Class 091007
... if we change the incline the total distance traveled by the ball changes, as does the total distance as a multiple of vertical descent. For what vertical descent is the ratio optimized? This tells us something about the energy loss at the interface between the ramps.
... estimate vertical descent of pendulum ...
`q001. Give your raw data for the experiment conducted Monday, with a ball traveling down one incline then up another, until it finally comes to rest.
Report you numbers in the form of an easily-table, including an explanation before or after the table of what the numbers mean.
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`q002. How much distance did the ball cover between the instant of release and the instant at which it came to rest? Give the distance, then explain how you got it from the data you reported in the preceding.
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`q003. The distance down the first ramp is about 30 cm, and the domino supporting it has a thickness of a little less than 1 cm. So the distance the ball traveled on the first ramp alone is more than 30 times the distance of its vertical descent.
Between release and coming to rest (after rolling back and forth for a time) the ball sometimes descends and sometimes ascends, but it ends up lower than it starting point by a vertical distance equal to the thickness of a single domino.
How many times the distance of its vertical descent is the total distance traveled by the ball?
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`q004. When the ball descends, does it potential energy increase or decrease? Does its kinetic energy increase or decrease?
When it ascends, does its potential energy increase or decrease? Does its kinetic energy increase or decrease?
Sketch a graph you think might represent the potential energy of the ball vs. clock time, from the instant of release to the instant it comes to rest.
Describe your graph:
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`q005. On the same graph, sketch the kinetic energy vs. clock time.
Describe this graph, and describe how the two graphs compare.
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`q006. Sketch two more graphs, on representing the ball's velocity vs. clock time, and another representing the ball's position vs. clock time.
Describe your graphs.
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`q007. The sketch you copied from the board in class depicts an incline rising from left to right at angle alpha with horizontal. The x axis is directed up the incline. The weight vector is vertical, and makes angle beta with the y axis. Theta is the angle from the positive x axis to the weight vector, as measured in the counterclockwise direction.
Sketch the same figure, but adjust your angles so that the angle of the incline with horizontal is 15 degrees. (If you divide a 90 degree angle into three equal angles, then take half of one of these angles, you will have a 15 degree angle. Using this method you should be able to hand-sketch a 15 degree angle within a couple of degrees.)
Sketch the projection lines for the weight vector, and estimate the lengths of the arrows representing the x and y components of this vector as percents of the length of the arrow representing the weight vector.
Give the following numbers below, in the specified order, with one number to a line:
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`q008. Repeat the preceding exercise for angle alpha = 10 degrees. You should identify all the quantities you did before, but report here only the x component of the weight as a percent of the weight:
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`q009. Repeat the preceding exercise for angle alpha = 5 degrees. You should identify all the quantities you did before, but report here only the x component of the weight as a percent of the weight:
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`q010. Explain how your pictures show why an increasing slope implies and increasing acceleration.
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`q011. Suppose the ball had a weight of 8 Newtons.
Using the percents you estimated previously:
What would be the x component of its weight on the 15 degree ramp?
What would be the x component of its weight on the 10 degree ramp?
What would be the x component of its weight on the 5 degree ramp?
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`q012. Suppose the ball had a mass of 3 kilograms. What would be its weight, in Newtons?
Using the percents you estimated previously:
What would be the x component of its weight on the 15 degree ramp?
What would be the x component of its weight on the 10 degree ramp?
What would be the x component of its weight on the 5 degree ramp?
Using Newton's Second Law, assuming that the net force on the ball is equal to the x component of its weight, what then would be its acceleration on the 15 degree ramp?
What would be its acceleration on each of the other ramps?
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`q013. Assume that the weight of the ball is 8 Newtons, as on one of the preceding problems.
Based on your estimates of the x component of the weight:
How much work would be done on the ball by this force if it rolled 5 meters down the ramp?
Assuming it started from rest, how fast would it be going after rolling 5 meters down the ramp?
Assuming it started from rest, what would be its kinetic energy after rolling 5 meters down the ramp?
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`q014. Assume that the weight of the ball is 8 Newtons, as before.
Based on your estimates of the x component of the weight:
How much work would be done on the ball by this force if it rolled 3 meters up the ramp?
Assuming it started with a speed of 6 meters / second, how fast would it be going after rolling 3 meters up the ramp?
Assuming it started with a speed of 6 meters / second, what would be its kinetic energy after rolling 3 meters up the ramp?
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`q015. Sketch the tension force for a pendulum which is displaced to an angle of 15 degrees with vertical.
Using an x-y coordinate system with the x axis horizontal, estimate the x component of the tension as a percent of the tension.
Repeat for angles of 10 degrees and 5 degrees.
