course Phy 201
ic_class_090914vvvv
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)
Trial 1: 450 degrees, 3 full cycles
Trial 2: 225 degrees, 2.25 full cycles
Trial 3: 405 degrees, 2.5 full cycles
Trial 4: 550 degrees, 4 full cycles
Trial 5: 225 degrees, 3 full cycles
Length of Pendulum: 14cm.
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Work out the average rate of change of rotational position (in degrees) with respect to clock time (in half-cycles of your pendulum).
Trial 1: rocAve = 450 degrees/3 full cycles= 150 degrees per full cycle
Trial 2: rocAve = 225 degrees/2.25 full cycles= 100 degrees per full cycle
Trial 3: rocAve = 405 degrees/2.5 full cycles= 162 degrees per full cycle
Trial 4: rocAve = 550 degrees/4 full cycles= 137 degrees per full cycle
Trial 5: rocAve = 225 degrees/3 full cycles= 75 degrees per full cycle
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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)
Trial 1: just paper clips time= 7 half cycles
Trial 2: paper clips with rubber band time= 4.5 half cycles
Distance traveled in both trials= 80 cm
Length of Pendulum= 3.5 cm
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Work out the average rate of change of position (in cm) with respect to clock time (in half-cycles of your pendulum).
Trial 1: 80cm/ 7 half cycles= 11.4 cm/ half cycle
Trial 2: 80cm/ 4.5 half cycles= 17.8 cm/ half cycle
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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)
Trial 1: Pendulum length= 22cm. First foot= 3.5 half cycles
Trial 2: Pendulum length= 16cm. Second foot= 7 half cycles
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Find the time spent on each ramp, seconds, using the approximate formula
· period = .2 sqrt(length).
Trial 1: period= .2 sqrt(22)= .94 sec
Trial 2: period= .2 sqrt(16)= .8 sec
( I’m not really sure if I’m doing this right.)
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Work out the average rate of change of position (in cm) with respect to clock time (in seconds) for the motion on each ramp.
Trial 1: rocAve= 12cm/.94 sec= 12.76 cm/s
Trial 2: rocAve=24cm/.8 sec= 30 cm/s
.94 s and .8 s are periods of the pendulum, not time intervals for motion down the ramp.
You know the period of the pendulum and the number of half-cycles for each ramp. So you can figure out the time it takes for that number of half-cycles, and that's the time down the ramp.
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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.
I must be doing something wrong cause the raw data I have really just doesn’t make any sense with what you want me to do.
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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:
Trial 1: North to South, from 0cm. to 22.5cm. Change in pos.=22.5cm Time= 3 half cycles
Trial 2: South to North, from 0cm. to 25cm. Change in pos.=25cm Time= 4 half cycles
(I know you’re not dumb I’m just trying to specify as much info as possible.)
Always assume your reader is clueless about the setup. Your data report is very good.
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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:
North
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Find acceleration for both trials:
Trial 1: aAve= (22.5cm -0cm)/3 half cycles= 7.5cm per half cycle
Trial 2: aAve= (25cm-0cm)/4 half cycles= 6.25cm per half cycle
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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 cm.
Distance of object to the floor: 40 cm.
Pendulum to floor: 64 cm.
Pendulum from wall: 14 cm.
Half cycles to contact: .5
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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):
aAve= (40cm – 0cm)/ .5 half cycles= 80cm/half cycle
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Opposing springs
Repeat the opposing-rubber-band experiment using springs.
Report your raw data:
Point 1: small rbnd= 0 to 10cm= 10 cm, large rbnd= 13cm to 35cm= 22cm. (10,22)
Point 2: small rbnd= 0 to 7.5cm= 7.5 cm, large rbnd= 12cm to 28cm= 16cm. (7.5, 16)
Point 3:small rbnd= 0 to 5.5cm= 5.5 cm, large rbnd= 8.5cm to 22.5cm= 14cm. (5.5, 14)
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Report the average slopes between the points on your graph:
Slope between P1and P2= (16-22)/(7.5-10)=-2.4
your numerator and denominator are both negative so your slope would be positive
Slope between P2and P3= (14-16)/(5.5-7.5)=-1
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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 slope between the points would probably have a positive slope and not so steep. Since the springs are ‘strechier’ the points would most likely be spread out, though I can’t exactly say what the slope would be. (Honestly, I can’t comprehend how to go about explaining this.)
