course Phy: 201
Class 090914Synopsis of what you should know to date, and the questions you need to get used to asking yourself:
At this point in your course everything you do is done in terms of the concept of an interval, using a few definitions (acceptable abbreviations for these definitions are included)
• the definition of average rate of change (abbreviation: rocDef)
• the definition of average velocity (abbreviation: vAveDef)
• the definition of average acceleration (abbreviation: aAveDef)
• the definition of a graph trapezoid (abbreviation: trapDef)
applied using a few principles
• the principle that all quantities with units must always be expressed in terms of units (abbreviation: unitsPrin)
• the principle that the change of a quantity on an interval is found by subtracting its initial value on that interval from its final value on that interval (abbreviation: changePrin)
expressed in terms of some basic mathematics
• basic arithmetic (including the arithmetic of fractions, and especially multiplication and division of fractions)
• the basic rules of algebra (rudimentary factoring (mostly of monomials), adding same quantity to both sides of an equation, multiplying both sides by same quantity)
• very basic geometry, especially the geometry of a trapezoid (average altitude at midpoint, average altitude average of initial and final altitudes, equal-area rectangle, slope of a line segment, area of a rectangle)
When considering an interval you should ask the following questions:
• What events define the beginning and end of the interval?
• What quantities are known at the beginning and end of the interval?
• Do we know the change in any quantity from the beginning to the end of the interval?
• Which of the known quantities are rates of change, and if any are, what is the definition of each rate of change?
If we represent the known information on a velocity vs. clock time trapezoid, that trapezoid has two altitudes, a width or base, a slope and an area. If we know any three of these quantities, we can figure out the rest.
• What are the units of each of these quantities?
• Do we know either, or both, of the quantities given by the 'graph altitudes' of the trapezoid?
• Do we know the quantity given by the slope of the trapezoid?
• Do we know the quantity given by the width, base or 'run' of the trapezoid (all three words refer to the same quantity)?
• Do we know the quantity given by the altitude of the equal-area rectangle (the same as the quantity represented by the average altitude of the trapezoid)?
• In summary, do we know three of the five quantities, and if so how do we find the other two?
If we consider the v0, vf, `dt trapezoid we get the equations of uniformly accelerated motion. If we consider the definitions of average velocity and average acceleration, we can verbally and/or algebraically work out the equations of uniformly accelerated motion. These equations allow us, if we know three of the quantities v0, vf, `dt, a and `ds, to find the other two.
• Of the five quantities v0, vf, `dt, a and `ds, which do we know?
• Which of the four equations of uniformly accelerated motion include three of these five quantities (depending on which three quantities we know, there might be one or two equations that include all three)?
• For every equation which includes three of the five quantities, there is a single unknown quantity. For each such equation, what is that quantity?
• Using basic Algebra I, solve each such equation for the unknown quantity (you may prefer to avoid solving the third equation for `dt, since that equation is quadratic in `dt and the solution might at this stage be confusing).
• Substitute your known values into the solved expression (including units with every quantity) and simplify (this includes the expressions for the units).
At this stage of the course, a few students completely understand these representations, and a few more are well on their way to doing so. The majority of the class could, however, benefit from checking their work on any problem involving motion against this list. With practice the puzzle of uniformly accelerated motion will fit together. When it does, you will have a good foundation for the rest of the course.
You will be asked this week to apply these ideas to a series of problems and experiments, while we also continue to work with the idea of forces.
Brief Experiments
Rotating Strap
Rotate a strap on top of a die and see through how many degrees it rotates (within +- 10 degrees, which you can easily estimate) and how long it takes to coast to rest (accurate to within 1/4 of a cycle of the fastest pendulum you can reasonably observe).
Do this for at least five trials, with as great a range as possible of rotational displacements.
