course 201
9/8 3
Class 090902vvvv
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)?
Notebook length=11 inches=27.94 cm or about 28 cm
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)?
3 ramps, or 30 cm, is the change in position over the first half of the time interval (about 2.75 cycles)
Answer the same for the second half of the interval.
7 ramps, or 70 cm, is the change in position over the second half of the time interval (again, about 2.75 cycles)
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.
Change in position= final initial= 4-3= 1 ramp= 30 cm, change in clock time= final-initial= 3-2.5=.5 cycles of my 28 cm pendulum
Mark the point of the graph that corresponds to the ball's first contact with the seventh ramp. Give the coordinates of that point.
(3.9, 6)
Do the same for the ball's last contact with the seventh ramp.
(4.25, 7)
What is q_rise between these points (recall that q_rise stands for 'the quantity represented by the rise')?
Rise= change in y= 7-6=1
&#This quantity has units, which must be specified with the quantity. &#
What is q_run between these points?
Run= change in x=4.25-3.9=.35
&#You need to include the units. &#
What therefore is q_slope between these points?
Slope=rise/run=1/.35=2.86
&#You need to specify the units of this quantity, and also think about what these units tell you about the meaning of the quantity, and help you identify errors. &#
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, 1.25, .8
1, .75, 1.33
1, .5, 2
1, .5, 2
1, .4, 2.5
1, .5, 2
1, .35, 2.86
1, .35, 2.86
1, .4, 2.5
1, .25, 4
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?
Rise= change in y= final velocity- initial velocity= 8 m/sec, which represents the change in velocity over that time interval
What is the run of the trapezoid and what does it represent?
Run= change in x= 4 seconds, which represents the time interval
What is the slope of the trapezoid and what does it represent?
Slope= rise/run=8/4=2, which represents the change in velocity divided by the change in clock time, which is the definition of the average rate of acceleration
I don't know that your wording is really in error, but to avoid ambiguity in your thinking it's best to use just the term 'average acceleration', as opposed to 'average rate of acceleration'. Acceleration is already a rate, so the use of the term in the latter phrase is redundant.
What are the dimensions of the equal-area rectangle and what do they represent?
The base value of 4 will stay the same. To find the new altitude, you average the initial and final velocities, arriving at (13-5)/2=4. The new dimensions are then 4x4, which represents a situation where there is not a change in the velocities, instead, the average velocity stays constant over the same period of time.
Careful: you don't subtract when averaging two quantities.
Also be sure to include units with every quantity.
What therefore is the area of the trapezoid and what does it represent?
The area is 16, representing the distance traveled over time
Your average velocity wasn't correct, and that affected your answer.
Also be sure you use units.
In terms of just the symbols v0, vf and `dt:
What expression represents the rise?
Vf-V0
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
this isn't how you find the average of two velocities
What expression therefore represents the area of the trapezoid?
`dt*{(Vf-VO)/2}
right idea, but this is based on an incorrect expression for the average velocity
What is the meaning of the slope?
The slope represents the average rate of acceleration, as it finds the rate of change between the average velocity and clock time
What is the meaning of the area?
The area represents the distance travelled over time, which is more obvious when the calculation is done and all the units cancel out except the one representing distance
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 would give the change in velocity
the change in clock time on this interval
`dt should represent the change in clock time, as it is a time interval
the average velocity on this interval, assuming a straight-line v vs. t graph
(Vf-V0)/2 will give the average velocity
&#See my previous related notes. &#
the average acceleration on this interval
the average acceleration is represented by the slope, and can be found using the `d velocity/time interval, or (Vf-V0)/`dt
&#See my previous related notes. &#
the change in position on this interval
The change in position can be calculated by finding the area of the graph trapezoids equal area rectangle. This is represented by `dt*{(Vf-V0)/2}
&#See my previous related notes. &#
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)/2
&#See my previous related notes. &#
the average acceleration on this interval
(Vf-V0)/`dt
&#See my previous related notes. &#
the change in position on this interval
`dt*{(Vf-V0)/2}
&#See my previous related notes. &#
How are your answers to the above questions related to the v0, vf, `dt trapezoid?
