course 201
9/13 7pm
Class 090909vvvv
Rubber band measurements
You have data that allows you to determine the length of the thin rubber bands and the length of the thick rubber band, for each setup.
Report the span of your hand on the 'ruler' you used to measure the rubber band system, as well as your height in inches. Report as two numbers separated by commas:
35.5, 67
Report you raw data below (this will be the quantities you actually measured in class; note that you didn't measure the lengths of the rubber bands but the positions of their ends, so your raw data will not include the lengths). :
One large rubber band, one small rubber band:
Relaxed:
Large band from 5 cm-16.5 cm
Small band from 21 cm-30.5 cm
Stretched a bit:
Large band from 5cm – 18cm
Small band from 22.75 cm-36.5 cm
Stretched more:
Large band from 5 cm-19cm
Small band from 23.75cm-41cm
One large rubber band, three small rubber bands together:
Relaxed:
Large band from 5 cm-17.5cm
3 Small bands from 22 cm-32.5 cm
Stretched a bit:
Large band from 5 cm-19.5 cm
3 Small bands from 24 cm-35.5 cm
Stretched more:
Large band from 5 cm – 22.5 ccm
3 Small bands from 27 cm-41cm
* In the first setup you had a single 'thin' rubber band opposing the single 'thicker' rubber band. You observed this system for three different degrees of stretch.
Beginning in the line below, report the lengths of the 'thin' rubber band and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
(lengths of 'thin' then 'thick' bands, first trial (least stretch)):
9.5, 11.5
(lengths of 'thin' then 'thick' bands, second trial (medium stretch)):
13.75, 13
(lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)):
17.25, 14
Now sketch a graph of y vs. x, where y = length of 'thin' rubber band and x = length of 'thick' band. Your graph will consist of three points.
What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
The slope is 1.7
Slope= rise/run=change in y/change in x= (final y value-initial y value)/(final x value-initial x value)= (13.75-9.5)/(13-11.5)=4.25/2.5= 1.7
What is the slope of the graph between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
The slope is 3.5
Slope= rise/run=change in y/change in x= (final y value-initial y value)/(final x value-initial x value)= (17.25-13.75)/(14-13)=3.5/1=3.5
&&&& Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
I believe that the difference in slopes is due to the actual behavior of the rubber bands. It makes sense to me that on the first stretch from relaxed, both rubber bands stretch almost equally, but on the second stretch, it would be easier for the thin rubber band to stretch farther than the thick. The slope tells us this, that the y value, the length of the thin rubber band, is increasing 1.7 units for every 1 unit the thick rubber band increases in the first set of points, and that in the second trial the thin stretched 3.5 units for every 1 unit the thick stretched. This seems reasonable, as it obviously takes less effort to stretch a thin rubber band a long distance than it does a thick one.
* In the second setup you had three 'thin' rubber bands, all stretched between the same two paper clips, opposing the single 'thicker' rubber band. You observed this system for three different degrees of stretch.
Beginning in the line below, report the lengths of the 'thin' rubber bands and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
(lengths of 'thin' then 'thick' bands, first trial (least stretch)):
10.5, 12.5
(lengths of 'thin' then 'thick' bands, second trial (medium stretch)):
11.5, 14.5
(lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)):
14, 17.5
Now sketch a graph of y vs. x, where y = common length of 'thin' rubber bands and x = length of 'thick' band. Your graph will consist of three points.
What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
The slope is 1/2
Slope= rise/run=change in y/change in x= (final y value-initial y value)/(final x value-initial x value)=(11.5-10.5)/(14.5-12.5)=1/2
What is the slope between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
The slope is
Slope= rise/run=change in y/change in x= (final y value-initial y value)/(final x value-initial x value)=(14-11.5)/(17.5-14.5)=2.5/3=.833
Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
Again, I believe that the difference in slopes is due to the actual behavior of the rubber bands. In this situation, it is no longer extremely easy for the thin rubber band to stretch more because there are three together, which strengthens the band and makes it more difficult to pull a long length. The slope shows this, that on the first trial the thin bands were actually changing lengths one unit for every 2 units that the thick rubber band was stretching. This leads me to believe that it was actually easier to stretch the thick band than the three thin ones together. The second trial confirms this, with a slope less than one, suggesting that once again the thin rubber bands covered less distance than the thick in this trial.
