**
course phy 201
20090920 2330
Rotating StrapRotate 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 - 12 half cycles - 270 degrees of rotation
T2 - 16 hc - 590 dor
T3 - 27 hc - 875 dor
T4 - 21 hc - 750 dor
T5 - 19 hc - 720 dor
pendulum length - 9.45 cm
Work out the average rate of change of rotational position (in degrees) with respect to clock time (in half-cycles of your pendulum).
&&&&
Ave Rate of change - angular velocity = roc of rot pos wrt to ct -- ang. v = (270+590+875+750+720(all in rot pos, deg))/
(12+16+27+21+19(in half cycles)) = 33.74 rot pos/ct
there's a different ave rate of change for each trial; one trial has no effect on the others
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) &&&& ***Was unable to
record all data for this experiment but will record what was obtained (and remembered)****
Trial 1 - 4 cycles - x paperclips on string wrt pully system
T2 - 2.75 cycles - x+y paperclips
T3 - 2 cycles - x+2y paperclips
I assume the y = 1 , meaning 1 paperclip was added throughout the experiment.
Pendulum L= 9.45 cm
Work out the average rate of change of position (in cm) with respect to clock time (in half-cycles of your pendulum). &&&&
ave roc = pos (cm) wrt ct (half-cycles) -- ave roc = (80cm)/(8+5.50+4)(half cycles) = 4.57 cm/half-cycles
there's a different ave rate of change for each trial; one trial has no effect on the others
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) &&&&
Visual - A-----ramp 1-----B-----ramp 2-----C
Point AB (first ramp) - 30 cm travelled - 3.75 cycles
Point AC (second ramp) - 30 cm travelled - 2.25 Cycles
Pendulum length - 20.9 cm
Find the time spent on each ramp, seconds, using the approximate formula
* period = .2 sqrt(length).
&&&&
T = .2sqrt(L) = .2sqrt(20.9cm) = .914 s (?)
Work out the average rate of change of position (in cm) with respect to clock time (in seconds) for the motion on each ramp. &&&&
Ramp 1 = velocity = `ds/`dt -- ave vel. = 30(cm)/(3.75cycles*.914s) = 8.753 cm/s
Ramp 2 = ave vel = `ds/`dt -- ave vel. = 30(cm)/(2.25cycles*.914s) = 14.588 cm/s
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. &&&&
Ramp 1 = ave accel = `dv/`dt -- ave accel. = ((8.753cm/s - 0cm/s) / 2))/(3.75cycles*.914s) = 1.277 cm/s^2
((8.753cm/s - 0cm/s) / 2))/(3.75cycles*.914s) represents ((average velocity - initial velocity) / 2) / `dt, which in this case turns out to be half the average velocity divided by the time interval
ave accel is change in vel / time interval
Ramp 2 = ave accel = `dv/`dt -- ave accel. = ((14.588cm/s - 8.753cm/s) / 2))/(2.25cycles*.914s) = 1.419 cm/s^2
Hotwheels car
the change in velocity for a run of an object down an incline isn't the difference between the average velocity on the last incline and the average velocity on this incline
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 - 4 (1/2 cycles) - `ds = 70.1 cm
T2N - 4 (1/2 cycles) - `ds = 62.1 cm
T3N - 3.5 (1/2 cycles) - `ds = 61 cm
Trial 1 South - 3 (1/2 cycles) - `ds = -41 cm
T2S - 4 (1/2 cycles) - `ds = -66cm
T3S - 2.5 (1/2 cycles) - `ds = -27 cm
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 - towards front of class (in relation to x-axis) , South - towards back of classroom
Find acceleration for both trials: &&&& Ave accel from northbound trial = 24.14 cm/s^2 --- ave accel from southbound trial = -23.69
cm/s^2 (obtained from finding average vel from each trial, multiplying by two to obtain v0. then solving each for ave accel -
`dv/`dt, which seconds obtained by multiplying full cycles by 1.21 seconds)
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 height ~ 100 cm
Pendulum length - 42.4 cm
Measurement was not obtained for dime drop but given the height of the pendulum and it's length, released around a 45 degree angle,
it is fair to say the distance the dime fell was approximately = (pendulum height - pendulum length)/45 degree angle which is the
midpoint of angular position of 90 degrees + (PLUS) pend. h - pen length= ((100 cm - 42.4 cm)/2) + (100 cm - 42.4 cm) = 86.4 cm <--
~height of dime when dropped --- also could be obtained from (100cm - 42.4 cm) * 1.5
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 = 43.2 cm/half-cycles^2 (ave vel = `ds/`dt - ave vel = 86.4cm / half-cycle) ave accel = ((vf-v0)/2)/`dt
Opposing springs
Repeat the opposing-rubber-band experiment using springs.
