ic_class_090916

Looks good overall.

Check the details of your solutions against the document appended below:

Class 090916

Note:  When answering these questions, give your answer

Thanks.

Calibrate Rubber Band Chains:

Calibrate a rubber band chain (i.e., find its length

as a function of the force exerted to stretch it) using 1, 2, 3, 4 and 5

dominoes.  Give your raw data below in five lines, with number of dominoes

and length of chain separated by a comma, and an explanation following in

subsequent lines:

&&&&

Graph chain length vs. number of dominoes, and calculate

graph slope between each pair of points.  Give your results below. 

Table form would be good, with columns for length and number of dominoes,

rise, run and slope.  However as long as you include an explanation,

any format would be acceptable.

A table might appear as follows:

# dominoes	length	rise	run	slope
1	106
2	136	30	1	30
3	150	14	1	14
4	162	12	1	12
5	172	10	1	10

A more detailed solution is given for the next

question.

&&&&

Double the chain and calibrate it using 2, 4, 6, 8 and 10

dominoes.  Give your raw data below, in the same format as before:

For example the data might be

# dominoes	length (cm)
2	53
4	68
6	75
8	81
10	86

 

&&&&

Graph length of doubled chain vs. number of dominoes, and

calculate graph slope between each pair of points.

The the graph of length vs. number of dominoes might

look like the following:

Note that the above graph depicts the data points and

a smooth curve which fits the general trend of the points without actually

passing through any of the points.  It is reasonable to expect some

error in measurement, so that the data points are scattered around the curve

which represents the 'real' lengths of the chain.

A connected-line graph of the data points would look

like the following:

The above graph doesn't have the smooth curvature we

would expect from the actual rubber band chain, and it is also subject to

measurement uncertainties.  However, it does provide a reasonable

approximation to the behavior of our rubber band.

Based on this graph, we insert vertical lines

representing our 'graph altitudes', and label them.  The 'graph

altitudes' represent chain lengths in cm.

The rise corresponding to the first trapezoid is 68 -

53 = 15, representing a change in length of 15 cm.  The run is 2,

representing a change of two dominoes.  Thus the slope is 15 / 2 = 7.5,

representing 15 cm / (2 dominoes) = 7.5 cm / domino.

The second trapezoid has slope 7 cm / (2 dominoes) =

3.5 cm/domino.

The third trapezoid has slope 6 cm / (2 dominoes) =

3.0 cm/domino.

The fourth trapezoid has slope 5 cm / (2 dominoes) =

2.5 cm/domino.

A table representing these calculations:

# dominoes	length	rise (cm)	run (dominoes)	slope(cm/domino)
2	53
4	68	15	2	7.5
6	75	7	2	3.5
8	81	6	2	3
10	86	5	2	2.5

&&&&

Rotate the strap using the chain

Suspend the strap from your domino chain, supporting

the strap at its center so it will rotate in (or close to) a horizontal

plane, sort of like a helicopter rotor.  Rotate the strap through a few

revolutions and then release it.  It will rotate first in one

direction, then in the other, then back in the original direction, etc.,

with amplitude decreasing as the energy of the system is dissipated. 

Make observations that allow you to determine the period of its motion, and

determine whether its period changes significantly.

Give your raw data and your (supported) conclusions:

&&&&

Double the chain and repeat. 

Give your raw data and your (supported) conclusions:

&&&&

How does period of the oscillation compare between the

two systems?

&&&&

'Bounce' the dominoes on the end of the chain

'Bounce' a bag of dominoes on the chain.  Is

there a natural frequency?  Does the natural frequency depend on the

number of dominoes?  If so how does it depend on the number of

dominoes?

You might not be able to give complete answers to

these questions based on your data from class.  Give your data, your

conclusions, and you hypotheses (i.e., the answers you expect to get)

regarding these questions.

&&&&

How would you design an experiment, or experiments, to

further test your hypotheses? 

&&&&

Repeat for doubled chain.  How are the

frequencies of doubled chain related to those of single chain, for same

number of dominoes?

&&&&

You might not be able to give complete answers to

these questions based on your data from class.  Give your data, your

conclusions, and you hypotheses (i.e., the answers you expect to get)

regarding these questions.

&&&&

How would you design an experiment, or experiments, to

further test your hypotheses? 