Question to think about (and answer as best you can): You you think that during the measurement process, the rubber bands were being held apart by force, power, energy or something else, or perhaps by all three? Answer beginning in the line below, and support your answers with your best reasoning. Don't worry about being wrong, but do give it some thought.
I think that the rubber bands were being held apart by force, the force of my muscles pulling the ends of the paperclips and making the rubber bands make up the distance by stretching out. If I were to suddenly lose my grip, and the force I was putting on the system was eliminated, the rubber bands would spring back into their relaxed, shorter position.
Acceleration of Toy Cars in Two Opposite Directions
You also measured the motion of a couple of toy cars, moving in two opposite directions. You were asked to obtain information that will tell you the acceleration of each car in each direction.
Give your raw data in starting in the line below. Be sure to include all information necessary to interpret your data.
10 lines of notebook paper pendulum, 96 cycles in 60 seconds
Positive direction:
3 half cycles, 13.75 inches traveled
4 half cycles, 24 inches traveled
Negative direction:
4 half cycles, 18.75 inches
3 half cycles, 20.5 inches
Pick 'north' or 'south' for your positive direction. State your choice:
North will be my positive direction
For the first trial in which the car moved in the northerly direction, explain how you reason out the acceleration. Show how you reason out your results, starting with the raw data, based on the definitions of rate of change, average velocity and average acceleration. You may assume that the acceleration is uniform, so that the v vs. t graph is in fact trapezoidal. Give you explanation starting in the line below:
To get to acceleration, we first have to have values for the change in velocity and the change in time. We already have the change in time in half cycles, which we can convert to seconds. To get a value for change in velocity, we need to calculate the change in position with respect to time.
First, to convert 3 half cycles of the pendulum to seconds: 3 half cycles=1.5 cycles, then set up a proportion to the known cycles in one minute, 1.5/x=53/60, x=1.7 seconds, which is `dtime
Then, to find the average velocity: average velocity is the average rate of change of position with respect to clocktime=`dposition/`dtime=13.75 inches/1.7 seconds=8.1 in/sec
Since this is the average velocity, and we know that final velocity was zero as the car stopped, it can be inferred that 16.2 was the maximum velocity.
To find the acceleration: acceleration is the average rate of change of velocity with respect to clock time=`dvelocity/`dtime=final velocity-initial velocity/`dtime= (16.2 in/sec-0 in/sec)/1.7 sec=9.5 in/sec^2
Repeat for the first trial in the southerly direction:
For the southern direction, position is considered negative.
To convert pendulum cycles to seconds: 4 half cycles= 2 cycles, 2/x=53/60, x=2.3 seconds
To find average velocity: average velocity is the average rate of change of position with respect to clocktime=`dposition/`dtime= -18.75 inches/2.3 seconds= -8.2 in/sec
This means -16.4 was the maximum velocity
To find the acceleration: acceleration is the average rate of change of velocity with respect to clock time=`dvelocity/`dtime=final velocity-initial velocity/`dtime= (-16.4 in/sec-0 in/sec)/2.3 sec= -7.1 in/sec^2
Find the acceleration for the next trial; however you may abbreviate your calculations and don't need to repeat the verbal explanations:
Trial 2, in the Northern (positive) direction:
4 half cycles=2 cycles, 2/ x=53/60, x=2.3 seconds
Average velocity=` dposition/`dtime= 24 inches/2.3 seconds=10.4 in/sec
Maximum velocity=20.8
Acceleration=`dvelocity/`dtime=final velocity-initial velocity/`dtime= 20.8 in/sec-0 in/sec)/2.3 sec= 9.0 in/sec^2
Trial 2, in the Southern (negative) direction:
3 half cycles=1.5 cycles, 1.5/x=53/60=1.7 seconds
Average velocity=`dposition/`dtime= -20.5 inches/1.7 seconds=-12.1 in/sec
Maximum velocity=-24.4
Acceleration=`dvelocity/`dtime=final velocity-initial velocity/`dtime= -24.2 in/sec-0 in/sec)/1.7 sec= 14.2 in/sec^2
Intervals
An interval runs from some initial event to a final event.
