Class 091019

Set up an Atwood machine using dominoes and clips.

• Choose a positive direction and start adding paperclips to the appropriate side until you get positive acceleration.  Measure the minimum possible positive acceleration, and note the number of clips that have been added.  Then add two more clips and repeat. 
• Remove all added clips and start adding clips to the negative side until you get the smallest possible negative acceleration, which you should measure.  Note the number of clips that have been added.  Then add two more clips to the negative side and repeat.
• Keep track of every 'move', including every mass added to either side of the system.

`q001.  Report your data for the Atwood machine.  Report all the raw data you will use in the analysis, in which you will figure out the four accelerations and the total effect of the gravitational force in accelerating each system.

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For our purposes here we will use the following data:

When 1, 2, 3 and 4 clips were added to the right-hand side the system did not accelerate.

When the 5th clip was added to the right-hand side the system descended 50 cm from rest in 4.5 cycles of a 36-cm pendulum.

When 2 more clips were added the system descended 50 cm from rest in 4.5 cycles of a 16-cm pendulum

Then the added clips were removed from the right-hand side and placed, one at a time, on the left-hand side, the system remained stationary when 1, 2 and 3 clips were added.

When the 4th clip was added on the left-hand side, the system descended 50 cm from rest in 4.5 cycles of a 25-cm pendulum.

After adding 2 more clips, the system descended 50 cm from rest in 4.5 cycles of a 14-cm pendulum.

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`q002.  Figure out the acceleration for each system, and report below, with a brief synopsis of how you proceeded from your raw data to your conclusions.

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When the 5th clip was added to the right-hand side the system descended 50 cm from rest in 4.5 cycles of a 36-cm pendulum. 

• The period of the pendulum is .2 sqrt(36) sec = 1.2 sec, so the time of descent was 4.5 * 1.2 sec = 5.4 sec.
• The acceleration associated with this descent from rest is about 3 cm/s^2 (calculation very approximate).

When 2 more clips were added the system descended 50 cm from rest in 3.5 cycles of a 16-cm pendulum

• The period of the pendulum is .2 sqrt(16) sec = .8 sec, so the time of descent was 4.5 * .8 sec = 3.6 sec.
• The acceleration associated with this descent from rest is about 9 cm/s^2 (calculation very approximate).

Then the added clips were removed from the right-hand side and placed, one at a time, on the left-hand side, the system remained stationary when 1, 2 and 3 clips were added.

When the 4th clip was added on the left-hand side, the system descended 50 cm from rest in 3.5 cycles of a 25-cm pendulum.

• The period of the pendulum is .2 sqrt(25) sec = 1.0 sec, so the time of descent was 4.5 * 1.0 sec = 4.5 sec.
• The acceleration associated with this descent from rest is about -5 cm/s^2 (calculation very approximate).

After adding 2 more clips, the system descended 50 cm from rest in 4.5 cycles of a 14-cm pendulum.

• The period of the pendulum is .2 sqrt(14) sec = .75 sec, so the time of descent was 4.5 * .75 sec = 3.4 sec.
• The acceleration associated with this descent from rest is about -10 cm/s^2 (calculation very approximate).

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`q003.  Find the total effect of the gravitational force on each system, as a multiple of the weight of a paper clip.  Paper clips added on one side exert positive force, clips added on the other side exert negative force on the system.

For example if you have 9 clips on one side and 5 clips on the other, the net gravitational force is equal to the weight of 4 clips.

Give a table of acceleration vs. total gravitational force.

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Assuming the direction to be positive when the right-hand side descends, the 5 additional clips added to this side result in a total gravitational force of +5 clip weights, and the 2 additional clips result in a total gravitational force of +7 clip weights.

The 4 additional clips added to the left side result in a total gravitational force of -4 clip weights, and the 2 additional clips result in a total gravitational force of -6 clip weights.

The table is therefore

wt  accel

-6    -10

-4     -5

+5   +3

+7   +9

where weights are in clip weights, and acceleration in cm/s^2.

