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course Phy 241
9/26 8
ExperimentsSpace 6 magnets, three on each side, symmetric ...
see how different you can make the symmetry and still get a balanced oscillation
For the magnets on the rotating strap:
`qx001. How fast was each of the magnets moving, on the average, during the second 180 degree interval? All the magnets had the same angular velocity (deg / second), but what was the average speed of each?
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For our data we used 4 mag, with the beam being 30.5 cm long a mag was placed a cm from the end on both sides. The measurements were mag #1 was placed at the 1cm to 5.5cm mark on the. 2nd was from 8cm to 12.5cm mark. 3rd was from 18cm to 22.5cm mark. 4th was from 25cm to 29.5cm mark. Note these measurement’s all lie on a scale from 0cm to 30.5cm which is distance from beginning to end of beam.
For 180 degree turn we found a time interval of 1.5 was best approx. of time required.
At 180 degrees in 1.5sec = 120(deg/sec)
Distance from center of beam to center of inner mag is 5cm and distance to outer mag’s center from center of the beam is 12cm.
To Find circumference:
Outer C = 2`pi*r = 2`pi*12cm = approx. 75.3982cm
Inner C = 2`pi*5cm = approx. 31.4159cm
!!!!!!!We only need half of circumference for 180 degree turn because total circumference is related to 360 degree turn!!!!!!
So average speed for Inner and Outer mag’s is half circumference divided by time interval
Outer ave speed = (75.3982cm/2)(Dist needed for 180 degree turn)/1.5sec(time required)
=approx. 25.1327(cm/sec)
Inner ave speed = [(31.4159cm/2)]/1.5sec = approx. 10.472 (cm/sec)
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`qx002. For each magnet, one of its ends was moving faster than the other. How fast was each end moving, and how fast was the center point moving?
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Inner mag:
End farthest from center, 7.25cm
C = 2`pi*r = 2`pi*7.25cm = approx. 45.5531cm
Ave speed = (45.5531cm/2)/1.5sec = approx. 15.1844 (cm/sec)
End closest to center , 2.75cm
C = 2`pi*r = 2`pi*2.75cm = approx. 17.2788cm
Ave speed = (17.2788cm /2)/1.5sec = approx. 5.7596 (cm/sec)
!!!Center has already been calculated for!!!!!
Outer mag:
End farthest from center, 14.25cm
C = 2`pi*r = 2`pi*14.25cm = approx. 89.5354cm
Ave speed = (89.5354cm /2)/1.5sec = approx. 29.8451 (cm/sec)
End closest to center , 10.75cm
C = 2`pi*r = 2`pi*10.75cm = approx. 67.5442cm
Ave speed = (67.5442cm /2)/1.5sec = approx. 22.5147 (cm/sec)
!!!Center has already been calculated for!!!!!
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`qx003. What was the KE of 1 gram of each magnet at its center, at the end closest to the axis of rotation, and at the end furthest from the axis of rotation?
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For KE = 1/2m*v^2
Inner mag:
End farthest from center, vAve = approx. 15.1844 (cm/sec)
KE = (1/2)1gram (15.1844 (cm/sec))^2
= approx. 115.283gram (cm/sec)^2
???????I am not sure if my solution of grams per (cm/sec)^2 is correct, if not what is correct notation???
End farthest from center, vAve = approx. 5.7596 (cm/sec)
KE = (1/2)1gram (5.7596 (cm/sec))^2
= approx.16.5865gram (cm/sec)^2
Center, vAve = approx. 10.472 (cm/sec)
KE = (1/2)1gram (10.472 (cm/sec))^2
= approx.54.8314gram (cm/sec)^2
Outer mag:
End farthest from center, vAve = approx. 29.8451 (cm/sec)
KE = (1/2)1gram (29.8451 (cm/sec))^2
= approx. 445.365gram (cm/sec)^2
End farthest from center, vAve = approx. 22.5147 (cm/sec)
KE = (1/2)1gram (22.5147 (cm/sec))^2
= approx.253.4559gram (cm/sec)^2
Center, vAve = approx. 25.1327(cm/sec)
KE = (1/2)1gram (25.1327(cm/sec))^2
= approx.315.8263gram (cm/sec)^2
???????We were to calculate for a mass of 1 gram, not the 50 grams you gave us in class correct?????