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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 slope would rise slower due to the stretchiness of the springs, and the points would be further spread out than before. However, the points would be bigger.
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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 ‘series combination’ because the strength of three springs combined greatly exceeds that of just a couple springs in a parallel line.
· Which do you think requires more energy, the parallel or the series combination?
The ‘series combination’
· Which do you think requires more power, the parallel or the series combination?
The ‘series combination’
Solving Equations of Motion
Solve the third equation of motion for a, explaining every step.
Step1: subtract Vo*’dT from both sides
Step2: multiply both sides by 2
Step3: divide both sides by ‘dT^2
{2[‘ds- (Vo*’dt)]}/’dt^2=a
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Solve the first equation of motion for `dt, explaining every step.
Step1: divide by
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Solve the fourth equation of motion for `ds, explaining every step.
Step1: subtract Vo^2 from both sides
Step2: divide by 2a
(Vf^2-Vo^2)/2a= ‘ds
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Solve the second equation of motion for v0, explaining every step.
Step1: multiply both sides by ‘dt
Step2: subtract Vf from both sides
Step3: multiply both sides by –1
-a*’dt +Vf= Vo
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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
m/s^2*s= m/s
acceleration’s units are m/s^2 and ‘dt’s units are seconds. When multiplied together the s^2 is canceled and is just plain s.
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1/2 a t^2
½ a t^2= ½[(Vf-Vo)/’dt]*t^2= 1/2(Vf-Vo)*t= The units are in meters
a equals [(Vf-Vo)/’dt] so you can subtitute a for [(Vf-Vo)/’dt] which cancels out ‘dt and t^2, leaving only 1/2(Vf-Vo)*t. V is velocity and its units are in m/sec where t’s is only in sec. So the seconds, after being multiplied together, cancel out leaving only meters, which is the final unit of this problem
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(vf - v0) / `dt
m/s^2
This is the equation for acceleration
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2 a `ds
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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 event at the beginning is the ball being released and the event at the end is the ball traveling past the end of the ramp after 2 seconds.
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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?
Vo= 0 cm/s
‘dt= 2 seconds
‘ds= 4 meters
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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?
The 3rd equation, a was not circled
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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.
Step1: subtract Vo*’dT from both sides
Step2: multiply both sides by 2
Step3: divide both sides by ‘dT^2
a={2[‘ds- (Vo*’dt)]}/’dt^2
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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.
a= {2[4m-(0m/s*2 sec)]}/2sec^2
a= {2[4m- 0m]}/2sec^2
a= {8m}/2sec^2
a= 4m/s^2
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* 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 beginning event is when it is dropped from rest and the ending event is when it hits the floor at 10 m/s^2
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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?
a= 10m/s^2
Vo= 0 m/s
‘ds= 2m
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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?
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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?
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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.
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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.
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you don't appear to have include Exercise 2 in your responses
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?
7* 5cm= 35cm, 90cycles/5cycles=18, 60s/18= 3.3 seconds
vAve= 35cm/ 3.3 seconds= 10.6cm/s approximately
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· If that object started from rest and accelerated uniformly, what was its average acceleration in cm/s^2?
aAve= (10.6cm/s- 0cm/s)/ 3.3seconds= 3.22cm/s^2 approximately
`dv/dt, not vAve / `dt
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· 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?
vAve=45cm/(2/3)s= 67.5 cm/s
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· Its acceleration is observed to be 5 dominoes / (half-cycle)^2. What is its acceleration in cm/s^2?
aAve= 25cm/(2/3)s^2= 56.25 cm/s^2
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There are a few errors so see my notes, but on the whole your work looks very good.
&#Let me know if you have questions. &#