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`q004.  Your table should include a fairly large 'gap' in the total gravitational force, where the acceleration is zero.  How wide is that 'gap'?  What does that 'gap' mean?

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A more complete table might read

wt  accel

-6    -10

-4     -5

-2      0

 0      0

+2      0

+4      0

+5   +3

+7   +9

where weights are in clip weights, and acceleration in cm/s^2.

The 'gap', where acceleration is zero, extends at least from -3 clip weights to +4 clip weights.

This 'gap' corresponds to the numbers of clips for which the system remains at rest.  To remain at rest some force other than the gravitational force must be acting.  In this case the most likely explanation seems to be a frictional force holding the pulley back.

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Suspend some number of dominoes, between 3 and 7, from your rubber band chain and measure its length.  Don't let anyone know how many dominoes you suspended.  Measure the length of the chain.  Measure once more with the chain supporting a single domino.  Then proceed as follows:

• Get together with 2 other people in the class and hook all three chains to the same paper clip. 
• Choose one person to provide the equilibrant force. 
• The other two will pull their chains back until they are at their previously measured lengths, keeping them at a right angle, while the third provides the equilibrant. 
• When the system is in this state, the person providing the equilibrant will measure the length of his or her chain, and data will be taken to determine the angles between the first two rubber bands and the equilibrant. 
• The two people whose rubber bands are at right angles will share the numbers of dominoes that resulted in the given length. 
• The lengths of all three rubber bands for this trial will be shared.

Then the experiment will be repeated twice more, so that it has been repeated once with each person providing the equilibrant.

If one or more groups of four people are required, adjust the experiment so that each individual provides the equilibrant force one time.

`q005.  You will find the magnitude of the resultant of the two forces exerted at right angles, the angle made by the equilibrant with these two forces, and the estimated force exerted by the equilibrant, based a two-point graph of number of dominoes supported vs. chain length.

Report the raw data you will use to find these quantities.

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Suppose the first person suspended 6 dominoes, and the second suspended 4 dominoes.  This is raw data.

Let's assume the first person's rubber band chain pulls along the negative x axis.  Its force is equal to the weight of 6 dominoes, so we say the force is -6 dominoes in the x direction.

Let's assume the first person's rubber band chain pulls along the positive y axis.  Its force is equal to the weight of 4 dominoes, so we say the force is +4 dominoes in the x direction.

The magnitude of the resultant force is found using the Pythagorean Theorem to be

F = sqrt( F_x^2 + F_y^2) = sqrt( (-6 dominoes)^2 + (4 dominoes)^2) = sqrt ( 52 dominoes^2) ) = 7.2 dominoes,

i.e., the resultant force is equal to the weight of 7.2 dominoes.

The angle of the resultant force is

theta = arcTan(F_y / F_x) = arcTan(4 dominoes / (-6 dominoes) ) + 180 deg = -34 deg + 180 deg = 146 deg, approx..

Now suppose the following data for the third person:

The rubber band chain, when exerting the equilibrant force, was observed to have a length of 80 cm.

This chain had a length of 60 cm when supporting 1 domino, and 71 cm when supporting 4 dominoes.

...

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`q006.  Find the magnitudes of the two forces, the angle of the equilibrant and the force exerted by the equilibrant, as estimated from your data.  Briefly indicate how you found these quantities.

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Very few students reported valid and complete data, and very few of those who did were able to follow these instructions.  We will address this experiment and its analysis upcoming classes.

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`q007.  Sketch the first two forces on a set of coordinate axes, and find the magnitude and angle of their resultant.  Give your results and your explanation of the calculation below.

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`q008.  Sketch the equilibrant on the same set of coordinate axes, and find its x and y components.

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`q009.  Compare the observed equilibrant with the resultant:

• To what extent is the equilibrant obtained from your data equal and opposite to the resultant of the other two forces? 
• By how much does its magnitude differ from that of the resultant?   
• By how much does its angle differ from that of the resultant? 
• By how much do its components differ from those of the resultant?

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`q010.  A ball rolls down an incline of length 30 cm with slope 0.1, and falls to the floor 1.05 meters below.

