)
• Do the same for the ball's last contact with the seventh ramp. &$&$
(8, 4 ¾)
• What is q_rise between these points (recall that q_rise stands for 'the quantity represented by the rise')? &$&$
9”
• What is q_run between these points? &$&$
1 oscillation
• What therefore is q_slope between these points?
.75 ft/ osc
• 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? &$&$
8 meters / second rep. the height of the equal area rectangle
• What is the run of the trapezoid and what does it represent? &$&$
4 sec.
• What is the slope of the trapezoid and what does it represent? &$&$
Slope = 13 – 5 = 8 meters/sec. 8m/sec. / 4 sec. = 2 m/sec^2
• What are the dimensions of the equal-area rectangle and what do they represent? &$&$
A rectangle that has the dimensions of alt 8 and base 4
• What therefore is the area of the trapezoid and what does it represent? &$&$
8 m/sec. * 4 sec. = 32 meters
In terms of just the symbols v0, vf and `dt:
• What expression represents the rise? &$&$
v0 – vf
• What expression represents the run? &$&$
‘dt
• What expression therefore represents the slope? &$&$
(vf – v0) / ‘dt
• What expression represents the width of the equal-area rectangle? &$&$
‘dt
• What expression represents the altitude of the equal-area rectangle? &$&$
(vf + v0) / 2
• What expression therefore represents the area of the trapezoid? &$&$
A = (vf + v0) / 2 * ‘dt
• What is the meaning of the slope? &$&$
• What is the meaning of the area? &$&$
velocity
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 – v0
• the change in clock time on this interval &$&$
‘dt
• the average velocity on this interval, assuming a straight-line v vs. t graph &$&$
A = (vf + v0) / 2 * ‘dt
• the average acceleration on this interval &$&$
Acc ave = (vf – v0) / ‘dt
• the change in position on this interval &$&$
vf – v0
• 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 &$&$
A = (vf + v0) / 2 * ‘dt
• the average acceleration on this interval &$&$
(vf – v0) / ‘dt
• the change in position on this interval &$&$
vf – v0
• How are your answers to the above questions related to the v0, vf, `dt trapezoid? &$&$
v0 = the beginning alt (normally the left side), vf = the ending alt (normally right), ‘dt = the change in time (the base)
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? &$&$
vf – v0 = 20 cm/sec – 50 cm/sec = -30cm/sec
• What is the altitude of the equal-area rectangle? &$&$
(vf + v0) / 2 = (20 cm/sec + 50 cm/sec) / 2 = 35 cm/sec
• Can you use one of your answers, with the given area, to determine the base of the trapezoid? &$&$
Yes. A = ‘dt (vf +v0)/2 , 140 cm = ‘dt (35 cm/sec), ‘dt = 4 sec
• Can you now find the slope of the trapezoid? &$&$
Yes. Slope= (-30 cm/sec)/ 4sec = -30/4 cm/sec^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): &$&$
• 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. &$&$
Ave vel = 60/ 1.5 = 40
This would be -60 cm / (1.5 s) = -40 cm/s.
• 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. &$&$
description?
• What is the car's initial velocity, its final velocity, and the change in its velocity on this interval? &$&$
Initial 60 final is 0
The only 60 in this problem is the -60 cm change in position. There is no velocity which has numeracal value 60.
The careful use of units might well have alerted you to this error.
• What therefore is its acceleration on this interval? &$&$
20 cm / 1.2 sec = 16 ¾ cm/sec
It's not clear where 20 cm came from.
1.2 sec is not a time interval relevant to 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. &$&$
As the ball approaches the magnet the magnetic force begins pulling on the ball causing it to speed up.
• Describe in words how the system is oriented when the magnetic force acting on it is slowing it down. &$&$
As the ball goes past the magnet the magnetic force pulls the ball back to it causing it to slow down dramatically.
• 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? &$&$
Just a few cm before the ball reaches the magnet. See if the magnetic pressure is strongest there.
• 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? &$&$
I would say that it has an almost equal effect but the slowing down has the most effect.
• Do you think the net effect of the magnetic force is to increase or decrease the frequency of the oscillation? &$&$
decrease
• Is it possible that the magnetic force slows the system down but increases its frequency of oscillation? &$&$
yes
• 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? &$&$
Yes, time it for a minute and repeat several times and see how constant the numbers you get are. We could expect that if we did everything the same to have pretty close numbers.
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. &$&$
I think that if the ball was to be slow while it passed the magnet that the magnet would begin to oscillate it.
We could test this by starting the ball closer to the magnet.
Very good, but you should take another look at the car problems sometime when you're fresh. No need to resubmit them; unless you have questions.