# Inertial Mass Measurement

While I was checking the worksheets at the end of class I was noticing that a few groups had “interesting” values for their cart masses on the final table of Unit 5.

• The mass values you put in the table derive from the accelerations you measure, not the spring scale measurements
• If you did things right then the Force column (the forces from the fan) should be the same for all cases where the same number of batteries were used.
• The masses you calculate this way are systematically wrong because you ignore friction.  But if they are wrong by more than about 20% then they are too wrong. (Later we’ll learn how to correct for the friction to get a better value.)
• If they are too wrong then check your method of calculation. The ratio of accelerations should be equal to the ratio of the total mass that is being accelerated. That means you have to do some algebra to solve for the mass of the fan-cart in terms of the bar’s mass.
• The question asking for g want the value implied by your data. If you measured the weight of the cart with a spring scale that weight would be in newtons. If you used to the spring scale and read the mass scale you have to use g = 9.8 N/kg to get the weight. Then you report another value of g  in the question which will be somewhat wrong because friction was ignored.
• If you find you just can’t get any reasonable values from your data then try to understand what’s wrong and explain your analysis of the issues. We don’t expect you to always get perfect results or to fudge your data and we don’t necessarily grade on the nearness of your answer to the “right” answer.

# Assignment Dates

There was an inconsistency in the assignment for Unit 5. Originally this unit was 3 sessions and the worksheets on force and motion were to be done over those three sessions.  Because of our short semester, and well-educated students, I have shortened unit 5 to 2 sessions but forgot to edit the assignment list and there was a reference to session 5-3. That’s fixed now and all the force-and-motion worksheets are due on Friday, Oct 1.

The general rule is that all the work from a unit is due on the first day of the next unit. The course schedule has the due dates on it too.  The only exception is that there was one extra day allowed for the 11 kinematics problems of Unit 4. The other thing is that on WebCT we recommend that you should do some of the assignments before the due date in order to be able to ask questions in class and to avoid starting too late.

This may be a confusing because normally SFU classes have one assignment per week. I apologize, but this is the most logical way to do things for this course. We have a shorter semester than many universities in North America and we need to compress the schedule to cover all that needs to be covered in a semester.

# Another time, another place

When you try to fit real data with the Fundamental Kinematic Equation (FUNKE) you may run into an issue trying to estimate two parameters, x0 and v0, if the segment of data you’re trying to fit starts a long time after t=0.  That’s because x0 and v0 are the position and velocity the cart would have had at t=0 if the acceleration had been constant all the way back to t=0. So these values may be a little weird.  For example if you wait 5 seconds before starting the cart at the origin with a constant acceleration of +1 m/s/s, then x0 would be 12.5 m and v0 would be −5 m/s! — because that’s where the cart would have had to start, and the velocity it would have had, if it had started 5 seconds earlier and travelled with a constant acceleration all the time.

 Position vs time graph of a cart starting at x=0 at t=5s. The actual motion is in blue, the motion extrapolated back to t=0 is in red.

It might be easier to estimate the parameters if one expresses the FUNKE in a different way: use t0, which is the time at which the parabola reaches its extreme value (either maximum or minimum) and xp, the position at that time. In other words, the time and the position of the parabola’s apex. Then the FUNKE becomes

$$x(t) = x_p+ \tfrac{1}{2}a(t-t_0)^2$$
Your fitting parameters are $x_p$,  $t_0$ and $a$. Notice that $v_0$ is not in this equation because at the apex the velocity is zero.
You can do your modelling assignment with this equation. It is another correct way of expressing motion with constant acceleration. As long as you explain what you’re doing you’ll get full credit if it’s right and partial credit if it’s wrong and we understand what you’re getting at.
There are many right ways of doing most problems. Unfortunately, there are many more wrong ways.
Can you find the mathematical relationship between x0, v0a and xp, t0, a ?

Expand and collect terms and you get

$x_0=&space;\left(&space;x_p&space;+\tfrac{1}{2}at_0^2\right)\;&space;{\rm&space;and}\;v_0=-&space;2at_0$
For example if a = 1 m/s2 and the curve reaches the minimum xp = 0 at t0 = 5 s.
What is x0 and v0?

x0 = 12.5 m
and v= −10 m/s

# the FUN-damental Equation?

In class you used a formulaic method to convince yourselves that the “fundamental kinematic equation” makes sense.  You used the derivative of a polynomial to get the velocity equation from the position equation. You are supposed to instantly recognize the velocity equation that you get as self-evident:

$$\frac{d}{dt}x(t) = \frac{d}{dt}\left( x_0 + v_0t + \tfrac{1}{2}at^2\right)$$

Do the derivatives to get the self-evident equation:
$$v(t) = …$$

(I’m leaving the result out because you’re supposed to think it through yourself.)

