ic_class_091014

The last three points lie pretty close to a straight line, so it isn't completely implausible that the trend is linear.

Right procedure, but the mass of the system is more than 20.2 kg.

Good; however the mass isn't right so your final result for velocity isn't right. This is easily corrected.

See appended solutions to correct this. I can't tell where you went wrong here.

Your answers should be in units of the vector magnitudes, not in percents.

The Atwood machine

The Atwood machine used in class consists of a domino

suspended from each side of a pulley.  When we add paper clips to one side

the system accelerates.  The more clips, the greater the acceleration.

If you weren't in class to get data today, you may assume

the following:

• for 1 added paper clip, the system moved 50 cm from
  rest in 6.5 swings of a 13 cm pendulum
• for 2 added paper clips, the system moved 50 cm from
  rest in 4.5 swings of a 16 cm pendulum
• for 3 added paper clips, the system moved 50 cm from
  rest in 3.5 swings of a 10 cm pendulum
• for 4 added paper clips, the system moved 50 cm from
  rest in 2.5 swings of a 12 cm pendulum

`q001.  Give your raw data for the Atwood machine. 

This includes all directly observed quantities used in calculating your results.

****

Using the information to be assumed by students who

weren't present:

• for 1 added paper clip, the system moved 50 cm from
  rest in 6.5 swings of a 13 cm pendulum
• for 2 added paper clips, the system moved 50 cm from
  rest in 4.5 swings of a 16 cm pendulum
• for 3 added paper clips, the system moved 50 cm from
  rest in 3.5 swings of a 10 cm pendulum
• for 4 added paper clips, the system moved 50 cm from
  rest in 2.5 swings of a 12 cm pendulum

&&&&

`q002.  For one of your trials show in detail how you

use your raw data to obtain the acceleration of the system for that trial.

****

Using the data for 2 added paper clips:

The system moved 50 cm in 3.5 swings of a 16 cm

pendulum.

A 16 cm pendulum has period .2 sqrt(16) sec = .8 sec.

3.5 swings therefore covers 4.5 * .8 sec = 3.6 sec.

average velocity is therefore 50 cm / (3.6 sec) = 15

cm/s, approx.

initial velocity is 0 so assuming uniform acceleration

(linear v vs. t) final velocity is 30 cm/s

acceleration is therefore 30 cm/s / (3.6 s) = 8

cm/s^2, approx..

(alternatively the equations of uniformly accelerated

motion, with v0 = 0, `ds = 50 cm and `dt = 2.8 s, give the same result)

&&&&

`q003.  Give a table of acceleration vs. number of

clips, one line at a time representing one trial at a time, with the number of

clips then the acceleration for the trial separated by four spaces.  At the

beginning or the end of the table, insert another line giving the units of each

column.  Alternatively, you can copy a table made using a spreadsheet.

acceleration vs. number of clips:

****

1, 4.6

2, 8

3, 20

4, 33

The number of clips is just a pure number, while the

accelerations are in cm/s^2.

&&&&

`q004.  Is it possible to fit a reasonable straight

line to the data?  How much 'leeway' do you think you have in where to fit

the line?

You're not trying to hit any of the points, and

generally your best line won't hit any.

You're basically trying to get as close as possible on

the average, while keeping the maximum vertical distance between your point

and the line small.

(Strictly you're trying to minimize the square of the

distance).

****

<h3>The second, third and fourth points lie along a nearly

straight line.  The first point distorts the picture significantly, pulling

the best-fit straight line away from the line of the last three points in such a

way as to raise its y intercept and reduce its slope.

A reasonable straight-line approximation for this

particular data might be close to the best-fit line, which is near y = 10 x - 8. 

However the fact that the points are not tightly grouped about this straight

line indicates a fair amount of uncertainty in both slope and intercept.

The rise of the graph is the change in acceleration, in

cm/s^2.

The run is the number of clips.

So the slope is (change in acceleration) / (change in

number of clips) = rate of change of acceleration with respect to number of

clips.

This is interpreted as the additional acceleration we

expect for each added paper clip.

For the given data the slope is 10; we therefore expect an

additional 10 cm/s^2 of acceleration for each added clip.

However recall the previous note about the uncertainty in

the acceleration.</h3>

&&&&

`q005.  Give the coordinates of two points on the

straight line you think comes as close as possible, on the average, to the

points of your graph.  Use one point near each end of your line, rather

that two points right next to one another.