Report your raw data: &&&& raw data was not written for the initial and final displacments, which was later discovered this was
inappropriate //
Initial, minimal strech:
thick spring - 12.4 cm, thin spring - 15.2 cm
Second, medium strech:
thick - 13.1 cm, thin - 18.8 cm
final, greatest stretch recorded:
thick - 14.2 cm, thin - 24.2 cm
Report the average slopes between the points on your graph: &&&&
on a thin vs thick graph, sloped averaged at 1.47 --- adding 1.23 (1st trial), 1.44 (2nd trial), and 1.70 (3rd trial) then dividing
by 3 to obtain average slope
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? &&&& Just as observed in the rubber band
experiment, I would imagine the slope to become more linear.
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): &&&& Assuming this chain not only consists of
the 3 stretchier springs, but also the larger spring - a graph would be assembled measuring the average of the smaller (stretchier)
springs vs. the larger spring. I would still expect to see the graph to increase in slope with the more force that is applied in the
pull of the chain. if this chain only consisted of the 3 small springs, the slope would appear linear.
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: *&*&*Going to assume you mean springs and not rubber bands for these questions*&*&*
* Which do you think requires more force, the parallel or the series combination? parallel combination
* Which do you think requires more energy, the parallel or the series combination? series combination
* Which do you think requires more power, the parallel or the series combination? parallel comination
Solving Equations of Motion
Solve the third equation of motion for a, explaining every step. &&&&
`ds = v0`dt+ ((a`dt)/2)`dt <--- 3rd equation
`ds - v0`dt = (a`dt^2)/2 <--- subtract v0`dt from both sides and apply square
a = (2(`ds - v0`dt) / (`dt^2) <--- multiply 2 and divide `dt^2 from both sides
Solve the first equation of motion for `dt, explaining every step. &&&&
`ds = ((vf+vo)/2)`dt <--- 1st equation
`ds = ((vf+vo)`dt)/2 <--- rearrange
`dt = (2`ds) / (vf+v0) <--- multiply 2 and divide (vf+v0) from both sides
Solve the fourth equation of motion for `ds, explaining every step. &&&&
vf^2 = v0^2 + 2a`ds <--- 4th equation
vf^2 - v0^2 = 2a`ds <--- subtract v0^2 from both sides
`ds = (vf^2 - v0^2) / 2a <--- divide 2a from both sides
Solve the second equation of motion for v0, explaining every step. &&&&
a = (vf-v0)/`dt <--- 2nd equation
a`dt - vf = -v0 <--- multiply `dt and subtract vf from both sides
-(a`dt - vf) = -(-v0) <--- divide both sides by -1 to make v0 positive
v0 = -`dta + vf
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 <-- multiplying by s (for seconds) cancels out an s in s^2 of the demoninator
1/2 a t^2 &&&&
.5 * m/s^2 * s^2
m/2 (or .5m) <-- s^2 cancel out when multiplying (ms^2/2s^2 - s^2 divide out)
(vf - v0) / `dt &&&&
(m-m) / s
v isn't in meters
ZERO <-- answer, m's subtract to 0 and dividing a 0 by anything is 0, thus meaning this equation would not have a unit associated
2 a `ds
you don't subtract the units; these are like terms and the units factor out of the subtraction
&&&&
2 m/s^2 m
2m^2/s^2 < -- meter's combine (multiply) to make m^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 time the ball is released at the top of the ramp
is the beginning and the end is when the ball reaches the end of the ramp (which for most will be when the ball falls off the ramp
and hits the table - just a little inaccuracy that easier to measure)
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? &&&&
v0, `dt, `ds
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 being 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. &&&&
`ds = (vf+v0)/2 * `dt ---> vf = (2`ds-`dtv0) / `dt
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)/2s - vf = 8m/2s - 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? &&&& the point in which the ball is let go (or dropped)
is the beginning adn the end is when the balls comes to a stop (hitting 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? &&&& vf, a, `ds
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 + 2a`ds with vf not being 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 + (a`dt/2)`dt with
`dt not circled
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`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 = v0^2 + 2a`ds --- vf^2 = 0m/s^2 + 2(10m/s^2)2m --- vf^2 = 40m^2/s^2 --- vf = 40 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? &&&&
ave vel - 1.4 dominoes/half-cycle --- sorry... the pendulum would be 1.5cycles per second or 3 half cycles per second OR .667
sec/cycle or 1.33sec/halfcyle and 7 dominoes * 5 cm = 35 cm // ave vel = 35 cm / (5 half cycles * 1.33sec/halfcycle) = 5.26 cm/s
* If that object started from rest and accelerated uniformly, what was its average acceleration in cm/s^2? &&&& ave acceleration
= (vf-v0)/`dt where vf is obtained from double the average velocity // accel. = ((2*5.26cm/s)-0)/6.65s = 1.58cm/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? &&&& ave vel = (9 dominoes*5cm) / (1.33 sec/1-halfcycles) ---> ave vel = 45cm/1.33sec --> ave vel = 33.83 cm/s
* Its acceleration is observed to be 5 dominoes / (half-cycle)^2. What is its acceleration in cm/s^2? &&&& acceleration of 5
dominoes / (half-cycle)^2 .. (guessing) (5*5cm) = 25 cm for `ds & (1.33s)^2 = 25cm/1.7689s^2 --> ave a = 14.13 cm/s^2"
&#Good responses. Let me know if you have questions. &#