Right. Just talking about 1 gram close to the specified point.
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`qx004. Based on your results do you think the KE each magnet is greater or less than the KE of its center?
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I would think the KE for the entire magnet to be greater than what we just calculated for due to the formula uses 1 half mass times velocity. Therefor calculations just done for 1 gram of mass should be lower than calculations done for entire mass of magnet.
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`qx005. Give your best estimate of the KE of each magnet, assuming its mass is 50 grams.
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Inner mag
[15.1844 (cm/sec) + 5.7596 (cm/sec)]/2 =10.472(cm/sec) (???Ave velocity of both ends of magnet??)
KE = (1/2)50gram*(10.472(cm/sec))^2
=2741.5696gram(cm/sec)^2
Outer mag
[29.8451 (cm/sec) + 22.5147 (cm/sec)]/2 =26.1799(cm/sec) (???Ave velocity of both ends of magnet??)
KE = (1/2)50gram*(26.1799 (cm/sec))^2
=17,134.6791gram(cm/sec)^2
Based on your calculations for 1 gram at each of three positions, do you think this is an overestimate or an underestimate?
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`qx006. Assuming that the strap has a mass of 50 grams, estimate its average KE during this interval.
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C = 2`pi*15.25cm = approx. 95.8186
95.8186/2 = approx. 47.9093/1.5 = 31.9395 (cm/sec)
KE = (1/2)50gram(31.9395)^2
= 25,503gram(cm/sec)^2
Good.
Based on patterns you have seen earlier in this set, do you think this is an overestimate or an underestimate?
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`qx007. (univ, gen invited) Do you think the KE of 1 gram at the center of a magnet is equal to, greater or less than 1/2 m v_Ave^2, where v_Ave is the average velocity on the interval?
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????This is the formula I used to calculate KE so I guess this was incorrect, what is the correct formula?????
You have to use calculus to answer your question.
However based on patterns you have seen so far you can answer the question that was posed. 'equal' is one possible answer, but of course you want to justify that.
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For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
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10sec to go up and return to starting position(which is period)
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`qx009. Was the period of the unbalanced vertical strap uniform?
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???I’m confused about this question, are you asking about using the mag to spin the strap?????
If so,
at 3cm apart(them mag on the strap and the mag used to spin the strap) it took 4 sec to spin 360 degree
we didn't do this one ... expected to get to it but everyone was having way too much fun with the balances
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`qx010. What is the evidence that the average magnitude of the rate of change of the angular velocity decreased with each cycle, even when the frequency of the cycles was not changing much?
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Because it would take a longer time interval to travel the same distance as the initial sep of the mag increased.
????Again I am not sure I am talking about the right data??????
interpretation of this question is still open
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For the experiment with toy cars and paperclips:
`qx011. Assume uniform acceleration for the trial with the greatest acceleration. Using your data find the final velocity for each (you probably already did this in the process of finding the acceleration for the 09/15 class). Assuming total mass 100 grams, find the change in KE from release to the end of the uniform-acceleration interval.
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With 5 small and 3 large paper clips my car traveled 50 cm in approx. 1 and (1/3) seconds.