Example: f an object's velocity is increasing at it coasts down a ramp, we might speak of the interval between first reaching velocity 2 meters / second and first reaching velocity 5 meters / second.
• The initial event would be the event of reaching 2 meters / second for the first time.
• The final event would be the event of reaching 5 meters / second for the first time.
In the case of the given information:
• 2 meters/second would be the initial velocity, which we would designate by v0.
• 5 meters/second would be the final velocity, which we would designate by vf.
Furthermore we can speak of initial and final positions and clock times, which we might or might not know:
• The initial event would occur at some initial clock time, which could be denoted by t0.
• The final event would occur at some initial clock time, which could be denoted by tf.
• The time interval corresponding to this interval would be the change in clock time, `dt = tf - t0.
• The initial event would occur at some position, which could be denoted by s0.
• The final event would occur at some final position, which could be denoted by sf.
• The change in position on this interval, also called the displacement, would be `ds = sf - s0.
In an experiment, any of the quantities t0, tf, `dt, s0, sf or `ds might be directly measured using a timing device (e.g., a pendulum, a stopwatch, a computerized timer) or a position-measuring device (e.g., a meter stick or a ruler). There would of course be some uncertainty in the measurement. We would not generally measure all these quantities, but we could.
Depending on the object, we might be able to obtain some measure of v0, vf or `dv. More commonly, we would infer these quantities by measuring a series of positions or clock times.
What might be the initial event and the final event in each of the following situations, and what quantities are given for each?
A drag racer completes a 1/4 mile time trial in 12 seconds.
Initial clock time=0 seconds, final clock time = 12 seconds, initial position=0 miles, final position =.25 miles
A muon created in the upper atmosphere spends 12 milliseconds in the atmosphere before disintegrating.
t0=0 milliseconds, tf= 12 milliseconds
After receiving a handoff at his own 10-yard line, with the game clock reading 3:42, a fullback gains 30 yards before being tackled with the game clock reading 3:37.
s0= 0 yards, sf=30 yards, t0=0 seconds, tf=5 seconds
A mosquito hatched at 9:00 a.m. is eaten by a bat at 3:42 p.m. of the same day, and a point 40 feet north of where it hatched.
S0=0 feet, sf=40 ft, t0=0 minutes, tf=402
Deriving the equations of uniformly accelerated motion from the definitions of rate, velocity and acceleration:
Recall from previous classes that the definitions of rate, velocity and acceleration, along with the assumption of uniform acceleration, lead us to the following equations for uniformly accelerated motion on an interval:
• Equation 1: `ds = (vf + v0) / 2 * `dt
• Equation 2: a_ave = (vf - v0) / `dt.
Both of these equations are easily derived by finding the slope and area of a v vs. t trapezoid with altitudes v0 and vf, and base `dt.
The five quantities which appear in these equations are:
• v0, the velocity at the beginning of the interval
• vf, the velocity at the end of the interval
• `dt, the change in clock time during the interval
• a_ave, the average acceleration on the interval
• `ds, the change in position, or displacement, on the interval.
It's worth noting that each of the two equations includes four of these five quantities.
We can derive two additional equations as follows:
Solve the second equation for vf and plug the resulting expression into the first equation, eliminating vf and obtaining the equation
• Equation 3: `ds = v0 `dt + .5 a `dt^2
Solve the second equation for `dt and plug the resulting expression into the first equation, eliminating `dt and obtaining the equation
• Equation 4: vf^2 = v0^2 + 2 a `ds.