****

A reasonable straight line for the data used here might

pass through the points (1 clip, 0 cm/s^2) and (5 clips, 40 cm/s^2). 

Neither is a data point.

&&&&

`q006.  Between the two points you specified in your

preceding answer, what is the rise, what is the run and what therefore is the

slope?  Be sure you specify the units of each of these quantities.

****

(1 clip, 0 cm/s^2) and (5 clips, 40 cm/s^2) is (40 cm/s^2

- 0 cm/s^2) / (5 clips - 1 clip) = 10 (cm/s^2) / clip.

&&&&

`q007.  How plausible is it that the actual

acceleration vs. number of clips is in fact well represented by a straight-line

graph, with the deviations of the individual points from the straight line being

due mostly to experimental uncertainties?

****

The fact that the points are cannot be tightly grouped

about a single straight line indicates a fair amount of uncertainty in both

slope and intercept.  However the dispersion about the straight line is not

extreme, so the slope is still expected to be somewhat accurate.

&&&&

`q008.  Specify the positive direction you chose for

your system.  This can be specified by stating which mass goes which way in

your chosen positive direction, or by stating whether the pulley rotates in the

clockwise or counterclockwise direction when the system is moving in your chosen

positive direction.

****

We will assume here that the greater mass is on the right,

and the positive direction will be the direction in which the mass on this side

descends while the mass on the left ascends.

The pulley will in this case rotate in the clockwise

direction.

(note for later reference that the positive direction for

rotation is generally regarded as the counterclockwise directions, for reasons

you will soon see)

&&&&

`q009.  Suppose that each side of the pulley has a

mass of 20 kg, and that each paperclip has a mass of about .2 kg (these masses

are not realistic for the system we observed, being much greater than the masses

we used in class).   Pretending that these are the actual masses in

your system:

How much force is exerted by gravity on each side of the

system when 1 paperclip is added to one side?

Assuming the absence of friction, what therefore is the

net force on the system?

What is the mass of the system?

What therefore should be its acceleration?

****

The side with the 20 kg mass will experience a downward

gravitational force of 20 kg * 9.8 m/s^2 = 196 Newtons.

The side with the paperclip will have a mass of 20.2 kg

and will experience a downward gravitational force of 20.2 kg * 9.8 m/s^2 = 198

Newtons.

The 196 Newton force is in the direction previously chosen

as negative, the 198 Newton force in the direction chosen as positive.

The net force is therefore

• F_net = 198 N + (-196 N) = 2 N.

The system consists of two 20 kg masses and the .2 kg

clip, all accelerating together.  So the mass of the system is 40.2 kg.

Its acceleration is therefore

• a = F_net / m = 2 N / (40.2 kg) = .050 m/s^2.

&&&&

`q010.  Find the acceleration for the system in the

preceding for 3 paperclips, and for 5 paperclips, added to the same side as

before.

Sketch a graph of acceleration vs. number of paperclips

and fit your best straight line to the graph.

How straight do you think your line is?

What is the slope of your line?  (be sure you include

units)

****

The systems will have net forces of 6 N and 10 N,

respectively, and masses of 40.6 kg and 41 kg, respectively.  So the

accelerations will be

a = F_net / m = 6 N / (40.6 kg) = .148 m/s^2 and

a = F_net / m = 10 N / (41 kg) = .244 m/s^2

Thus we have acceleration vs. number of clips, according

to the following table:

1, .048

3, .148

5, .244

with the first column representing number of clips,

the second accel in m/s^2.

A straight line fits these points very well, though not

exactly.

• The slope of the line is very close to .050 m/s^2.
• The interpretation of the slope is that the
  acceleration changes by 050 (m/s^2) for each additional clip.  So the
  slope could be expressed as .050 m/s^2 / clip.

&&&&

`q011.  For this series of examples, what is the mass

of a single paperclip, as a percent of the mass of the entire system?

What percent of the acceleration of gravity is the slope

of the graph you made for the preceding problem?

****

The mass of the clip is .2 kg, the mass of the entire

system varies from 40.2 kg to 41 kg.