Report your raw data (including data sufficient to determine the length of your pendulum in centimeters) &&&&
1. 450°, 3 cycles
2. 225°, 3 cycles
3. 405°, 2.5 cycles
4. 550°, 4 cycles
5. 225°, 3 cycles
Work out the average rate of change of rotational position (in degrees) with respect to clock time (in half-cycles of your pendulum). &&&&
Average rate of change is the change in A with respect to the change in B; A is going to be rotational position; B is clock time. So the average rate of change will be change in A divided by change in B.
1. 450°/ 6 half-cycles = 75° per half-cycle
2. 225°/ 6 half-cycles = 37 ½ ° per half-cycle
3. 405°/ 5 half-cycles = 81° per half-cycle
4. 550°/ 8 half-cycles = 68.75° per half-cycle
5. 225°/ 6 half-cycles = 37 ½ ° per half-cycle
Atwood machine
Your instructor will operate the apparatus and tell you the displacement of the system, and the number of excess paperclips. You time it for each trial. The displacement of the system is 80 cm from start to stop.
Report your raw data (including data sufficient to determine the length of your pendulum in centimeters) &&&&
Started toward the right side, ended on left side, 5 half-cycles
Work out the average rate of change of position (in cm) with respect to clock time (in half-cycles of your pendulum). &&&&
Average rate of change is the change in A with respect to the change in B; A is going to be change in position; B is clock time. So the average rate of change will be change in A divided by change in B.
80cm/5 half-cycles = 16 centimeters per half-cycle
Ball down two ramps
Set up a two-ramp system, the first with a 'two-quarter' slope and the second with a 'one-domino' slope.
Time the system from release at the start of the first ramp to the end of the first ramp, determining the time interval as accurately as possible, using synchronization between your pendulum and the initial and final events for each interval.
Do the same for the interval from release at the start of the first ramp to the end of the second ramp.
Report your raw data (including data sufficient to determine the length of your pendulum in centimeters) &&&&
Ramp is 1 foot long for the first ramp, it took 3 ½ half-cycles, ramp is 2 feet long for both ramps, it took 7 half-cycles
Find the time spent on each ramp, seconds, using the approximate formula
• period = .2 sqrt(length).
&&&&
For first ramp: period = .2 sqrt(22) = 0.93 seconds
For second ramp: period = .2 sqrt (16) = 0.8 seconds
Work out the average rate of change of position (in cm) with respect to clock time (in seconds) for the motion on each ramp. &&&&
Average rate of change is the change in A with respect to the change in B; A is going to be change in position; B is clock time. So the average rate of change will be change in A divided by change in B.
For first ramp: 30.48 centimeters / 0.93 seconds = 32.77 centimeters per second
For second ramp: 60.96 centimeters / 0.8 seconds = 95.25 centimeters per second
Work out the average rate of change of velocity (in cm/s) with respect to clock time (in seconds) for the motion on each ramp. &&&&
For first ramp: 32.77 centimeters per second/ 0.93 seconds = 109 centimeters.
For second ramp: 60.96 centimeters per second/ 0.8 seconds = 76.2 centimeters
(cm/s) / s is not cm
ave rate of change of velocity with respect to clock time is not average velocity / change in clock time; 32.77 cm/s is an average velocity
60.96 cm/s doesn't appear anywhere in your analysis
Hotwheels car
The hotwheels car will be passed along from one group to the next. Make at least one good observation of the displacement and time required in both the north and south directions.
Report raw data: &&&&
North to South: 3 half-cycles, 22.5 centimeters; South to North: 4 half-cycles, 25 centimeters.
Indicate your choice of north or south as the positive direction, and stick with this choice for the rest of the analysis of this experiment: &&&&
South
Find acceleration for both trials: &&&&
(22.5 centimeter/ 3 half-cycles) = 7.5 centimeters per half-cycle
(25 centimeters/ 4 half-cycles) = 6.25 centimeters per half-cycle
these are average velocities, not accelerations
Dropped object timed using pendulum
Drop an object to the floor at the same instant you release a pendulum whose equlibrium position is the wall.
