Class Notes 100927
Questions and notes will be added to this document, and it will be noted when the document is complete.
Testing
If possible complete the document at at Practice Major Quiz (qa) by next class. If this is not possible, try to at least read the document and see if your thinking is consistent with the given solutions.
You should plan to complete the first test, entitle the Major Quiz by midweek next week. The procedure is to go to the Learning Lab, print out the current version of the test, have it signed by the Learning Lab staff and follow their instructions for completing and submitting the test. Tests are posted at http://vhmthphy.vhcc.edu/tests/ , and the tests change every 5 minutes. You can look at as many versions of the test as you wish.
Text Assignments
General College Physics:
Chapter 2 problems should be completed by next Monday.
1-5, 6, 9, 11, 16, 22, 27
34, 37, 43, 52, 55, 58, 63
Chapter 4 problems for the sections previously assigned include the following. You should read over them and give some thought as to how to solve them.
2, 4, 5, 6, 7, 8, 9
11, 13, 17, 19, 20
Give a quick read to Chapter 6, Sections 1-6
University Physics:
Chapter 2 problems should be completed by next Monday.
54, 57, 60, 63, 66
72, 77, 82, 87, 90
Chapter 4 problems for the sections previously assigned include the following. You should read over them and give some thought as to how to solve them.
36, 38, 39, 41, 42
45, 46, 48, 49, 52
Give a quick read to Chapter 6
Important Resources for Test Preparation: Introductory Problem Sets and other recommended resources:
At the Assignments page for distance students (http://vhcc2.vhcc.edu/ph1fall9/frames_pages/assignments.htm) you will see the following:
In some of Assignments 1-8, there are Randomized Problems. The first couple include instructions. These problems are of the same form as many problems that occur on the Major Quiz, so it is recommended that you submit any of these problems that aren't completely obvious to you, using the same procedures you have used to submit other documents. Insert your responses in the appropriate places. I do ask that you mark insertions before and after with #### so I can quickly locate them, but these are short documents and if you happen to forget it won't be a big deal.
Also you will see the link http://vhmthphy.vhcc.edu/ph1introsets/default.htm, which is a link to the Introductory Problem Sets. You should take a quick read through Sets 1 and 2, just to see what's there. Ideally you will understand every problem, and will understand completely the explanations given in the provided solutions, generalized solutions, and figures.
The Introductory Problem Sets are designed to stand alone as an introduction to very basic problems and procedures. One important thing, from your perspective, is that after the Major Quiz, problems from the Introductory Problem Sets frequently appear on tests.
Notes from Class
Suppose a 120 gram toy car has acceleration of magnitude 15 cm/s^2, as a result of friction.
How much force does friction exert on the car?
F_net = m a.
We are given the mass m = 120 grams and the acceleration 15 cm/s^2.
The net force is therefore F_net = 120 grams * 15 cm/s^2 = 1800 gram cm/s^2.
The unit g cm/s^2 is not an SI unit, but being the product of a mass and an acceleration, it is a perfectly valid unit of force.
Since the 15 cm/s^2 is said to be the result of friction, the net force is the frictional force.
Suppose the car coasts to rest while traveling 40 cm. How much work is done by the frictional force?
`dW = F * `dx.
We found the force to be 1800 g cm/s^2, and we are told that `dx = 40 cm.
The work is therefore
`dW = 1800 g cm/s^2 * 40 cm = 72 000 g cm^2/s^2.
Notes about the sign of the work:
- There are two frictional forces, the frictional force exerted on the car and the frictional force being exerted by the car. It isn't specified which frictional force is doing the work, so we can't determine the appropriate signs of F and `dx. The solution we have given is therefore just the magnitude of the work.
- If we are talking about the work done by the frictional force acting on the car then, since the direction of this force is opposite that of the car's displacement, the work would be negative.
- If we are talking about the frictional force exerted by the car, it is in the direction of the displacement and the work is therefore positive.
The car could coast 40 cm to rest while doing 72 000 g cm^2/s^2 of work if its original KE was 72 000 g cm/s^2. How fast would it have to be moving to have this much KE?
KE = 1/2 m v^2.