• The mass of the clip as a proportion of the mass of
  the system is between .2 / 40.2 = .00498 and .2 / 41 = .00493; so the mass
  lies between .0050 and .0049 of the mass of the system. 
• In terms of percents, the mass of a clip is between
  .5% and .49% of the mass of the system.
• The acceleration of gravity is 9.8 m/s^2.  The
  slope of the line is about .050 m/s^2.  This is .050 m/s^2 / (9.8
  m/s^2) = .0051, or .51%.

It is worth noting that this result is very close to the

.49% or .50% we obtained for the mass of a clip as a percent of the mass of the

system.

So, for this example at least, the mass of a clip is about

.5% the mass of the system, and the slope of the graph is about .5% the

acceleration of gravity.

Remembering the the slope of the graph represents the

additional acceleration expected from the addition of 1 clip, the similarity in

these results should be striking.

&&&&

`q012.  From the slope of the graph you made for your

experiment, can you conjecture the mass of a paperclip as a percent of the mass

of the system?

****

The additional acceleration corresponding to 1 single

paperclip is .5% the acceleration of gravity.

• If gravity acted unimpeded on the mass of the entire
  system, then the net force on the system would be its weight, and it would
  acceleration with 100% the acceleration of gravity.
• If the net force on the system was half the force
  exerted by gravity (i.e., if the net force was equal to half its weight), it
  would acceleration with half that acceleration.
• To accelerate with 10% the acceleration of gravity,
  the net force would have to be 10% of weight of the system.
• If the system accelerates with 2% the acceleration of
  gravity, then the net force must be 2% of the weight of the system.

Based on the graph the system accelerates with .50% the

acceleration of gravity. 

• So according to the graph, the additional mass of a
  single paperclip would corresponds to .50% the mass of the system.

&&&&

`q013.  On your graph, what is the horizontal

intercept of your straight line (i.e., if the line is extended, where does the

line cross the x axis)? 

What are the units and the meaning of this point?

****

Using preceding results, the horizontal intercept was at

(1, 0), indicating that a mass difference equivalent to 1 paperclip corresponded

to 0 acceleration. 

• To the left of the intercept our straight line
  indicates a negative acceleration.  So a mass difference equivalent to
  less than 1 paperclip would correspond to a negative acceleration.  We
  did not make any observations in this experiment related to negative
  acceleration. 
• To the right of the intercept our graph is positive. 
  So a mass difference equivalent to more than 1 paperclip would correspond to
  a positive acceleration.  This is consistent with our observations in
  this experiment.

&&&&

`q014.  If the frictional force on the system is

increased, would the acceleration of the system increase, decrease or stay the

same?

What effect would this have on the points of your graph?

What effect would this have on the straight line that

approximates your points?

What effect would this have on the x intercept of the

straight line?

****

More friction would 'hold the system back' more. 

This would have two effects on the system:

• It would require more clips on the 'positive' side
  before the system would accelerate in the positive direction.
• A given number of clips would result in less
  acceleration that if friction was absent or negligible.

The corresponding effects on the graph will be as follows:

• The acceleration for any given number of clips will
  be reduced as a result of friction.  Since acceleration is graphed as
  the 'vertical' quantity, this will lower each point of the graph.  The
  straight-line approximation will therefore be lowered as well.
• The x intercept, which corresponds to the number of
  additional clips required to begin accelerating the system in the positive
  direction, would increase due to friction (i.e., the x intercept of the
  graph would be further to the right).
• If the force exerted by friction is the same
  regardless of velocity, then the frictional force will reduce the
  acceleration for each trial by the same amount.  This will have the
  effect of lowering each point of the graph by the same amount.  This
  will affect the position, but not the slope, of the straight-line
  approximation to the graph.

It is well worth noting that if the frictional force is

constant (e.g., if it is independent of position and velocity), the graph will

be lower than the 'ideal' no-friction graph, but the slope of the graph will not

be affected.

&&&&

Energy considerations

`q015.  Going back to the example problem where each

mass is 20 kg and each clip has mass .2 kg, let's assume that three clips are

added to the mass on the left, so that the system accelerates in the

counterclockwise direction. 

We want to analyze the energy situation if the system

moves .7 meters in our chosen positive direction.

What downward force is exerted on a 20 kg mass by gravity?

By how much does the gravitational potential energy of the

20 kg mass on the right therefore change as the system moves +.7 meters?

Answer the same for the 20 kg mass on the left.

Answer the same for the three clips.