The details of the algebra are given below:
Derivation of Equation 3:
Start with Equation 2, and solve for vf:
a_Ave = (vf - v0) / `dt. Multiply both sides by `dt:
a_Ave * `dt = (vf - v0) / `dt * `dt. The right-hand side rearranges to (vf - v0) * (`dt / `dt), and `dt / `dt = 1 so
a_Ave * `dt = vf - v0. Add v0 to both sides
a_Ave * `dt + v0 = vf - v0 + v0. Since -v0 + v0 = 0 we get
a_Ave * `dt + v0 = vf. Switch right- and left-hand sides of the equation and reverse the order of addition:
vf = v0 + a_Ave * `dt.
Now plug this expression into Equation 1:
`ds = (vf + v0) / 2 * `dt. Substituting v0 + a_Ave * `dt for vf we have
`ds = (v0 + a_Ave * `dt + v0) / 2 * `dt. Simplify the numerator of the right-hand side by adding the two v0 terms:
`ds = ( 2 v0 + a_Ave * `dt) / 2 * `dt. Applying the distributive property of multiplication over addition we have
`ds = (2 v0) / 2 * `dt + a_Ave * `dt / 2 * `dt. Since (2 v0) / 2 = (2 / 2) * v0 = 1 * v0 = v0, `dt * `dt = `dt^2, and 1/2 = .5 we have
`ds = v0 * `dt + .5 a_Ave * `dt^2.
This is Equation 3.
Derivation of Equation 4:
We first solve Equation 2 for `dt.
Starting with
a_Ave = (vf - v0) / `dt, we multiply both sides by `dt to obtain
a_Ave * `dt = vf - v0 (this was also done and fully explained in the derivation of Equation 3 above).
Now divide both sides by a_Ave to get
`dt = (vf - v0) / a_Ave.
Now plug this expression into Equation 1:
`ds = (vf + v0) / 2 * `dt. Substituting (vf - v0) / a_Ave for `dt we get
`ds = (vf + v0) / 2 * (vf - v0) / `a_Ave. This gives us
`ds = (vf + v0) * (vf - v0) / (2 * a_Ave).
Note that by the distributive law (vf + v0) * (vf - v0) = vf * (vf - v0) + v0 * (vf - v0) = vf * vf - vf * v0 + v0 * vf - v0 * v0 = vf^2 - vf * v0 + vf * v0 - v0^2 = vf^2 - v0^2, so our equation becomes
`ds = (vf^2 - v0^2) / (2 * a_Ave) .
We choose to solve this equation for vf^2.
We first multiply both sides by 2 * a_Ave:
`ds * (2 * a_Ave) = (vf^2 - v0^2) . Then add v0^2 to both sides to get
2 * a_Ave * `ds + v0^2 = vf^2 - v0^2 + v0^2. Since -v0^2 + v0^2 = 0, after switching left-and right-hand sides we have
vf^2 = v0^2 + 2 a_Ave `ds.
This is Equation 4.
Using the Equations of Uniformly Accelerated Motion
The four equations of uniformly accelerated motion, as derived above, are:
• Equation 1: `ds = (vf + v0) / 2 * `dt
• Equation 2: a_Ave = (vf - v0) / `dt
• Equation 3: `ds = v0 * `dt + .5 * a_Ave * `dt^2
• Equation 4: vf^2 = v0^2 + 2 a_Ave * `dt^2.
Since acceleration is uniform, we don't really need the subscript _Ave in our symbol a_Ave. Instead we can just write a.
Equation 2 is fine the way it appears, but is often more convenient to use rearranged to the form vf = v0 + a_Ave * `dt (starting with Equation 2 as expressed above, multiply both sides by `dt, then add v0 to both sides and switch right- and left-hand sides).
So the equations are more commonly expressed as
• Equation 1: `ds = (vf + v0) / 2 * `dt
• Equation 2: vf = v0 + a * `dt
• Equation 3: `ds = v0 * `dt + .5 * a * `dt^2
• Equation 4: vf^2 = v0^2 + 2 a * `dt^2.