If we multiply both sides of the equation by 2, then divide both sides by m and finally take the square root of both sides we get
v = +- sqrt( 2 KE / m).
Plugging in KE = 72 000 g cm/s^2 and m = 120 g we get
v = +- sqrt( 2 * 72 000 g cm/s^2 / 120 g) = +- sqrt(1200 cm^2 / s^2) = +- 35 sqrt( cm^2 / s^2)
sqrt(cm^2 / s^2) = cm/s, so our final result is
v = +- 35 cm/s.
If we assume the displacement to be + 40 cm, then the initial velocity is +35 cm/s.
Note that all these results were obtained using the units g, cm and s. Since all units were compatible, none of the calculations required conversion of units, and while these units could have been used none of our solutions required units like Newtons, dynes, Joules or ergs.
How long does it take the car to come to rest?
Alternatively, if we again assume acceleration to be uniform, we can work this out from the original information, which tells us that | a | = 15 cm/s^2, vf = 0 and `ds = 40 cm:
University Physics Notes
A pendulum or a mass on a spring is influenced by a net force which is proportional to distance from equilibrium, acting in the direction of equilibrium. A force of this nature is described by
The buoyancy-stabiliized balance is characterized by restoring torque of similar nature. More about this later, but the mathematics for the two systems are nearly identical.
Now starting with F_net = - k x we use Newton's Seconds law to rewrite this as
Since a = dv/dt and v = dx/dt, we conclude that a = x '', where ' indicates derivative with respect to t. Our equation becomes
To solve this equation we need to find a function x(t) which satisfies this equation.
The function must have the characteristic that its second derivative is a constant multiple of the itself. We know three functions with this characteristic:
An exponential function of the form x(t) = A e^( c t ), with c a constant, has second derivative c^2 A e^ (c t). Thus x '' = c^2 * x.
A sine function of the form x(t) = A sin(c t) has second derivative - c^2 A sin(c t). Thus x '' = - c^2 x.
A cosine function of the form x(t) = A cos(c t) has second derivative - c^2 A sin(c t). Thus x '' = - c^2 x.
If we use the exponential function in our equation x '' = -k / m * x, we get the equation
c^2 A e^(c t) = -k / m * A e^(c t), which simplifies to
c^2 = - k / m, with solution
c = +- sqrt(k / m) * i, where in this case i = sqrt(-1).
The imaginary solution would seem to be an obstacle, but in fact it isn't. I've sketched a few details below, but they aren't necessary for your work right now (worth knowing for your upcoming diff eq course). These details can be used as an overall reference later in this course:
Euler's formula
- e^(i theta) = cos(theta) + i sin(theta)
is obtained by doing a Taylor expansion of the exponential (e^x = 1 + x + x^2 / 2! + x^3 / 3! + ... ), and substituting (i theta) for x. When you simplify the resulting expression and separate the real and imaginary parts the Taylor expansions for cosine(theta) pop right out at you.
Substituting c = sqrt(k/m) * i into x = A e^(c t) we get x = A (cos( sqrt(k/m) t) + i sin(sqrt(k/m) t).
Either the real or the imaginary part of the solution will work for the equation. So will any linear combination of the two.
Conclusion:
- x(t) = B cos(sqrt(k/m t) + C sin(sqrt(k/m t)
is the general solution to the equation.
This solution can be put into the form x(t) = A sin( omega t + phi), where A and phi can be any constants. A is interpreted as amplitude, phi as the starting phase of the motion.
If we use the cosine function in the equation x '' = -k/m * x we get the equation
-c^2 A cos( c t) = -k/m A cos(ct)
so that -c^2 = -k/m and c = +-sqrt( k/ m).
Our solution becomes x(t) = A cos( sqrt(k/m) * t)
This solution can be interpreted in terms of the circular model of the sine and cosine functions. This description goes beyond what you are required to know right now, but can serve as a reference for present and future work with simple harmonic oscillators:
x = A cos(theta) is the x component of the point on a reference circle of radius A, where the radial vector from the center of the circle to the point makes angle theta with the positive x axis.
x(t) = A cos( omega * t) is therefore the x component of the point on a reference circle of radius A, where the radial vector from the center of the circle to the point makes angle theta = omega * t with the positive x axis.