Give your three estimates below:
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`q016. Based on your answers to the preceding, would you expect the acceleration of the pendulum to increase, decrease or remain the same as it swings back to its equilibrium position? Be sure to explain your reasoning.
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`q017. Explain why it would or would not be a good idea to use the equations of uniformly accelerated motion to model the motion of a freely swinging pendulum.
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`q018. You should be able to do these review problems in a couple of minutes each. If not, you still have some work to do on projectile motion:
If an object falls freely from rest through a vertical distance of 2 meters, while traveling at a constant horizontal velocity of 250 cm/second, then how far does it travel in the horizontal direction as it falls?
If another object falls freely from rest through the same distance, and in the process of falling travels 40 cm in the horizontal direction, then what is its horizontal velocity?
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... analyze pendulum-projectile system
... text, equations of motion
`q018. Your textbook uses a version of the equations of motion which is different from the one we use.
In our version we model motion on an interval, with `dt, `ds, a, v0 and vf denoting time interval, displacement, acceleration, initial velocity and final velocity.
Now, suppose we have an object moving along the x axis, starting at initial position x0 and reaching final position xf. Suppose its initial velocity is v0 and its final velocity is denoted v. We assume that the initial clock time is 0, and we will use t for the final clock time. The acceleration is denoted a.
In terms of the symbols a, x0, xf, v0, v and t:
What is the value of `ds?
What is the value of `dt?
What is vf?
Our first equation of uniformly accelerated motion is `ds = (vf + v0) / 2 * `dt. How is this equation therefore written in terms of a, x0, xf, v0, v and t?
The author of your text gives the equations of motion as follows:
v = v0 + a t
x = x0 + v0 t + 1/2 a t^2
v^2 = v0^2 + 2 a ( x - x0)
vAve = (v + v0) / 2.
These equations have their advantages and disadvantages.
One disadvantage is that they are expressed in terms of more than five variables.
One advantage is that their meaning is not confined to motion on a single interval.
Here is the equivalence between these equations and the equations we are using, as applied to motion on an interval:
Our equations apply to motion on an interval, so we express them in terms of `dt, the change in clock time. The author uses t, which has the disadvantage that it can be interpreted ambiguously (does it mean clock time, or does it mean time interval? could be either, and doesn't distinguish). The use of t has a number of advantages as well, as we will see shortly.
The textbook's symbol v means the same thing as the symbol vf.
The textbook's vAve means the same thing as the vAve we use; however instead of the book's equation vAve = (v + v0) / 2, we multiply by `dt, so our equation is `ds = (vf + v0) / 2 * `dt.
x is the position at the end of the interval, and x0 the position at the beginning, so (x - x0) is simply the change in position, which we call `ds. So the author's equation v^2 = v0^2 + 2 a ( x - x0) means the same as our fourth equation vf^2 = v0^2 + 2 a `ds.
The author's second equation is x = x0 + v0 t + 1/2 a t^2. If we subtract x0 from both sides we get
x - x0 - v0 `dt + 1/2 a `dt^2.
questions about the force diagram picture (estimate forces, work, energy changes, etc.)
... Think of two rubber band chains both exerting force on a paper clip (all supported on a tabletop which exerts no appreciable frictional force on the system). If no other force acts on the paper clip, then the chains will be perfectly lined up. The forces exerted on the paper clip will be equal and opposite. If we displace the paper clip slightly so that the forces are no longer lined up, then release the clip, it will accelerate ....
questions about analyzing previous experiments (qa with solns)
For the 090930 class you were asked to analyze the motion of a pendulum you released from its equilibrium position.
One of the questions was typically not answered, mostly because your instructor didn't clearly designate a place for you to answer it. Since the results obtained by most students were not consistent with their reported data, this question (an a request for one bit of raw data) is repeated below.
Give the length and pullback of the pendulum for one of your trials, and the distance the pendulum fell to the floor:
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Using your raw data show how you find the following, for the interval between the event of letting go of the string and the event of the washer's first contact with the floor.
Give your explanation below:
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Another question that was often not answered concerns the velocity of the pendulum if dropped from rest, through a distance equal to that of its vertical descent.
How far did you estimate the washer descended, in the vertical direction, as it swung back from release to equilibrium?
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Optional problem: If an object dropped .7 cm from rest, accelerating downward at 980 cm/s^2, how fast would it be traveling at the end of the interval? (Hint: you know the initial velocity, the displacement and the acceleration. Use the appropriate equation(s) of motion to answer the question. You should get a result between 30 and 40 cm/s.)
If it isn't completely obvious to you how to get this result, then you need the practice and should show your work:
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Now, how fast would the washer be traveling if it dropped from rest through the vertical descent you estimated?