&&&&

If you swing the chain like a pendulum, does its

length change?  Describe how the length of the pendulum might be

expected to change as it swings back and forth.

&&&&

Slingshot a domino block across the tabletop

Use your chain like a slingshot to 'shoot' a domino

block so that it slides along the tabletop.  Observe the translational

and rotational displacements of the block between release and coming to

rest, vs. pullback distance.

Give your results, in a series of lines.  Each

line should have pullback distance, translational displacement and

rotational displacement, separated by commas:

&&&&

Describe what you think is happening in this system

related to force and energy.

&&&&

Complete analysis of systems observed in previous class

Rotating Strap: 

For last time you calculated the average rate of

change of position with respect to clock time for each of five trials on the

rotating strap.  This average rate of change of position is an average

velocity.  Find the average rate of change of velocity with respect to

clock time for each trial.  As always, include a detailed explanation:

For example, if you observed that the strap rotated

through 300 degrees, and required 5 seconds to complete a rotation, then the

average rate of change of its angular position would be

ave roc of angular pos wrt clock time = (change in

position) / (change in clock time) = 300 deg / (5 sec) = 60 deg / sec.

We call this the average angular velocity.

We know that the final angular velocity is zero and

the average angular velocity is 60 deg/s.  A velocity vs. clock time

trapezoid will reveal that the initial angular velocity is 120 deg / s, so

that the change in angular velocity is (0 deg / s - 120 deg / s) = -120 deg

/ s, and the average rate of change of angular velocity with respect to

clock time is (-120 deg / s) / (6 s) = -20 deg / sec^2.

**** error ****

&&&&

(Note: Since the system is rotating its positions,

velocities and accelerations are actually rotational positions,

rotational velocities and rotational accelerations.  They are

technically called angular positions, angular velocity and angular

accelerations, because the position of the system is measured in units

of angle (e.g., for this experiment, the position is measured in

degrees).  These quantities even use different symbols, to avoid

confusion between rotational motion and translational motion (motion

from one place to another).  So technically the question above

doesn't use the terms 'position', 'velocity', etc. quite correctly. 

However the reasoning and the analysis are identical to the reasoning

we've been using to analyze motion, and for the moment we're not going

to worry about the technical terms and symbols.)

Atwood Machine:

Find the average rate of change of velocity with

respect to clock time for each trial of the Atwood machine.

Most students correctly found the average rate of

change of position with respect to clock time, obtaining vAve = `ds / `dt.

At this point there is enough information to sketch

the v vs. t trapezoid, and this should have been done to avoid the most

common errors:

• Some identified this quantity as the average rate
  of change of velocity with respect to clock time, which is incorrect.
• A number of students also divided vAve by `dt and
  identified this as the rate of change of velocity with respect to clock
  time.   This is also incorrect.  Should be change in
  velocity divided by change in clock time, and vAve is not the change in
  velocity.

The v vs. t trapezoid would form a triangle, with

initial 'graph altitude' 0 cm/s (the system was released from rest) and

midpoint 'graph altitude' equal to the average velocity.  The final

velocity would correspond to the final 'graph altitude', and the change in

velocity to the 'rise' of the graph.

&&&&

Hotwheels car:

For the Hotwheels car observed in the last class,

double-check to be sure you have your signs right:

• You pushed the car in two different directions on
  your two trials, one in the direction you chose as positive, and one in
  the direction you chose as negative.
• You will therefore have one trial in which your
  displacement was positive and one in which it was negative.
• Your final velocity in each case was zero. 
  In one case your initial velocity was positive, in the other it was
  negative.  Be careful that your change in velocity for each trial
  has the correct sign, and that the corresponding acceleration therefore
  has the correct sign.

There were numerous errors on this problem, of the

same type as those made on the Atwood Machine.  Subsequent assignments

have requested corrected and/or verified calculations for these two

situations.

&&&&

New Exercises

Exercise 1: 

A ball rolls from rest down each of 3 ramps, the

first supported by 1 domino at one end, the second by 2 dominoes, the

third by 3 dominoes.  The ramp is 60 cm long, and a domino is 1 cm

thick.  The motion is in every case measured by the same simple

pendulum.