There’s another way to justify the FUNKE (That stands for FUNdamental Kinematic Equation.) Start from the idea that displacement is the area beneath the v(t) graph. Draw the graph for a typical case of constant acceleration and figure out the displacement Δx:

 From my Phys100 webpage.
Now I’ll colour-code the equation so you can see where each term comes from.
$$\Delta x = \color{green}{v_1\Delta t} + \color{red}{\tfrac{1}{2}a(\Delta t^2)}$$
You only need to do some simple substitutions in notation to transform it to the form of the FUNKE at the start:
• Δx = x(t) – x0
• Δt = t (lazy notation)
• v1 = v0

Now our graph that helped us get the equation was just to aid our thinking. It shows a positive a. But nothing in the derivation of the equation assumed that acceleration was positive so the result is general and holds for negative acceleration too.

As an exercise draw the figure for a negative a and see how it works out.

What about negative initial velocity and positive a?

There’s one more case. What is it?

# π, Exhaustion, Method of

When you study calculus they usually do differentiation and then integration. Historically the order was different.

In school you learned what the number π stands for and you probably memorized its value to a few digits.  But could you figure out what it was if you had to?

The story of how people figured out the value of π is the story of integration.  We can start in ancient Egypt where an approximate value was got my measuring circles. It was a Greek guy, Eudoxus of Cnidus, who starting thinking about how to figure it out exactly. He developed the Method of Exhaustion to find the areas of various shapes and another Greek dude, Archimedes,  used the Method of Exhaustion to nail down the value of pie (oops, π) to a small range:

3 10/71 < π < 3 1/7.

 Square-Circle-Square (Yes it’s a circle)

To understand how he did it imagine a circle with radius r. Draw a square outside it with sides of length 2r. Now draw another square  inside whose diagonal has length 2r. The sides of this square have length 2r/√2.
The area of the outer square, 4r2, is obviously larger than that of the circle. The area of the inner square, 2r2, is obviously smaller than the circle’In t. Therefore,

2 < π < 4.

We can narrow down the range by doing the same thing with hexagons, octagons and regular polygons with more and more sides:

Archimedes continued this until he got to a 96-sided polygon and that’s how he computed the result above.  After that he was exhausted.

In the second session of Unit 4 we go through a similar argument to show that the displacement for something that’s constantly accelerating can be got from the area under the velocity vs time graph just like you can do for a constant velocity. The same argument can then be used for any shape of v(t) curve.

Most of the time we just say that it’s obvious. Obvious is the most dangerous word in the English language. Mathematicians teach us to be careful and to prove things using small self-evident steps so that we can be sure.
Look at these two graphs.
You might agree that the average velocity is given by $\langle v \rangle = \frac{v_1 +v_2}{2}$  in both cases.  But is that true for both (a) and (b)?
Go back to the original definition of <v>. You’ll find it’s only true for one case.

# One-question physics test

There’s a one-question physics test that lets me know if we’re making progress. It’s the last question of Unit 3.
“Consider the ball toss carefully. Assume that upward is the positive direction. Indicate in the table that follows whether the velocity is positive, zero or negative during each of the three parts of the motion. Also indicate if the acceleration is positive, zero or negative.”
Why?
Learning physics requires revising preconceptions we come to the class with.  The meaning of acceleration illustrates one such preconception. As commonly used, the word acceleration means moving faster. The reasoning goes like this:

If you’re not moving then you can’t be accelerating. Zero velocity means not moving doesn’t it? — therefore if v = 0 then a = 0.

Right?

Wrong!
The precise definition of acceleration in the physics context is the rate of change of velocity: a= dv/dt
The velocity can be instantaneously at zero, but if the velocity is changing then acceleration is not zero.
If you understand this then you’re changing your preconceptions based on your physics lessons. If not then either you’re not listening, or you hear but don’t want to accept something that violates your sense of what’s right.
The reason we need to get this is that forces lead to acceleration and physics is all about forces and what they do to things.
If you think you got it then try this question:

Mike jumps out of a tree and lands on a trampoline. The trampoline sags 0.5 m before launching Mike back into the air. At the very bottom, where the sag is the greatest, Mike’s acceleration is:

A. Upward B. Downward C. Zero

# Troubleshooting

The concepts Friday were easy, but sometimes the equipment didn’t cooperate. If you’re having trouble with the equipment, it’s probably a good idea to check out a few of the usual suspects before you panic. It might take us instructors a while to get over to help you and a little easy troubleshooting can save time.