A slope of 0.1 can be associated with a rise of 0.1 and a run of 1.0.  A vector with a rise of 0.1 and a run of 1.0 has a y component of 0.1 and an x component of 1.0, so it makes angle

theta = arcTan(y comp / x comp) = arcTan (0.1 / 1.0) = 5.7 degrees

with the positive x axis, as measured in the counterclockwise direction.

A ramp slope of 0.1 can also be associated with a vector whose rise is either +0.1 or -0.1, and whose run is either +1.0 or -1.0.

• What possible angles might the vector therefore make with the positive x axis?

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In every case the vector makes an angle of 5.7 deg with the positive x axis.

However the vector can point into any quadrant, and its angle as measured counterclockwise from the positive x axis will depend on that quadrant.  The 5.7 deg angle can be either 'above' or 'below' the either the positive or negative x axis, as follows:

• If the rise and run are +0.1 and +1.0, then the vector is directed into the first quadrant and its angle is 5.7 degrees.
• If the rise and run are +0.1 and 11.0, then the vector is directed into the second quadrant and its angle is 180 deg - 5.7 deg = 174.3 deg.
• If the rise and run are -0.1 and -1.0, then the vector is directed into the third quadrant and its angle is 180 deg + 5.7 degrees.
• If the rise and run are -0.1 and +1.0, then the vector is directed into the fourth quadrant and its angle is 360 deg - 5.7 degrees = 354.3 deg.

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`q011.  The ball comes off the bottom of the incline with its velocity in the direction of the ramp.  What therefore are the possible angles, in the coordinate plane, of the velocity vector with a horizontal x axis?

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The possible angles are 5.7 deg, 174.3 deg, 185.7 deg and 354.3 deg.

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`q012.  If the ball requires 1.1 seconds to roll the length of the ramp, what is the magnitude of its velocity vector as it comes off the end of the ramp?

What therefore are the x and y components of its velocity vector at this instant?

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The ball's average velocity in the direction down the incline is 30 cm / (1.1 sec) = 27 cm/s, approx..  Its initial velocity is zero and we assume constant acceleration, so its final velocity is about 54 cm/s down the ramp.

The ball rolls down the incline, so that its velocity at the instant it leaves the ramp has a negative y component.  So the angle of its velocity vector is either 174.3 deg or 354.3 deg.

You could use either vector for your calculation.

If the angle is 174.3 deg, then the components of the velocity are

• v_x = 54 cm/s * cos(174.3 deg) = -53 cm/s, approx..
• v_y = 54 cm/s * sin(174.3 deg) = 5.4 cm/s, approx.

If the angle is 354.3 deg then we get

• v_x = 54 cm/s * cos(354.3 deg) = +53 cm/s, approx.
• v_y = 54 cm/s * sin(354.3 deg) = -5.4 cm/s, approx..

The initial x velocity is either positive or negative, depending on the side from which you view the ramp.

Either way the initial y velocity is negative.

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`q013.  The initial vertical velocity of the ball is equal to the y component of its velocity as it comes off the ramp.  Its acceleration in the vertical direction is that of gravity. 

• Use this information and the distance of the fall to determine the time required for the ball to fall to the floor.

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Choosing the upward direction as positive, consistent with the standard orientation of the x-y coordinate system, we find that we know v0, a and `ds for the vertical motion:

• initial vertical velocity = -5.4 cm/s
• vertical acceleration = -980 cm/s^2
• vertical displacement = -105 cm

The fourth equation of motion tells us that the final vertical velocity is about -450 cm/s, so the average vertical velocity is about (-5.4 cm/s + (-450 cm/s)) / 2 = - 230 cm/s.

The time of fall is about -105 cm / (-230 cm/s) = .46 sec, approx..

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`q014.  The ball's horizontal velocity is constant, and equal to the horizontal component of its velocity as it comes off the ramp.  Using this with the time interval determined previously, how far does it travel in the horizontal direction as it falls?

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The ball's horizontal velocity is either -53 cm/s or +53 cm/s, depending on how you view the system.