It requires 6 half-cycles to roll down the first,

4 half-cycles to roll down the second and 3 half-cycles to roll down the

third.

Assuming constant acceleration on each ramp, find

the average acceleration on each.  Explain the details of your

calculation:

The average velocities are respectively

• 60 cm / (6 half-cycles) = 10 cm / half-cycle
• 60 cm / (4 half-cycles) = 15 cm / half-cycle
• 60 cm / (3 half-cycles) = 20 cm / half-cycle

A sketch of the graph trapezoids will show that

the changes in velocity are respectively 20 cm/half-cycle, 30

cm/half-cycle and 40 cm/half-cycle.

The accelerations are respectively

• (20 cm/half-cycle) / (6 half-cycles) = 3.3
  cm/half-cycle
• (30 cm/half-cycle) / (4 half-cycles) = 7.5
  cm/half-cycle
• (40 cm/half-cycle) / (3 half-cycles) = 13
  cm/half-cycle.

&&&&

Find the slope of each ramp.

The ramp slopes are respectively 1/60 = .017, 2/60

= 1/30 = .033 and 3/60 = 1/20 = .05.

&&&&

Graph acceleration vs. ramp slope.  Your

graph will consist of three points.  Give the coordinates of these

points.

The coordinates are (.017, 3.3 cm/half-cycle),

(.033, 7.5 cm/half-cycle), (.05, 13 cm/half-cycle).

The graph (which you should have hand-sketched)

should look something like this:

 

&&&&

Connect the three points with straight line

segments, and find the slope of each line segment.  Each slope

represents a average rate of change of A with respect to B. 

Identify the A quantity and the B quantity, and explain as best you can

what this rate of change tells you.

The rise and run between the first and second

point are respectively 7.5 cm/half-cycle - 3.3 cm/half-cycle = 4.2

cm/half-cycle, and .033 - .017 = .016, so the slope is (4.2

cm/half-cycle) / (.016) = 260 cm/(half-cycle).

The slope of the second segment is  410

cm/(half-cycle).

&&&&

Exercise 2:  A ball rolls down two consecutive

ramps, starting at the top of the first and rolling without interruption

onto and down the second.  Each ramp is 30 cm long.

The acceleration on the first ramp is 15 cm/s^2,

and the acceleration on the second is 20 cm/s^2.

For motion down the first ramp:

What event begins the interval and what even

ends the interval?

What are the initial velocity, acceleration

and displacement?

Using the equations of motion find the final

velocity for this interval.

Using the final velocity with the other

information about this interval, reason out the time spent on the

first ramp.

The beginning event is the release of the ball

at the top of the ramp, the terminating event is the ball reaching

the end of the first ramp.

The initial velocity is zero, since the ball

starts from rest.  Choosing the positive direction to be down

the ramp, the given acceleration is 15 cm/s^2 and the given

displacement is 30 cm.  Thus we know v0, a and `ds.

We can use the third equation to find vf =

+-sqrt(v0^2 + 2 a `ds).  We end up with a final velocity in the

neighborhood of 30 cm/s.

We can then reason out or use the equations to

find `dt, which is in the neighborhood of 2 seconds.

&&&&

For motion down the second ramp:

What event begins the interval and what even

ends the interval?

What are the initial velocity, acceleration

and displacement?

Using the equations of motion find the final

velocity for this interval.

Using the final velocity with the other

information about this interval, reason out the time spent on the

first ramp.

The beginning event is the ball reaching the

top of the second ramp, which it does as the same instant it reaches

the end of the first.  The terminating event is the ball

reaching the end of the second ramp.

The initial velocity is about 30 cm/s, since

the ball reaches the beginning of the second ramp at the same

instant it reaches the end of the first.  The given

acceleration on this ramp is 20 cm/s^2 and the given displacement is

30 cm.  Thus we know v0, a and `ds.

We can use the third equation to find vf =

+-sqrt(v0^2 + 2 a `ds).  We end up with a final velocity in the

neighborhood of 46 cm/s.

We can then reason out or use the equations to

find `dt, which is in the neighborhood of 1.3 seconds.

 

Challenge Exercise: 

The first part of this exercise is no more

challenging than the preceding problem.  It uses the result of that

problem: 

A ball accelerates uniformly down a ramp of length

60 cm, right next to the two 30-cm ramps of the preceding exercise. 