What therefore is the PE change of the system?

****

Gravity exerts a force of 20 kg * 9.8 m/s^2 = 196 N on a

20 kg mass, and a force of 2 N on a .2 kg mass.

As the mass on the right descends, gravity acts downward,

so the gravitational force on the right-hand mass is in the same direction as

its displacement.  Gravity therefore does positive work, equal to 196 N *

.7 m = 137 Joules, on the left-hand mass.  The PE of this mass, being equal

and opposite to the work done by the conservative force acting on it, therefore

decreases by 137 J.  For this mass we say that `dPE = -137 J (if we prefer

to use different symbols for the PE change of each component, we might

distinguish this one by stating that `dPE_rh_mass = -137 J).

As the mass on the left ascends, gravity acts downward, so

the gravitational force on the left -hand mass is in the direction opposite its

displacement.  Gravity therefore does negative work, equal to -196 N * .7 m

= -137 Joules, on the left-hand mass.  The PE of this mass, being equal and

opposite to the work done by the conservative force acting on it, therefore

increases by 137 J.  For this mass we say that `dPE = +137 J (if we wish to

distinguish our symbols we might say that `dPE_lh_mass = +137 J).

The three clips have total mass .6 kg.  They are on

the right-hand side of the system.  The analysis is similar to that of the

first paragraph, and we conclude that the PE of the clips decreases by .6 kg *

9.8 m/s^2 * .7 m = 4.1 Joules.  For the mass of the clips we say that `dPE

= -4.1 Joules (or to distinguish symbols, `dPE_clips = -4.1 J)..

The PE change of the system is equal to that of its

components:

`dPE_system = -137 J + 137 J - 4.1 J = -4.1 J.

(if we prefer to distinguish our symbols we might write

`dPE_system = `dPE_rh_mass + `dPE_lh_mass + `dPE_clips =

-137 J + 137 J - 4.1 J = -4.1 J).

&&&&

`q016.  Assuming that no nonconservative forces act

on the system, what therefore is its change in kinetic energy?

****

If no nonconservative forces act on the system then `dW_NC_ON

= 0 so that our energy conservation law

`dW_NC_ON = `dKE + `dPE implies that

`dKE + `dPE = 0

we conclude that

`dKE = - `dPE = - (-4.1 J) = +4.1 J.

&&&&

`q017.  The kinetic energy of the system is 1/2 m

v^2, where m is the mass of the system.

Assuming it started from rest, its KE at the end of the

interval will be equal to its change in KE.

What therefore is its KE at the end of the interval?

What is the mass of the system?

What therefore is its velocity at this point?

****

Since, starting from rest, its KE at the end of the

interval is equal to `dKE we see that at the end of the interval

KE = 4.1 J.

The mass of the system with 3 clips is 40.6 N.

Since by the definition of kinetic energy we have

KE = 1/2 m v^2

we can find the velocity of the system by solving this

equation for v. 

Multiplying both sides of the equation by 2 / m we have

2 / m * KE = 2 / m * (1/2 m v^2) so that

2 KE / m = v^2 and

v = +- sqrt( 2 KE / m).

Substituting 4.1 J for the KE and 40.6 kg for m we obtain

v = +- sqrt( 2 * 4.1 J / (40.6 kg) ) = +- .14 m/s,

approx..

Our velocity could therefore be either +.14 m/s or -.14

m/s.  The choice of which depends on our original choice of positive

direction.

In this case we specified the positive direction to be

that in which the greater mass descends.  Since for this particular trial

it is the greater mass that descends, we conclude that its final velocity is

v = +.14 m/s.

&&&&

`q018.  How would your answers to the last two

questions change if there is a frictional force of 3 N acting on the system?

****

A 3 Newton frictional force would act in the direction

opposite motion, and would therefore do negative work.  The frictional

force is nonconservative.  Assuming no additional forces act on the system,

we would therefore have

`dW_NC_ON = - (3 N) * (.7 m) = -2.1 Joules.

Our energy conservation equation would therefore be

`dW_NC_ON = `dKE + `dPE

-2.1 J = `dKE + (-4.1 J) so that

`dKE = -2.1 J + 4.1 J = 2 J.

The velocity of the system would therefore be

v = +- sqrt( 2 KE / m) = +- sqrt( 2 * 2 J / (40.6 kg) ) =

+- .10 m/s.