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Note that the pendulum descends as a result of the gravitational force being exerted on it. Its vertical displacement is downward, and gravity exerts a downward force on it, so the gravitational force acting on it does positive work. It follows that its potential energy decreases, which should be accompanied by an increase in its kinetic energy.
Explain this in your own words:
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Definition of force
Recall the F_net = m a, where the fact that F_net and a are both vector quantities, having magnitude and direction.
Note that vector quantities are written here in boldface; when handwritten they are indicated by an arrow over the symbol. If a vector, say F_net, is written in plain type (in this case as F_net) then this simply refers to the magnitude of the vector, with no regard for its direction.
We can verify that, for a fixed mass, F_net is proportional to a by applying a variety of known net forces to a fixed mass, taking data to determine the acceleration for each trial.
(match with motion of ball down ramp; determine final velocity of ball the projectile)
If F_net = m a, then if various forces were applied to the same mass, would we expect that greater F_net would be associated with greater or lesser acceleration a? Would a graph of (F_net vs. a) therefore be increasing or decreasing?
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If F_net = m a, then what would a graph of (F_net vs. a) look like?
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We test whether, for a fixed net force, the product m * a is constant. We can do this by applying the same net force to a variety of different masses (e.g., find a way to exert the same net force on an object, and just keep piling on masses).
Would you expect the same net force to result in greater or lesser acceleration, if the mass was increased?
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What would you therefore expect a graph of a vs. m to look like?
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If your data allow you to determine m and a for a number of trials, then how could you test whether the product m * a is constant?
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Definition of work and KE
Recall that substitution of a = F / m into the fourth equation of uniformly accelerated motion results in the work-kinetic energy theorem
F_net `ds = `d(KE), where KE = 1/2 m v^2 is the kinetic energy of mass m moving at velocity v.
F_net and `ds are vector quantities; both have magnitude and direction.
`dKE has no direction, and neither does the product F_net `ds. (Formally this product is what is called a 'dot product', with which you might be familiar. In Phy 121 and 201 we don't use the formality of the dot product, so if you're in one of these courses you won 't to even know the name 'dot product'. However if you do know about the dot product (which should be covered in Mth 164 or equivalent but usually isn't) you may use it. The following paragraph is equivalent to the definition of the dot product).
If F_net and `ds both have the same direction, then their product is positive. If they have opposite directions, then their product is negative. (You don't have to worry about this just yet, but note that if they don't both act along the same line, we use the component of F_net along the line of `ds; this component can be in the same direction as `ds or in the direction opposite `ds, and the product would be positive or negative accordingly).
Definition of impulse and momentum
Recall that substituting a = F / m into the second equatino of uniformly accelerated motion results in the impulse-momentum theorem:
F_net * `dt = `d ( m v),
where m v is the momentum of the mass m moving with velocity v.
F_net * `dt is called the impulse of the force acting through time interval `dt.
Note that both impulse and momentum are vector quantities. Both have magnitude and direction. The direction of the impulse is the same as the direction of the force, and the direction of the change in momentum is the same as the direction of the change in velocity. More about this later.
Definition of PE
Start with an example.
If I lift a steel ball from the tabletop to the higher end of a ramp, I do work on it. Then if I release it, gravity does work on it. This work is equal and opposite to the work I did, and of course the ball speeds up. Its kinetic energy therefore increases.
When I lifted the ball I gave it the potential to increase its kinetic energy; when it was released this potential was realized and its kinetic energy increased.
When we lift things we change their position in such a way that the work we have done can in a sense be 'returned' to us (not always conveniently; consider lifting a heavy rock and dropping it on your foot).
Now think about the work done by gravity during this process:
When I lift the object I displace it upward, while gravity exerts a downward force on it. The gravitational force is therefore in the direction opposite the displacement, and the work done by the gravitational force ON the object is therefore the product of a positive quantity and a negative quantity. (For example if 'up' is regarded as positive, the displacement is positive and the gravitational force negative; if 'down' is regarded as positive, the displacement is negative and the gravitational force positive). Either way the product of force and displacement is negative, and we conclude that gravity did negative work.
We also understand that the potential energy of the object was increased.
We therefore define the change in potential energy (denoted `d(PE) or just `dPE) in this example to be equal and opposite to the work done on the object by gravity. If gravity does negative work, `dPE is positive. If gravity does negative work, `dPE is negative.
Other forces, like rubber band tensions, have the potential to 'return' at least some of the work we do on them, and we will soon generalize the definition of `dPE to include such forces.
For now, regard the object being lifted as the 'system' on which work is being done, and be sure you understand every detail of the following statement:
`dPE is equal and opposite to the work done by gravity ON the system.