The ball is released from rest at the same instant as the ball in the

preceding exercise.

What is its acceleration if it reaches the end of

its ramp at the same instant the other ball reaches the end of the

second ramp?

The ball required 2 seconds on one ramp and 1.3

seconds on the other, so it required 3.3 seconds to travel the combined

length of the two ramps.

A ball which travels down a 60 cm ramp in 3.3

seconds has an average velocity around 18 cm/s and an acceleration of

about 22 cm/s^2.

&&&&

The second part is pretty challenging:

The 60 cm ramp is made a bit steeper, so that its

acceleration is decreased by 5 cm/s^2.  The experiment is repeated. 

How far will the ball on this ramp have traveled when it passes the

other ball?

The acceleration of the second ball is now about

17 cm/s^2.

We first have to figure out on which ramp the

first ball will be when the second passes it.

The first ball remains on the first ramp for 2

seconds.

In a 2-second interval the second ball,

accelerating from rest at about 17 cm/s^2, will

reach a velocity of about 17 cm/s^2 * 2 s = 34 cm/s, so its average

velocity will be about 17 cm/s and it will travel 34 cm.  It will

therefore reach the 30 cm position before the two seconds has elapsed

and the second ball will therefore be ahead of it at this point. 

(Note that since both balls accelerate from rest, the one with the

greater acceleration will remain ahead throughout this first interval,

so we really didn't need to do those calculations.  However the

information we have gained, that the first ball is about 4 cm behind

when it reaches the end of its first ramp, helps us visualize the

motion).

Before the acceleration of the second ball was

decreased, the two balls both reached the 60 cm position simultaneously. 

So it should be clear that the first ball will now reach the 60 cm

position first.

The second ball will therefore pass the first ball

sometime while the first ball is on its second ramp.

Summarizing the situation at the end of 2 seconds:

The first ball is at the end of its first ramp and

the beginning of its second ramp.  It is moving at about 30 cm/s

and will begin accelerating at 20 cm/s^2.

The second ball is 34 cm down its ramp, moving at

34 cm/s, and continues to accelerate at 17 cm/s^2.

We will now analyze the interval whose initial

event is the first ball reaching the beginning of the second ramp, and

which ends when the two balls are at the same position.

We don't know how long this will take, so we

consider the time interval `dt to be unknown.

For the first ball the initial velocity on this

interval is about v0_1 = 30 cm/s and acceleration is a_1 = 20 cm/s^2. 

Its displacement during this interval will be `ds_1 = v0_1 `dt + .5 a_1

`dt^2.

The second ball is moving at initial velocity 34

cm/s and accelerates at 17 cm/s^2.  Its displacement during the

same interval will be `ds_2 = v0_2 `dt + .5 a_2 `dt^2.

The two will be at the same position when the

first ball has moved 4 cm further than the second (recall that the

second ball begins this interval 4 cm ahead of the first).

Thus `ds_1 = `ds_2 + 4 cm, so that

v0_1 `dt + .5 a_1 `dt^2 = v0_2 `dt + .5 a_2 `dt^2

+ (4 cm).

This equation is quadratic in `dt, so we will put

it into the standard form a x^2 + b x + c = 0 and solve using the

quadratic formula.  The equation we get is

.5 (a_2 - a_1) `dt^2 + (v0_2 - v0_1) `dt + (4 cm)

= 0.

Substituting our known quantities we get

.5 (17 cm/s^2 - 20 cm/s^2) * `dt^2 + (34 cm/s - 30

cm/s) `dt + 4 cm = 0

Simplifying:

-1.5 cm/s^2 `dt^2 + 4 cm/s * `dt + 4 cm = 0

and we find that

`dt = [ - 4 cm/s +- sqrt( (4 cm/s)^2 - 4 * (-1.5

cm/s^2) * (4 cm) ) ] / (2 * (-1.5) cm/s^2)

 = (-4 cm/s +- sqrt(40 cm^2 / s^2) ) / (-3

cm/s^2)

 = (-4 cm/s +- 6.4 cm/s) / (-3 cm/s^2)

 = 3.5 s or -.8 s.

 

 

We therefore need to find the time interval `dt at

which both balls have the same position along their respective ramps.