Adjust the length of the pendulum and/or the height of the object until the pendulum reaches equlibrium at the same instant the ball reaches the floor.
Report your raw data, including pendulum length and distance to floor (including distance units): &&&&
Pendulum length 27 centimeters, distance dime travelled 40 centimeters, pendulum from floor 64 centimeters, pendulum from wall 14 centimeters.
Figure out the acceleration of the falling object in units of distance (using whatever distance unit you specified above) and clock time (measured in number of half-cycles): &&&&
Average acceleration is the change in velocity with respect to the change in clock time. Average velocity is the change in position with respect to clock time. So:
Velocity = 40 centimeters/ 2 half-cycles = 20 centimeters per half-cycle; therefore the Acceleration = 20 centimeters/half-cycle/2 half-cycle = 10 centimeters
acceleration is not ave velocity / change in clock time
Opposing springs
Repeat the opposing-rubber-band experiment using springs.
Report your raw data: &&&&
0, 10, 13, 35
0, 7 ½, 12, 28
0, 5 ½, 8 ½, 22 ½
Report the average slopes between the points on your graph: &&&&
1, 12/5
If you were to repeat the experiment, using three of the 'stretchier' springs instead of just one, with all three stretched between the same pair of paper clips, what do you think would be the slope of your graph? &&&&
The slopes of the graph are going to become steeper because the lengths are going to be larger because we can stretch the springs more.
terminology note: for future reference we will use the term 'parallel combination' to describe the three rubber bands in this question
If you were to repeat the experiment using three of the 'stretchier' springs (all identical to the first), this time forming a 'chain' of springs and paper clips, what do you think would be the slope of your graph? There are different ways of interpreting this question; as long as your answer applies to a 'chain', as described, and as long as you clearly describe what is being graphed, your answer will be acceptable (this of course doesn't imply that it will be correct): &&&&
The slopes would be almost identical to the first slopes because we are creating a chain and not applying multiple springs. We can only stretch the springs to the ability of the first. Therefore, the lengths of the springs will be relatively close to the first; therefore the slopes will be relatively close.
terminology note: for future reference we will use the term 'series combination' to describe the three rubber bands in this question
We haven't yet defined force, energy and power, so you aren't yet expected to come up with rigorously correct answers to these questions. Just answer based on your current notions of what each of these terms means:
If each of the 'stretchier' springs starts at its equilibrium length and ends up stretched to a length 1 cm longer than its equilibrium length, then:
• Which do you think requires more force, the parallel or the series combination?
The parallel because there is only one of them so it is going to take more force to stretch it compared to the series combination which gives in a little more.
• Which do you think requires more energy, the parallel or the series combination?
The series combination because you take the amount of energy of the parallel and think of it as three times more.
• Which do you think requires more power, the parallel or the series combination?
The parallel because the series combination will give a little more so the parallel will require more power.
Solving Equations of Motion
Solve the third equation of motion for a, explaining every step. &&&&
For this equation, first I subtract both sides by v0 ‘dt to get ds - v0 ‘dt = 0.5a ‘dt^2. Then I divided both sides by 0.5 ‘dt to get ds – v0 ‘dt/ 0.5 ‘dt^2. Therefore the final equation is a = ds – v0 ‘dt / 0.5 ‘dt^2.
Solve the first equation of motion for `dt, explaining every step. &&&&
For this equation, first I divide both sides by (vf + v0)/2. Or in other words, multiply by the reciprocal 2/ (vf + v0). And the final equation will be ‘dt =‘ds[(vf+v0)/2].
Solve the fourth equation of motion for `ds, explaining every step. &&&&
For this equation, first I subtracted v0^2 from both sides to get vf^2 – v0^2 = 2a ‘ds. Then I divided both sides by 2a to get (vf^2 – v0^2) / 2a. The final equation is ‘ds = [(vf^2 – v0^2) /2].