1. Check the power:
• Is the block plugged into powerbar?
• Is the powerbar switch on?
• Make sure that the little power plug is pushed all the way into the LabPro interface. (If you push it in and you hear the LabPro jingle, then it wasn’t pushed in enough.)
• If there’s power, then there should be some lights on inside.
2. Check the USB cable. The USB cable should be plugged into the LabPro and the other end to the computer. It might go to the USB port in the monitor instead of the computer, but people tell me this is less reliable. And make sure it’s going to the USB port of your computer.
3. Make sure the sensor is plugged into the correct port. For example the Motion Sensor should normally be plugged into the DIG/SONIC 1 port for the setup file that’s recommended in unit 3.  (It can work in the other port but you have to change the setting in LoggerPro software under “Setup Sensors…”.)
4. If you lose your LoggerPro tool bar, click on the little button in the upper right-hand corner of the Logger Pro window.
5. If the sensors were not plugged in when you loaded the setup file then try reloading the setup file again after the sensors are plugged in. Many of the sensors are auto-detected, but not all. Later on you’ll have to learn how to tell LoggerPro by hand which sensor is on what port. That’s under the menu item “Setup Sensors…”.
6. If you’re getting noisy motion data, then clear out all extra stuff from nearby your track like pencil boxes, bags, books etc. Keep your hands away from the track while measuring. Try adjusting the tilt of the detector and put it on narrow beam to measure cart motion.

If these steps don’t solve the problem then you have to go into advanced troubleshooting. This is based on a divide and conquer strategy.  Try to isolate where the trouble is by testing each component of the chain using a sort of search algorithm:

1. Figure out if the problem is with the computer or with the interface+sensor. You can do this by trying your interface on another computer. If it still doesn’t work then it’s the interface if it does then it was the computer.
2. Now let’s assume you found that it was the interface.  Do each of the following in sequence until you find the component that’s at fault.
• Try changing the LabPro with another one.
• Try another sensor.
• Try another cable.
• Try another power supply.

Similarly if you had found that the computer was a fault you’d have to figure out whether it was a software or hardware issue and then subdivide using the same process.

Troubleshooting skills are important in almost any technical field you might be involved in. Get good at them and you’ll be indispensable.

# New Problems Proposed

Last year I changed Unit 2 but did not write replacements for problems 2-4 and 2-5.  Here are two PROPOSED problems. They are not assigned this year but I post them here for the record and discussion.

SP2-4

Make a table showing the probabilities that the sum of the roll of 3 dice are  3, 4, 5, 6, 7, …18. You have already done 11 in the activity guide. Explain your method of calculation. Hint: You can use a symmetry argument to shorten the calculations. The probability for getting 3 is the same as for 18, the probability of getting 4 is the same as for 17, etc.

SP2-5

Instead of measuring the background level in one run of 80 1-minute intervals you do 2 experiments, measuring 30 1-minute intervals and then 50 1-minute intervals. The results are as follows:

30 intervals: average = 11.8 counts/minute, SD = 4.2  counts/minute

50 intervals: average = 12.2 counts/minute, SD = 3.8 counts/minute

Use the formulas for the average and Standard Deviations to determine what the average and SDs would have been if you had done this experiment as a single experiment of 80 1-minute intervals.

# Level of Confidence?

Here is a question I received from a student by e-mail.

“Could you clraify exactly what it means to estimate the level of confidence of our conclusion. What is our conclusion on (which part of the guide) and how are we to estimate the level of confidence of the conclusion if it is a written statement.”

“You are trying to answer the question:

‘Is there a significant difference between the number of counts/minute detected when the nu-salt is present compared to when the nu-salt was not present?’

You need to use the statistical quantities we measured for the number of counts/minute both with and without the nu-salt present: average, standard deviation, number of counts and standard deviation of the mean. The level of confidence will depend on how far apart the averages of the two experiments are compared to the standard deviations of the means.”

In other words, if the means differ by about 1 SDM then you can conclude that the difference is real to a 68% level of confidence. If they differ by 2 SDM then they differ to a 95% level of confidence.  If they differ by something between 1 SDM and 2 SDM  then you conclude that the difference is real to something between 68% and 95% level of confidence.

# Unit 2 Bugs

On WebCT there is an Activity 2-11 upload but there is no Activity 2-11.  That was something I forgot to delete after I revised Unit 2 last year. It is not there now and is not due — don’t worry about it.   Furthermore, there is no SP2-4 and SP-5, yet.

(I was thinking about assigning a problem on calculating the probabilities of all possible outcomes of throwing three dice, but I wonder if that would be too hard?)

This is a copy of a class email.