For 50 cm traveled in approx. 1 and (1/3) seconds
vAve = 50cm/(1.3334sec) = approx. 37.4981 (cm/sec)
37.4981 (cm/sec)*2 = approx. 74.9963 cm/sec = vf(when v0 = 0)
`dKE = (1/2)*100grams*(74.9963 cm/sec)^2 - (1/2)*100grams*(0 cm/sec)^2
= 281,222.2507grams (cm/sec)^2 – 0
=281,222.2507grams (cm/sec)^2
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For the experiment with toy cars and magnets:
`qx012. For the experiment with toy cars and magnets, assume uniform acceleration for the coasting part of each trial, and assume that the total mass of car and magnet is 100 grams. If the car has 40 milliJoules of kinetic energy, then how fast must it be moving? Hint: write down the definition of KE, and note it contains three quantities, two of which are given. It's not difficult to solve for the third.
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KE = (1/2)m*v^2
therefor ,
40 milliJoules = (1/2)100grams*v^2
40 milliJoules = 50grams*v^2
v^2 = 40 milliJoules/50grams
v = `sqrt(.8 milliJoules/gram) = approx. .8944 milliJoules/gram
?????How do you convert this into a standard velocity??????
That would be .8944 sqrt(milliJoules/gram).
sqrt(mJ/ g) = sqrt( (milli kg m^2 / s^2) / gram) = sqrt( (g m^2 / s^2) / g ) = sqrt(m^2/s^2) = m/s.
Based on fact that milli kilo is .001 * 1000 = 1, definition of Joule as N * m (from `dW = F_net `dx), and N as kg * m/s^2 (from F = m a).
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`qx013. Based on the energy calculations you did in response to 09/15 question, what do you think should have been the maximum velocity of the car on each of your trials? You should be able to make a good first-order approximation, which assumes that the PE of the magnets converts totally into the KE of the car and magnet.
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`qx014. How is your result for KE modified if you take account of the work done against friction, up to the point where the magnetic force decreases to the magnitude of the (presumably constant) frictional force? You will likely be asked to measure this, but for the moment assume that the frictional force and magnetic force are equal and opposite when the magnets are 12 cm apart.
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???I’m not sure I totally understand the question????
I would think it would make the result for KE lower because instead of calculating for an interval 60cm long you would be calculating for an interval 12 cm long
check modified class notes for 100920, which now contain notes relevant to this
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`qx015. If frictional forces assume in the 9/15 document were in fact underestimated by a factor of 4, then how will this affect your results for the last two questions?
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If frictional force was underestimated by a factor of 4 then the KE used to counter that fractional force is a underestimate as well.
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`qx016. What did you get previously for the acceleration of the car, when you measured acceleration in two directions along the tabletop by giving the car a push in each direction and allowing it to coast to rest?
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To left:
68cm in 3sec gives an acceleration of 22.6667 cm/sec^2
To right
38cm in 2sec gives an acceleration of 19 cm/sec^2
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`qx017. Using the acceleration you obtained find the frictional force on the car, assuming mass 100 grams, and assuming also a constant frictional force.
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???Using formula given in class of “Any mass * accel = Force” give the following results. Is this the correct approach?????
To left:
Frictional force = 100grams*22.6667 cm/sec^2
=2266.7 grams (cm/sec^2)
To right:
Frictional force = 100grams*19 cm/sec^2
=1900 grams (cm/sec^2)
?????What exactly is being said with a solution in grams (cm/sec), if I’m even doing that correctly??????
that's a unit of mass * acceleration, so it's a force unit
it's 1/1000 * 1/100, or 1 / 100 000 of a kg m/s^2, which is Newton.
the g cm/sec^2 has a name, the dyne. 1 dyne = 10^-5 N.
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`qx018. Based on this frictional force
How long should your car coast on each trial, given the max velocity just estimated and the position data from your experiment?
... this could be done with an inclined air track ...
... collision: release two cars simultaneously, one carrying two magnets and the other carrying one
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It should be equal to the time interval for which our data reflects, this would have to be true because coasting time was used to find aver acceleration which was then used to calculate everything else.
you should be able to use the frictional force and the mass to find the acceleration of the coasting car
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The following questions are for university physics students, though all but one are accessible to general college physics students, who are invited but by no means required to attempt them. Questions of this nature will be denoted by (univ; gen invited). Questions which actually require calculus are denoted (univ; calculus required). General College Physics students with a calculus background are invited to attempt these questions.
