course 201
9/16 1 am
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)
Pendulum length: 10 lines of notebook paper
Trials:
#1- 160 degrees in 4 half cycles
#2- 185 degrees in 8 half cycles
#3- 85 degrees in 4 half cycles
#4- 380 degrees in 12 half cycles
#5- 60 degrees in 3 half cycles
#6- 110 degrees in 5 half cycles
Work out the average rate of change of rotational position (in degrees) with respect to clock time (in half-cycles of your pendulum).
#1- `ds/`dt=(sf-s0)/(tf-t0)= (160 degrees- 0 degrees)/(4 half cycles-0 half cycles)=160 degrees/4 half cycles=40 degrees/half cycle
#2-`ds/`dt=185 degrees/8 half cycles=23.1 degrees/half cycle
#3-`ds/`dt=85 degrees/4 half cycles=21.3 degrees/half cycle
#4-`ds/`dt=380 degrees/12 half cycles=31.7 degrees/half cycle
#5- `ds/`dt=60 degrees/3 half cycles=20 degrees/half cycle
#6-`ds/`dt=110 degrees/5 half cycles=22 degrees/half cycle
On average, the system had an average rate of acceleration of (40+23.1+21.3+31.7+20+22)/6= 26.4 degrees/half cycle
Your preceding calculations are average accelerations.
Your last calculation was not requested and while harmless, you should understand why it was unnecessary.
The reason is there is no point in averaging the average accelerations for a series of different time intervals.
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)
Pendulum length: 10 lines of notebook paper
First trial with system as is: 10 half cycles of pendulum
First trial with rubber band added to system: 7 half cycles
Second trial with rubber band added to system: 7 half cycles
Work out the average rate of change of position (in cm) with respect to clock time (in half-cycles of your pendulum).
&&&&
First trial with system as is: `ds/`dt= 80 cm/10 half cycles= 8 cm/half cycle
First(and second) trial with rubber band added to system: `ds/`dt= 80 cm/7 half cycles=11.4 cm/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 system was 2 dominos at first ramp, then one domino at second ramp
Attempting to time the ball at the end of the first ramp as close as possible to the pendulum reaching equilibrium:
20 cm pendulum had 6 half cycles, not at equilibrium
25 cm pendulum had 5.5 half cycles, yes for equilibrium
Attemping to time the ball at the end of the second ramp as close as possible to the pendulum reaching equilibrium:
20 cm pendulum had 9 half cycles, not at equilibrium
25 cm pendulum had 9 half cycles, not at equilibrium
30 cm pendulum had 8 half cycles, not at equilibrium
15 cm pendulum had 10.5 half cycles, yes for equilibrium
Find the time spent on each ramp, seconds, using the approximate formula
period = .2 sqrt(length).
Time spent on first ramp would be: period=.2(25)^1/2, period = 1 second, with 5.5 periods completed: 5.5(1 second)= 5.5 seconds
Time spent on ramps one and two: period=.2(15)^1/2, period=.77 seconds, with 10.5 periods completed: 10.5(.77 seconds)=8.1 seconds
Therefore time spent on the second ramp alone would be 8.1 seconds-5.5 seconds=2.6 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.
First ramp (30 cm): `ds/`dt=30 cm/5.5 seconds= 5.5 cm/second
Second ramp: `ds/`dt=30cm/2.6 seconds= 11.5 cm/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.
First ramp: Since the average velocity is 5.5 cm/s, and we stopped our marble with a washer at the end of the ramp, making its final velocity 0 cm/s, it can be inferred that the maximum velocity is twice that, or 11 cm/s
Acceleration=`dvelocity/`dtime=final velocity-initial velocity/`dtime= (11cm/sec-0 cm/sec)/5.5 sec=2 cm/sec^2
Second ramp: Applying the same reasoning as above, maximum velocity would be 2(11.5cm/s)=23 cm/s
Acceleration=`dvelocity/`dtime=final velocity-initial velocity/`dtime= (23cm/sec-0 cm/sec)/2.6 sec=8.8 cm/sec^2
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:
10 lines of notebook paper pendulum, 96 cycles in 60 seconds
Positive direction:
3 half cycles, 13.75 inches traveled
Negative direction:
4 half cycles, 18.75 inches
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 is my positive direction
Find acceleration for both trials:
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
it's final velocity that's zero; this will reverse the sign of your acceleration
the same will happen with your result for the opposite 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
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):
For a 36 inch drop of a dime:
6 inch pendulum, pendulum hit wall first
12 inch pendulum, pendulum first
14 inch pendulum, pendulum first
16 inch pendulum, pendulum first
18 inch pendulum, very close pendulum likely hit first
20 inch pendulum, dime hit ground first
19 inch pendulum, same time
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):
Acceleration = `dvelocity/`dtime, so we must first find velocity.
Velocity=`ds/`dt=36 inches/.5 half cycles=72 inches/half cycle
Since this is the average velocity, and the final velocity of the dropping dime was 0, it can be inferred that the maximum velocity was 144 inches/half cycle
Acceleration=`dvelocity/`dtime= (vf-v0)/`dt= (144 in/half cycle-0 in/halfcycle)/.5 half cycles=144 in/halfcycle /.5 half cycles=288in/halfcycle^2
Opposing springs
Repeat the opposing-rubber-band experiment using springs.