If t is interpreted as time, then theta = omega * t gives the angle of the radial vector whose angular velocity is omega. The projection of this vector onto the x axis models the position of the oscillating object as a function of time.
In our solution function x(t), which involves cos(sqrt(k/m) * t), omega is just sqrt(k/m). That is:
- The restoring force constant k from the force function F_net = - k x, and the mass of the oscillating object, determine the angular velocity of the reference-circle point by the simple relationship
omega = sqrt( k / m ).
In a nutshell:
x(t) is the x projection of the vector r(t) of constant magnitude A whose angular position is given by the function theta = omega * t.
Consequences:
The reference point has constant speed omega * A, and it at any instant moving in the direction tangent to the circle.
The direction tangent to the circle is perpendicular to that of the position vector r(t). So the velocity of the reference point is described by the vector v(t) whose magnitude is omega * A and whose direction is perpendicular to the r(t) vector.
The reference point has an acceleration toward the center of the circle, whose magnitude is a = v^2 / r, where v and r are the magnitudes of the velocity and position vectors v(t) and r(t). Those magnitudes are omega * A and A, so the magnitude of the acceleration can also be describe by a = omega^2 * A.
Thus we can sketch the reference circle, with r being the radial vector from center to the reference point, v perpendicular to r representing the velocity of the reference point and a perpendicular to v directed toward the center of the circle.
The projection of the r vector in the x direction is the corresponding position of the oscillating point, the projection of the v vector in the x direction is the corresponding velocity of that point, and the projection of the vector a in the x direction is the corresponding acceleration of that point.
These projection vectors will be identical to v_x(t) = x ' (t) and a_x(t) = x '' (t). Remember that x(t) = A cos(omega * t).
This gets even better. We can project the r, v and a vectors in any fixed direction we choose. Think of projecting these vectors onto a directed line through the center--any directed line through the center. (If that line happens to be the y axis, then we will have y(t) = A sin(omega t), and the velocity and acceleration functions will be the projections v_y(t) = y ' (t) and a_y(t) = y '' (t). If the line happens to be in some other direction, then the position function will be a linear combination of the sine and cosine functions, as will the velocity and acceleration functions).
Lab activities
Notes:
Include concise explanation: Whether specifically requested or not, all responses should include a brief explanation or description beginning on the line following other requested information. Abbreviations and incomplete grammar are acceptable; if you go too far with this I'll let you know, but I don't want to keep typing demands reasonable.
Reporting data:
Explaining your analysis:
1. Projecting point on CD onto paper on tabletop.
`qx001. Your points will lie along (or close to) an x axis perpendicular to the line you sketched on your paper. With the origin at the center point, what were the positions of your points corresponding to theta = 0, 30, 60, 90, 120, 150 and 180 degrees? Report as 7 numbers separated by commas in the first line, with brief explanation starting in the second line.
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`qx002. What were the coordinates of your points corresponding to theta = 180, 210, 240, 270, 300, 330 and 360 degrees?
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`qx003. Suppose the disk rotated with a constant angular velocity, with an actual object moving along the tabletop just below the point on the disk. How would the velocity of that object change as the disk rotated through one complete revolution? Sketch (on your paper) and describe (below) a graph representing the velocity vs. clock time behavior of that point. Include an explanation connecting your results to your data.
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`qx004. For the same object as above, sketch a graph representing the acceleration vs. clock time behavior of that point. Desribe your graph and include an explanation connecting your results to your data.
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`qx005. For the same object, sketch a graph representing the net force on the object vs. clock time for one revolution of the disk. Describe your graph and include an explanation connecting your description to your data.
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... describe r, v and a vectors ...
2. Quick collision experiment
`qx006. In the first line below give the landing positions of the 'straight drop', uninterrupted steel ball, the marble, and the steel ball after it collides with the marble, separated by commas.
In the second line, report the horizontal displacement of the uninterrupted steel ball, the marble, and the steel ball after it collides with the marble, separated by commas.
Starting in the third line give the units of your measurements and a brief explanation.