The work ON is capitalized for emphasis. It is natural in some cases to think in terms of the work being done ON a system. In other cases it is more natural to think in terms of the work being done BY the system. The word 'ON' or 'BY' applies to the word 'system', and the word applied to the word 'system' will be of critical importance in our understanding of energy and energy conservation.
Note that the word 'by' is applied to 'gravity', not to 'system'. In this use it's not capitalized and it's not particularly important.
... would Newton have discovered the laws of motion more quickly if he had had Powerpoint? ... Einstein with Mathematica? (I think 'no' and 'yes' respectively) Would any education major without rigorous content knowledge understand what I'm talking about or its implications for education? (I think 'very few')
Forces on an incline
A ball on a steel incline experiences the downward force of gravity, and an elastic response of the incline to that force. The elastic force acts in the direction perpendicular to the incline (think of the example of the nickel and the bent meter stick). This elastic force is called the normal force. (note that normal forces are not always elastic; the object on the incline might compress the material of the incline, and the incline therefore exert an equal and opposite compressive force, which also acts perpendicular to the incline; either way we call this a normal force, and because of the nature of elastic and compressive forces they are always exerted perpendicular to the incline).
These two forces, the gravitational and normal forces, do not act along the same line. So one cannot balance the other. If no other forces are present, then the combination of these two forces will constitute a force in the direction down the incline and the ball will accelerate accordingly along the incline.
Suppose the incline rises as we move from left to right. Then the normal force acts 'up and to the left'. The steeper the incline the less the normal force will act 'upward' and the more 'to the left'. If we sketch the normal force and the gravitational force on a set of x-y axes, in the standard 'vertical-horizontal' orientation, the gravitational force will be represented by an arrow directed along the negative y axis, and the normal force by an arrow directed into the second quadrant. This sketch was presented in class and you should either remember it or have it in your notes.
The gravitational force is already along the y axis. The normal force can be broken into its components along the x and the y axes. This was also presented in class.
The angle of the incline was around 20 degrees. So the normal force would have made an angle of about 20 degrees with the y axis. If you don't have the sketch in your notes, or if you drew it at an angle much different from the 20-degree angle presumed here, make yourself a quick sketch, based on the 20-degree angle. Include the projection lines and components of this force, and give an estimated answer the following:
What percent of the normal force is its x component? (more specifically: What is the length of the arrow representing the x component of the normal force as a percent of the length of the arrow representing the normal force?)
What percent of the normal force is its y component?
Note that these percents will add up to more than 100%, for the same reason that the shortest distance between two points is a straight line and more specifically because of the Pythagorean Theorem. (the maximum possible sum is a little over 140%)
What are your two estimated percents?
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It is also possible to 'rotate' the x-y axes so that the x axis is directed along the incline, so that the y axis becomes perpendicular to the incline. This has the advantage that the motion of the ball, being on the incline, is along the x axis, while the normal force is along the y axis.
Sketch the same forces as before, but with the x and y axes rotated counterclockwise 20 degrees, so that the x axis is up the incline and the y axis perpendicular to the incline.
In your sketch the gravitational force, being straight down, should now be in the third quadrant of your rotated coordinate system. This is consistent with the figure presented in class.
The normal force is already along the y axis, so doesn't need projection lines. Sketch the projection lines for the gravitational force, and sketch arrows representing the x and y components of this force.
Estimate the x and y projections as percents of the gravitational force. Give your results below:
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Will the x component of the gravitational force increase, decrease or remain the same as the ball rolls down the incline?
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How does the force situation change when the ball rolls off the end of the incline and begins falling to the floor?
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A sketch was also presented of a pendulum at about a 20 degree angle with vertical. The gravitational force vector was again directed downward, and the tension force was represented by a vector parallel to the pendulum string (as sketched on the board, the tension force was up and to the left).
Using x and y axes in standard orientation, with the y axis downward, sketch the components of the tension vector and estimate the x and y components as percents of the tension force:
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Make a similar sketch representing the pendulum at a 10 degree angle from vertical, and again estimate the x and y components as percents of the tension force.
What are your new estimates?
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Does the x component of the tension force increase or decrease as the pendulum swings back toward equilibrium?
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Does the gravitational potential energy of the pendulum increase or decrease as it swings back toward equilibrium? Does the KE of the pendulum increase or decrease?
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Does the gravitational potential energy of the ball increase or decrease as it rolls down the incline? Does the KE of the pendulum increase or decrease?
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Homework:
Your label for this assignment:
ic_class_091007
Copy and paste this label into the form.
Report your results from today's class using the Submit Work Form. Answer the questions posed above.
The topics we addressed today in class relate to q_a_'s #11 through 14.
Today q_A_12 and q_a_13 are assigned for the weekend.