We will use the equations in this form. (Note that your text has an equivalent but slightly different form of the equations; we'll explain the difference later).
You should verify the following properties of these equations for yourself:
• The equations are expressed in terms of the five variables v0, vf, `dt, a and `ds.
• Each equation includes exactly four of these five variables.
• Every possible combination of three of the five variables appears in at least one of the four equations (for example, v0, `dt and `ds is one possible combination of three of the five variables, and all appear in Equation 1; in addition all three appear in Equation 3).
It follows that if we know any three of the five quantities v0, vf, `dt, a and `ds, we can find at least one equation in which these quantities all appear; and furthermore since there are only four quantities in any one equation, that equation can be solved to find the value of the fourth quantity.
We will do a number of exercises as we learn to apply this scheme.
* Exercise 1: You have already analyzed your information for the race cars you observed in class. Now do the following:
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From your raw data you can determine `ds and `dt.
The initial and final events can be described as 'car leaves end of finger' and 'car comes to rest'. So you also know the value of which of the five quantities?
Final velocity
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know (i.e., circle vf, `ds and `dt in all four equations).
Write down the one equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled?
`ds = (vf + v0) / 2 * `dt , and v0 was not circled
Solve this equation for the non-circled variable and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but make your best attempt. We will be discussing this much more fully in our next class, but you at least need to get the wheels turning, and your instructor needs to know what you can and cannot do with the algebra.
V0=(2`ds/`dt)-vf
Having solved the equation as best you can, substitute the values of the three known quantities vf, `ds and `dt into that equation. Then simplify your expression to get the value of the unknown quantity. Again, this will take some practice and you might have made some errors, but do your best, for the same reasons outlined above.
V0=(2*13.75 in/1.7sec)=16.2 in/sec
* Exercise 2
A ball is released from rest on a ramp of length 4 meters, and is timed from the instant it is released to the instant it reaches the end of the ramp. It requires 2 seconds to reach the end of the ramp.
What are the events that define the beginning and the end of the interval?
S0=0 meters, sf=4 meters, t0=0 seconds, tf=2 seconds
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From the given information you know the values of three of the five quantities. What are the known quantities? `dt, `ds, v0
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know.
Write down an equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled?
`ds = (vf + v0) / 2 * `dt, vf was not circled
Solve this equation for the non-circled variable and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but as before make your best attempt.
Vf=(2`ds/`dt)-v0
Having solved the equation as best you can, substitute the values of the three known quantities into that equation. Then simplify your expression to get the value of the unknown quantity. Again, do your best.
Vf=(2*4meters/2 sec)-0
Vf=4 meters/sec
* Exercise 3
A ball is dropped from rest and falls 2 meters to the floor, accelerating at 10 m/s^2 during its fall.
What are the events that define the beginning and the end of the interval?
S0=2 meters, sf=0 meters,
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From the given information you know the values of three of the five quantities. What are the known quantities?
`ds=2 meters, v0= 0 m/s, a=10 m/s^2
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know.
Write down the one equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled?
`ds=v0*`dt+a*`dt^2/2, with `dt uncircled
There are two equations which each contain three of the five symbols. Write down the other equation and circle the three known symbols in the equation. Which equation did you write down, and which symbol was not circled?
Vf^2=v0^2+2a*`ds, with vf uncircled
One of your equations has `dt as the 'uncircled' variable. You want to avoid that situation (though if you're ambitious you may give it a try). Solve the other equation for its non-circled variable (which should be vf) and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but as before make your best attempt.
Vf=(v0^2+2a*`ds)^1/2
Having solved the equation as best you can, substitute the values of the three known quantities into that equation. Then simplify your expression to get the value of the unknown quantity. Again, do your best.
Vf=(0^2 +2(10m/s^2)(2 meters))^1/2
Vf=(40 meters^2/seconds^2)^1/2
Vf=6.3 meters/second
Excellent work.