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`qx007. Assuming that the time of fall was .4 seconds, what do you conclude was the velocity of each object at the instant it left the end of the last ramp? Report three numbers separated by commas in the first line, in the same order used in the preceding question. Units, explanation, etc. should start in the second line.
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`qx008. In the collision, the velocity of the steel ball changed, as did the velocity of the marble. What was the change in the velocity of each? Report number in the first line, brief explanation in the second.
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3. Motion of unbalanced vertical strap
`qx009. The original vertical strap system oscillated about an equilibrium position with one end lower than the other. Why do you think the equilibrium position had that end lower?
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`qx010. What changed about the behavior of the system when a couple of #8 nuts were added to the higher end? What is your explanation?
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`qx011. Would it have been possible to balance the system at a position where the end with the #8 nuts was higher? Would it have been challenging to do so? Explain.
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`qx012. Did the frequency of oscillation of the system appear to be constant?
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4. Balancing the styrofoam rectangle
`qx013. Was the styrofoam rectangle easiest to balance when the paperclip was inserted along an axis through the point below its center of mass, at a point above its center of mass, or at the point of its center of mass? Why do you think it was so?
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`qx014. At which positions of the paperclip did the system did the system oscillate? At which positions did it appear to oscillate with constant frequency? At which positions did it appear to oscillate with nonconstant frequency?
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5. Two cars with repelling magnets
`qx015. Why do you think the two cars traveled different distances when released?
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`qx016. Which car do you think exerted the greater force on the other?
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6. Graphs
`qx017. The graphs you were given in class depict coasting distance, in centimeters, vs. separation in centimeters, for a 120-gram toy car whose acceleration due to friction is 15 cm/s^2 (plus or minus an uncertainty of 10%). Sketch four tangent lines to the first curve, spaced equally from near one end of the curve and the other. Find, with reasonable accuracy, the coordinates of two points on each tangent line, and use these coordinates to find the approximate slope of each tangent line. In the first line below, report your four slopes. In the second line report the x and y coordinates of the two points used to find the slope of the third tangent line, reporting x and y coordinates of the first point then x and y coordinates of the second, using four numbers separated by commas.
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`qx018. Each centimeter of coasting distance corresponds to very roughly 2000 g cm^2 / s^2 of energy lost to friction. That energy came from the potential energy of the magnets at the given separation. So the vertical axis of your graph can be relabeled to represent the energy lost to friction, and hence the potential energy of the magnet system. For example, 20 cm on the vertical axis corresponds to 20 cm of coasting distance, each cm corresponds to 2000 g cm^2 / s^2 of potential energy, so the 20 cm coasting distance corresponds to 20 * 2000 g cm^2 / s^2 = 40 000 g cm^2 / s^2 of potential energy. The number 20 on the vertical axis can therefore be cross-labeled as 40 000 or 40 k, representing 40 000 g cm^2 / s^2 of PE. You should be able to quickly relabel the vertical axis of your graph.
Using the relabeled vertical coordinates, find the y coordinates of the two points you used to find the slope of your third tangent line, then report the x and y coordinates of those two points as four numbers in the first line below. In the second line report the rise and run between these points, and the slope. In the third line report the units of the rise, the units of the run and the resulting units of the slope. Starting in the fourth line explain what you think the rise means, what the run means, and what you think the slope means.
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`qx019. Report the slopes of all four of your tangent lines, in terms of your relabeled coordinates, as four numbers in the first line below. You can easily and quickly find these four slopes from the slopes you previously reported for the four tangent lines. Starting in the second line report very briefly how you found your slopes.
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`qx020. University Physics Students: Find the derivative of the given y vs. x function y = 88 x^1.083 (this is a simple power function with a simple rule for its derivative) and evaluate at each of the four tangent points. Give the derivative function in the first line, in the second line the values you got at the four points, and in the third line compare your values to the slopes you obtained previously.
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`qx021. University Physics Students: What is the specific function that describes PE vs. separation for the magnet system? What is the meaning of the derivative of this function?
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... questions related to class notes ...
balancing demo
vertical strap demo
opposing cars demo, question of balancing paper clips
... vf ' = v0 / vf
... ref circle in complex plane ...