Class 091005
There are twenty questions in this document. Most of this just covers what you saw in class, so many of the questions should be short and relatively easy to answer (or at least attempt). However that's still a lot of questions and a lot of ideas to think about, so only one q_a_ will be assigned instead of two.
Questions based on 090930 class
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.
`q001. 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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`q002. 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.
`q003. How far did you estimate the washer descended, in the vertical direction, as it swung back from release to equilibrium?
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`q004. 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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`q005. Now, how fast would the washer be traveling if it dropped from rest through the vertical descent you estimated (between release and equilibrium)?
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`q006. 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)
`q007. 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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`q008. 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).
`q009. Would you expect the same net force to result in greater or lesser acceleration, if the mass was increased?
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`q010. What would you therefore expect a graph of a vs. m to look like?
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`q011. 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.
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:
The word 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. It is this word which orients our thinking about the system.
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.
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).
On a non-horizontal incline, these two forces, the gravitational and normal forces, do not act along the same line. So one cannot balance the other.
Suppose the incline rises as we move from left to right. We can sketch the force situation. This sketch was presented in class and you should either remember it or have it in your notes. Observations on this sketch:
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.
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:
`q012. 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.
`q013. Estimate the x and y projections as percents of the gravitational force. Give your results below:
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`q014. 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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`q016. Using x and y axes in standard orientation, with the y axis downward, sketch the components of the tension vector for the pendulum, and estimate the x and y components as percents of the tension force:
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`q017. 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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`q018. Does the x component of the tension force increase or decrease as the pendulum swings back toward equilibrium?
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`q019. 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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`q020. 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_091005
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 only q_a_11 is assigned. This weekend #'s 12 and 13 will be assigned, in case you want to work ahead. Complete URL's for the upcoming qa's are included below.
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_11.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_12.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_13.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_14.htm