See above, as I have explained each in terms of the trapezoid
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?
Rise=change in y= Vf-V0=30 cm/sec, which represents the change in velocity
What is the altitude of the equal-area rectangle?
The altitude of the equal area rectangle is found by averaging the final and initial velocities= (Vf-V0)/2=35 cm/sec
Can you use one of your answers, with the given area, to determine the base of the trapezoid?
Yes. Since the value for area is given, and I know that the formula for area is `dt*average velocity, and I found average velocity to be 35, I can solve for `dt. 140=`dt*35, dividing both sides by 35 finds `dt=4
Can you now find the slope of the trapezoid?
Yes. Slope=rise/run=(Vf-VO)/4=30/4=7.5
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 arrived at the correct answer of 53.3 cm/sec^2. The following was my reasoning:
Slope is rise/run. Rise corresponds to the change in velocity. Knowing that our average velocity was -40 cm/sec, our starting velocity can be assumed to be twice that, or -80 cm/sec. To find the change in velocity, we subtract -80 cm/sec, the starting velocity, from 0 cm/sec, the ending velocity, to get 80 cm/sec. Run is given as 1.5 seconds. Therefore m=rise/run=80 cm/sec /1.5 sec=53 cm/sec^2
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.
When the system is rotating in a clockwise direction, with the extra strap laying on the table at about 3 oclock, as was demonstrated in class, the magnet is positioned between 1 oclock and 2:45 on the imaginary clockwise motion clock when the magnetic force attracts the magnet to the excess strap, pulling the system down in the direction it was originally moving- clockwise.
Describe in words how the system is oriented when the magnetic force acting on it is slowing it down.
In the same system described above, the magnet would be at 3:15 to about 5 oclock on the imaginary clock of motion when the magnetic forces then pulls the system in the opposite direction, up, counterclockwise.
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?
I would guess that this depends on the strength of the magnet. But assuming our magnet had only a slight attraction from more than 1.5 inches away, and then the force was much greater, I would think that when the magnet was at about 2 oclock on the imaginary clock of motion, magnetic force would be at its greatest, and thus speeding up the system the most. To test this, I would need some way to tell very small amounts of time- perhaps a video camera that I could then slow down the footage. I would test the system by starting the magnet more than 2 inches away, and seeing how long it took for the magnetic force to pull. I would follow with more trials, each time decreasing the amount of space between the magnet and the strap, to observe at what point the magnetic force pulled the system with the most speed.
excellent
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 overall effect is to slow the system down. When the system first comes in contact with the magnet, the magnetic force pushes the system in its original direction, speeding it up. However the magnetic force quickly catches the system again, this time pulling it in the opposite direction. Since it takes more force to pull the system in the opposite direction, the system then doesnt make it quite as much progress in the opposite direction, shortening its cycle. This continues, with the magnetic force pushing the magnet down a distance, and then pulling it back up (slowing it down) a shorter distance. Since the distance between the magnet and the strap is then decreasing, it is soon pulling the system to itself with only slight oscillations in both directions. Eventually it stops completely.
Do you think the net effect of the magnetic force is to increase or decrease the frequency of the oscillation?
Overall, the magnetic force decreases the frequency of the oscillations because it wastes energy by pulling the system backwards, slowing it down. Then it doesnt oscillate quite as far forward again, and eventually the oscillations become so small that they stop altogether.
Is it possible that the magnetic force slows the system down but increases its frequency of oscillation?
Possibly. It may take more energy to complete one full oscillation without the magnet, while a small oscillation involving the magnet happens easily.
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?
Since the cycles are not of a constant distance, happening indefinitely, I would say no. However, knowing the rate of decay in effect, of the cycles of the system, you could use it as a time device as it is constant to that pattern. I would anticipate that this would just be more difficult.
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.
???? Is this referring to the balls rolling down the incline to stop the system from moving forward? If so, the interaction is slight, and ruins the system by causing the bolt in the middle to become off center, and the systems forward movement slows because it is not moving in the center, it is rolling to one side. This could be tested by trials in class.
Good work, but you did make a couple of errors so be sure to see my notes. Nothing I don't expect you to understand, but if you do have questions please ask.