Class 091005

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.

• The time required to fall your observed vertical distance, starting with initial vertical velocity zero and accelerating downward at 980 cm/s^2.
• The displacement of the washer in the horizontal direction.
• The horizontal velocity of the washer.

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 incline

Forces on pendulum

Work-energy on incline

Work-energy on pendulum

Q_a_you_ready_for_the_ major_quiz_q

note:  organize everything in qa query etc into a single menu page

Write down your answers to the following questions on a piece of paper.  The correct answers are given later but whatever you do, don’t look at them.

What is the definition of average rate of change?

What is the definition of average velocity?

What is the definition of average acceleration?

Sketch a v vs. t trapezoid. 

What is the meaning of each of the following quantities?

the ‘left graph altitude’

the ‘right graph altitude’

the ‘left graph altitude’

the altitude of the equal-area rectangle

the width of the equal-area rectangle

the area of the equal-area rectangle

the rise

the run

the slope

If the graph altitudes are v0 and vf, and the width of the trapezoid `dt, then

what is the expression for the average velocity on the interval

what is the expression for the change in velocity on the interval

what is the expression for the slope

what is the expression for the area

what two equations of motion do we get from the expressions for slope and area

What are the mks units of each of the following quantities (mks units are expressed in terms of the basic units meters and seconds; e.g., m^3 / s^2 would be an mks unit; m^3 / s^2 however is not an answer to any of the questions below)?

position  

velocity 

acceleration 

clock time 

time interval

change in velocity

displacement

change in position

rate of change of position with respect to clock time 

rate of change of velocity with respect to clock time 

v0 

a 

`ds 

`dt

vf

v0 * `dt

`ds * `dt

`ds * a

v0 / `dt 

v0 * `dt^2  

a * `dt^2 

vf – v0 

What do you get when you do the following units calculations? 

(m/s) * s 

(m/s) / s 

(m/s^2) * s  

(m/s^2) / s 

(m/s^2) * m 

(m^2 / s) / s  

(m/s)^2

m * m/s

m * (m/s)^2

Class 090930

Horizontal range of a pendulum let loose at equilibrium

Hold a washer pendulum stationary at its equilibrium point and let it drop to the floor.  Mark or measure the position at which it strikes the floor.

Now, holding the pendulum so its equilibrium point is the same as before, pull it back and release it so that it swings back toward equilibrium.  The instant it reaches its equilibrium point, let go of the string and let it fall to the floor.  Record the distance the pendulum was pulled back from its equilibrium position, its length, its height above the floor, and the position at which it strikes the floor.

Repeat for a few trials.

Report your raw data below:

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Starting with the event of letting go and ending with the event of first contact with the floor, we assume that the washer is in free fall, the only force acting on it being the force of gravity (air resistance on the washer and on the thread will actually be present, but will be insignificant compared to our uncertainties in measurement).  We will analyze the data to determine its velocity at equilibrium.

If you hold the pendulum string in a fixed point, you can move the washer around a circle whose radius is equal to the length of the pendulum.  So as long as you are holding the string in a fixed position the pendulum will move along an arc of a circle. 

• As it swings back toward equilibrium, the direction of its motion is mostly horizontal but it also descends, so its motion has a downward vertical component. 
• As it approaches equilibrium it moves faster and faster, but due to the shape of the circle it descends more and more slowly. 
• After it passes equilibrium it continues for a time moving in the horizontal direction as it moves upward in the vertical direction. 

At equilibrium the washer is at its lowest point, neither rising nor falling, so at that instant it is moving entirely in the horizontal direction.

• If you let go of the string before the washer reaches equilibrium, then its fall will begin with a downward vertical velocity. 
• If you let go after equilibrium, then the initial vertical of its fall will be upward. 
• If you let go exactly at the equilibrium point, then the initial velocity of its fall will be entirely horizontal and its initial vertical velocity will be zero.

• Of course you can't let go exactly at the instant the washer reaches the equilibrium position, and the initial velocity of the fall won't in fact be entirely vertical.  The initial position of the washer won't be exactly at the equilibrium point, either. 
• Since in our analysis we will assume that the initial event occurs at the equilibrium point, with zero vertical velocity, there are multiple sources of uncertainty in this experiment.

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.

• The time required to fall your observed vertical distance, starting with initial vertical velocity zero and accelerating downward at 980 cm/s^2.
• The displacement of the washer in the horizontal direction.
• The horizontal velocity of the washer.

For each of your trials, report the pendulum length, the pullback distance, and the horizontal velocity of the falling washer.  Use one line to report the results of each trial:

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You need do this part for only one of your trials:

Suppose the pendulum was released a little early, so that the magnitude of its initial vertical velocity was 10% of the horizontal speed you just calculated.  (If you didn't get that part you can assume a horizontal speed of 60 cm/s)

• How long would it then take to reach the floor?  (you know the initial vertical velocity, vertical displacement and vertical acceleration)

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Estimate the vertical 'drop' as the pendulum swings to equilibrium

You need do this part for only one of your trials:

Draw a circle on your paper.  The diameter of your circle should be at least half the length of your paper.

Sketch the pendulum, hanging at equilibrium, as follows:

• The center of the circle represents the point at which you held the pendulum. 
• The radius of the circle represents the length of your pendulum. 
• The 'lowest' point on the circle, vertically below the center, will represent the center of the washer. 
• Sketch the washer, centered at this 'lowest' point.
• Sketch the string, which will extend from the center of the circle to the washer.

Now sketch the pendulum at its 'pullback' position (the 'held' end of the string will still be at the center of the circle).  Keep your sketch reasonably to scale.

Presumably you have observed the pullback and the length of the pendulum.  What is the pullback of the pendulum as a percent of its length? (e.g., if the length is 16 cm and the pullback 4 cm then pullback is 25% of the length).

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Since the radius of the circle represents the length of the pendulum, the number you just gave is also the pullback as a percent of the radius of the circle.

When the washer was pulled back in the horizontal, it was also raised in the vertical direction.  Estimate the distance it was raised as a percent of the distance it was pulled back.

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What therefore is your estimate of the distance the washer was raised as a percent of the pendulum's length?

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You measured the length.  Based on your preceding estimate, how much was it raised?

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If a coin was dropped from rest, and allowed to fall a distance equal to your previous result, how fast would it be going at the end of its fall? 

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Homework:

Your label for this assignment: 

ic_class_090930

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.

Do q_a_9 and q_a_10 .

In case the links don't work, the full addresses of the qa's are given below:

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_3.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_4.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_5.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_6.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_7.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_8.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_9.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm

Note also that you will receive a subsequent document with some alternative materials, and that you will be asked to complete a short portion of that document.