Report your raw data:
Relaxed:
Long tight spring from 1.25 inches to 7.25 inches
Short loose spring from 8.375 inches to 10.25 inches
Stretched:
Long spring from 1.25 inches to 7.75 inches
Short spring from 9.25 inches to 11.375 inches
Stretched more:
Long spring from 1.25 inches to 10 inches
Short spring from 11.25 inches to 14 inches
Report the average slopes between the points on your graph:
To find the slopes I need to first find the points, by calculating the length of each spring.
Relaxed:
Long: 7.25-1.25=6
Short:10.25-8.375=1.875
Stretched:
Long: 7.75-1.25=6.5
Short: 11.375-9.25=2.125
Stretched more:
Long: 10-1.25= 8.75
Short: 14-11.25=2.75
This makes my points (6, 1.875), (6.5, 2.125), and (8.75, 2.75)
Slope between first two points=rise/run=`dy/`dx=(2.125-1.875)/( 6.5-6)=.25/.5=1/2
Slope between second two points=`dy/`dx=(2.75-2.125)/(8.75-6.5)=.625/2.25=.278
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?
If stretchier refers to what Ive called shorter, then I would expect the long spring to carry even more of the pull, because the shorter springs already cannot stretch as far as the long. The two extra springs will make it even more difficult to force apart, possibly resulting in a slope of .25 or smaller.
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):
Since the force will be evenly distributed along the identical springs, the slope would be expected to be 1, because the comparison of length of spring x and length of spring y in each trial should result in the same lengths, making the slope one.
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?
I would think that the series combination would require more force, as it is making the work go into a shorter distance, with three springs all able to stretch to the same maximum
Which do you think requires more energy, the parallel or the series combination?
I would say that the series combination would require the most energy, for the same reason as above.
Which do you think requires more power, the parallel or the series combination?
Again, I would have to think that the series combination would require the most power.
Solving Equations of Motion
Solve the third equation of motion for a, explaining every step.
`ds=v0*`dt+.5*a*`dt
First, subtract v0*`dt from both sides
.5*a*`dt=`ds-v0*`dt
Then divide both sides by .5*`dt
A=(`ds-v0*`dt)/.5*`dt
Solve the first equation of motion for `dt, explaining every step.
`ds=.5(vf+v0) *`dt
Divide both sides by .5(vf+v0)
`dt=`ds/.5(vf+v0)
Solve the fourth equation of motion for `ds, explaining every step.
Vf^2=v0^2 + 2*a*`ds
First subtract v0^2 from both sides
2*a*`ds=vf^2-v0^2
Then divide both sides by 2*a
`ds=(vf^2-v0^2)/2a
Solve the second equation of motion for v0, explaining every step.
A=(vf-v0)/`dt
First multiply both sides by `dt
A*`dt=vf-v0
Then subtract vf from both sides
-v0=-vf+a*`dt
Then multiply both sides by -1 to remove the negative from v0
V0=vf-a*`dt
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
a=m/s^2, `dt=s
m/s^2*s=m/s
this is because the multiplication creates m*s/s^2, and the rules of exponents tell you to subtract when dividing, which gives m/s^-1, which is the same as m/s
1/2 a t^2
.5(m/s^2)(s^2)=m*s^2/s^2=m
This happens because .5 has no units and the s^2 in the top and bottom subtract out using the rules of exponents
(vf - v0) / `dt
(m/s-m/s)/s=(m/s)/s=m/s*s=m/s^2
I am assuming that subtracting the initial and final velocities does not give 0. What is happening here is m*s^-1*s^-1, and when you multiply exponents you add, giving m*s^-2, which is the same as m/s^2
2 a `ds
2 m/s^2*m=m^2/s^2
2 does not have units, so it can be disregarded. What is happening here is m*m/s^2, which is then expressed as m^2/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?
T0 and tf
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, with vf 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*4m/2s)-0, vf=4m/s
* 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?
S0 and sf
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=10m/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?
Vf^2=v0^2+2*a*`ds, vf not circled
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?
`ds=v0`dt+.5a`dt^2, with `dt 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
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 dominoes=7*5cm=35cm, 5 half cycles=2.5 cycles(90cycles/60seconds)=1.67 seconds
Velocity=`ds/`dt=35 cm/1.67 seconds=21cm/sec
If that object started from rest and accelerated uniformly, what was its average acceleration in cm/s^2?
Starting from rest means v0 was 0 cm/s, and with the average velocity of 21 cm/sec, it can be assumed then that vf=42 cm/sec
Acceleration=`dvelocity/`dt=(vf-v0)/`dt=(42cm/s-0cm/s)/1.67s=25.2 cm/s^2
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?
(9 dominoes/.5 cycle)(5cm/1domino)(90cycles/60seconds)= 135 cm/sec
Its acceleration is observed to be 5 dominoes / (half-cycle)^2. What is its acceleration in cm/s^2?
(5 dominoes/.5 cycle^2)(5cm/1 domino)(8100 cycle ^2/3600 sec^2)=112.5 cm/s^2
&#This looks very good. Let me know if you have any questions. &#