Solve the second equation of motion for v0, explaining every step. &&&&
For this equation, first I multiplied both sides by ‘dt to get a ‘dt = vf – v0. Then I subtracted vf from both sides to get a ‘dt – vf = -v0. Finally I have to divide by -1 to get positive v0 by itself. The final equation is v0 = (a ‘dt – vf)/ -1.
Units calculations with symbolic expressions
Using SI units (meters and seconds) find the units of each of the following quantities, explaining every step of the algebra of the units:
a * `dt &&&&
Acceleration is meters per seconds and ‘dt is seconds. Therefore I have m/s * s. The seconds are going to divide out and the units are going to be meters for the final answer.
acceleration is in m/s^2, not m/s
1/2 a t^2 &&&&
Acceleration is meters per seconds and ‘dt is seconds. Therefore I have ½ m/s * s^2. So I will have m*s^2/ s. The seconds will divide out or simplify. The final units are ½ m*s.
1/2 is a numerical factor and is not part of the unit
acceleration is in m/s^2, not m/s
(vf - v0) / `dt &&&&
Velocities are m/s and ‘dt is seconds. Therefore I have m/s *s. The seconds will divide out and the final units are meters.
2 a `ds &&&&
Acceleration is meters per seconds and ‘ds is meters. Therefore I have m/s *m. The meters multiply. The final units are meters^2 per seconds or m^2/s.
acceleration is in m/s^2
Identifying initial and final events and kinematic quantities
* Exercise 1: 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? &&&&
The ball is released from rest, the ball reaches the end of the ramp.
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, ‘dt, 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. &&&&
2(4 meters/ 2 seconds) – 0 meters/seconds = 2(2 meters/second) = 4 meters/second
* Exercise 2: 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? &&&&
The ball is dropped from rest, the ball hits the floor
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, v0, a_ave
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+0.5 * a_ave * ‘dt
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_ave* ‘ds, vf
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^2=v0^2 + 2a_ave * ‘ds
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 = 0^2 + 2(10 m/s^2) * 2 m = (20 m/s^2) * 2m = 40 m^2/s^2 = 6.32 m/s
Additional Exercises
* Exercise 3: A pendulum completes 90 cycles in a minute. A domino is 5 cm long.
There are four questions, with increasing difficulty. Based on typical performance of classes at this stage of the course, it is expected that most students will figure out the first one, while most students won't figure out the last (your instructor will of course be happy if the latter is an underestimate).
Here are the questions:
• If an object travels through a displacement of 7 dominoes in 5 half-cycles, then what is its average velocity in cm/s? &&&&
The pendulum will complete 1.5 cycles/second or 0.75 half-cycles/second. The object is going to travel 35 centimeters(a domino is 5 cm long and it travels through 7, 5*7), so therefore, 35 cm/ 3.75 seconds = about 9 centimeters/second.
• If that object started from rest and accelerated uniformly, what was its average acceleration in cm/s^2? &&&&
Average acceleration is change in velocity divided by change in clock time. Therefore the average acceleration is 9 centimeters/second/ 3.75 seconds = 2.4 centimeters
change in velocity / change in clock time, not average velocity / change in clock time
• From observations, the average velocity of the ball is estimated to be 9 dominoes per half-cycle. What is its average velocity in cm/sec? &&&&
Average velocity is change in position divided by clock time. Therefore the change in position is 45 centimeters and the change in clock time is 3.75 seconds. Therefore the average velocity is 45centimeters/ 3.75 seconds = 12 centimeters per second.
• Its acceleration is observed to be 5 dominoes / (half-cycle)^2. What is its acceleration in cm/s^2? &&&&
Change in position would be 25 ( 5*5) and change in clock time would be 3.75 seconds. Therefore the acceleration is 25centimeters /3.75 seconds = about 6.6 centimeters / seconds.
Homework:
Your label for this assignment:
ic_class_090914
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.
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Good work on most questions but see my notes for some important corrections.