Either way its speed is a constant 53 cm/s so the distance it travels is

distance = ave speed * time interval = 53 cm/s * .46 sec = 25 cm, approx..

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`q015.  Assuming the ball to be a sphere of density 8 grams / cm^3 and diameter 2.5 cm, find its kinetic energy at each of the following instants:

• when it is released
• when it leaves the end of the ramp
• when it reaches the floor

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We find the mass of the ball:

The volume of a sphere is 4/3 pi r^3, and the radius of the ball is 1/2 * 2.5 cm = 1.25 cm.  So its volume is approximately

• volume = 4/3 pi ( 1.25 cm)^2 = 8 cm^3, approx.

so its mass is about

• mass = density * volume = 8 grams / cm^3 * 8 cm^3 = 64 grams.

We find the ball's KE at the three specified points:

When released the ball is at rest so its KE is zero.

Moving at 54 cm/s its KE is

• KE = 1/2 m v^2 = 1/2 * (64 grams) * (54 cm/s)^2 = 100 000 g cm^2 / s^2 (this is about 100 000 ergs).

If we calculate its KE in terms of kg and meters we get

• KE = 1/2 m v^2 = 1/2 * .064 kg * (.54 m/s)^2 = .01 kg m^2 / s^2 = .01 Joule.

When it reaches the floor it has a horizontal speed of 53 cm/s and a vertical speed of about 450 cm/s, so its resultant speed is

• v_floor = sqrt( (53 cm/s)^2 + (450 cm/s)^2 ) = 452 cm/s

so its KE would be

• KE = 1/2 m v_floor^2 = 1/2 * .064 kg * (4.52 cm/s)^2 = .65 kg m^2/s^2 = .65 Joules

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`q016.  Continuing the preceding, assuming that the ball descends 3 cm on the ramp:

• What is its PE change while on the ramp, and what is its PE change while falling to the floor?
• How much work is therefore done by the frictional force between the ball and the ramp?

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PE change is equal and opposite to the work done by conservative forces.

On the ramp its PE change is equal and opposite to the work done by the conservative force of gravity acting on the ball.

• The gravitational force is .064 kg * 9.8 m/s^2 = .64 Newtons, directed downward.
• The displacement in the vertical direction is 3 cm = .03 m downward.
• The work done by the gravitational force is therefore positive, equal to .63 N * .03 m = .02 Joules, approx..
• The PE change being equal and opposite to the work done by the gravitational force we have
• `dPE = -.02 Joules, approx.

Previously we found that KE increases from 0 to .01 Joule, approx., an increase of about .01 Joule.

Since

• `dW_NC_ON = `dPE + `dKE

we find that

• `dW_NC_ON = -.02 J + .01 J = -.01 J.

The nonconservative force (in this case the frictional force) does about -.01 Joule of work on the ball.

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`q017.  Answer for your rubber band, for the number of dominoes you chose to suspend from it:

Assuming that each domino weighs .2 Newton, and making the approximation that the force vs. length curve for the rubber band chain is linear, what is the average force required to stretch the chain  from its 1-domino length to the length corresponding to your chosen number of dominoes?

How much work is therefore required to stretch it between these two lengths?

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Suppose that the 1-domino length of your chain is 60 cm when supporting 1 domino, and 72 cm when supporting 4 dominoes.

Then a graph of force vs. length will include the points

• (60 cm, 1 domino) and (72 cm, 4 dominoes).

If the graph is linear then its slope is

• slope = rise / run = (4 dominoes - 1 domino) / (72 cm - 60 cm) = 3 dominoes / (12 cm) = .25 dominoes / cm,

which can also be expressed in Newtons, with 1 domino weighing about .2 N, as

• .25 * .2 N / cm = .05 N / cm.

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Homework:

Your label for this assignment: 

ic_class_091019

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.

Work through and submit q_A_15 and q_A_16.  These qa's are on impulse and momentum, and collisions.

URL's of qa's 10-19:

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
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
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_15.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_16.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_17.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_18.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_19.htm