Again our choice of positive direction implies the

positive solution so we have final velocity

v = +.10 m/s.

&&&&

Vectors

`q019.  The figure on the board represented three

vectors, one of magnitude 10 units at 305 deg, one of magnitude 8 units at 158

deg and one of 4 units at 80 deg.

According to our estimates:

• the x and y components of the first force are
  respectively 45% and -86% of that force
• the x and y components of the second force are
  respectively -90% and 46% of that force
• the x and y components of the third force are
  respectively 20% and 95% of that force

Based on these estimates calculate the x and y components

of the three forces.

What is the sum of all the x components?

What is the sum of all the y components?

****

Let's call the three forces F_1, F_2 and F_3.

The x and y components of the three forces are

F_1x = .45 * 10 units = 4.5 units and F_1y = -.86 * 10

units = -8.6 units

F_2x = -.90 * 8 units = -7.2 units and F_2y = -.46 * 8

units = -3.7 units

F_3x = .20 * 4 units = .8 units and F_2y = .95 * 4 units =

3.8 units

Using F_tot for the total force, the sum of the x and y

components components are

F_tot_x = F_1x + F_2x + F_3x = 4.5 unit + (-7.2 units) + 

0.8 units = -1.9 units

F_tot_y = F_1y + F_2y + F_3y = -8.6 units + (- 3.7 units )

+ 3.8 units = -8.5 units.

&&&&

`q020.  The actual percents are given by the sine and

cosine functions as decimals.  For example if the percents are 45% and

-86%, the cosine function would give us .45 and the sine function would give us

-.86.

For an angle of 305 degrees, use your calculator to find

sin(305 deg) and cos(305 deg).  (make sure you calculator is in 'degree'

mode; using the sin/cos button find sin(305) and cos(305))

What do you get and how do the accurate values compare

with our estimates?

Using the accurate values of the sine and cosine of 305

degrees, what are the x and components of the first vector?

****

With a calculator in degree mode, as opposed to radian

mode, you get

• cos(305 deg) = 0.5735764363 (compare with our
  estimate .45)
• sin(305 deg) = -0.8191520442 (compare with our
  estimate -.86)

These results are reasonably close to those of our

ballpark estimate.

NOTE:  If you got -.26 and -.97, corresponding to

-26% and -97%, you didn't have your calculator in degree mode.  Those

results are what you would get if your calculator was in radian mode.

Using 2-significant-figure approximations the x and y

components of the first vector are

• F_1x = 10 units * cos(305 deg) = 10 units * .57 = 5.7
  and
• F_1y = 10 units * sin(305 deg) = 10 units * (-.82) =
  -8.2.

&&&&

`q021.  Use the same procedure to find the x and y

components of the second and third vectors.

Give you results below, and include a brief explanation of

your results.

****

Results for the sines and cosines are:

• cos(158 deg) = -0.9271838545 (compare with our
  estimate -.90)
• sin(158 deg) = 0.3746065934 (compare with our
  estimate .46)
• cos(90 deg) = 0.1736481776 (compare with our estimate
  .20)
• sin(80 deg) = 0.9848077530 (compare with our estimate
  .95)

So

• F_2x = 8 units * (-.93) = -7.4 units and F_2y  = 
  8 units * .46  =  3.6 units
• F_3x  =  4 units * .17  =  0.7
  units and F_3y  =  4 units * .98  =  3.9 units.

&&&&

`q022.  If you add up the x components of the three

vectors, what do you get?

If you add up the y components of the three vectors, what

do you get?

****

The x and y components of the three vectors are:

• F_1x =  10 units * .57  =  5.7 units
  and F_1y = 10 units * (-.82) = -8.2 units
• F_2x = 8 units * (-.93) = -7.4 units and F_2y  = 
  8 units * .46  =  3.6 units
• F_3x  =  4 units * .17  =  0.7
  units and F_3y  =  4 units * .98  =  3.9 units.

Thus

• F_x = F_1x + F_2x + F_3x = 5.7 units + (-7.4 units) +
  0.7 units = -.10 units and
• F_y = F_1y + F_2y + F_3y = -8.2 units + 3.6 units +
  3.9 units = -.7 units.

Good work. You appear, however, to have misconstrued the vector components.

See my notes and check against the solutions I've provided.