`qx019. (univ; gen invited) Looking at how v0 affects vf, with numbers: On a series of trials, a car begins motion on a 30 cm track with initial velocities 0, 5 cm/s, 10 cm/s, 15 cm/s and 20 cm/s. By analyzing the first trial in the standard way, the acceleration is found to be 8 cm/s^2. Using the equations of uniformly accelerated motion, find the symbolic form of the final velocity in terms of the symbols v0, a and `dx. Then plug the information common to all trials into this equation (i.e., plug in the values of `ds and a) to get an expression whose only unknown quantity is v0. Finally plug your values of v0 for the various trials into your expression, and obtain your values for vf. Sketch a graph of vf vs. v0 and explain as best you can, in terms of your direct experience with these systems, why the graph has the shape it does.
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vf^2 = v0^2+2a`dx
vf^2 = v0^2 + 2*8cm/sec^2*30cm
=v0^2 + 480 (cm/sec)^2
For v0 = 5 cm/s
vf^2 = (5 cm/s)^2 + 480 (cm/sec)^2 !!!Taking square root of entire equation!!!!!
vf = 5 cm/s + 21.9089 cm/s = 26.9089 cm/s
For v0 = 10 cm/s
vf^2 = (10 cm/s)^2 + 480 (cm/sec)^2 !!!Taking square root of entire equation!!!!!
vf = 10 cm/s + 21.9089 cm/s = 31.9089 cm/s
For v0 = 15 cm/s
vf^2 = (15 cm/s)^2 + 480 (cm/sec)^2 !!!Taking square root of entire equation!!!!!
vf = 15 cm/s + 21.9089 cm/s = 36.9089 cm/s
For v0 = 20 cm/s
vf^2 = (20 cm/s)^2 + 480 (cm/sec)^2 !!!Taking square root of entire equation!!!!!
vf = 20 cm/s + 21.9089 cm/s = 41.9089 cm/s
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`qx020. (univ; gen invited) Use your calculators to graph vf vs. v0, using the expressions into which you plugged your values of v0, and verify your graph.
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The graph seems to fit my solutions very well
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`qx021. (univ, calculus required) Should the derivative of vf with respect to v0 be positive or negative? Don't answer in terms of your function, your graph or your results. There is a good common-sense answer based on the behavior of the system and the nature of uniform acceleration.
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Positive, because the fn is increasing in a pos way. This means the rate at which vf is changing in regards to v0 is pos.
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`qx022. (univ; calculus required) What is the derivative of vf with respect to v0? What does this derivative function tell you about the behavior of the system?
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That the fn is increasing at a constant rate.
you need to calculate the derivative
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`qx023. (univ; gen invited) Using the velocity and position functions for uniform acceleration, and the resulting equations of uniform acceleration, can you get an expression for the time required to achieve velocity v_mid_x in terms of v0, a and `dx?
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For vf^2 = v0^2 + 2a`dx
We can get `dx = (vf^2 – v0^2)/2a
if we divide both sides by 2 we should be able to get v mid x. This is not in terms of v0, a and `dx though because vf also needs to be known.
????every equation has either vf or `dt in them, so is this not possible??????
we'll leave this open for now
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`qx024. (univ; gen invited) Using the velocity and position functions for uniform acceleration, and the resulting equations of uniform acceleration, can you get an expression for the velocity v_mid_t in terms of v0, a and `dx?
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For vf = v0 + a`dt, we can use formula vf^2 = v0^2 + 2a`dx and solve for vf
vf = `sqrt(v0^2 + 2a`dx), now plug this in for vf in 1st equation
We get,
`sqrt(v0^2 + 2a`dx) = v0 + a`dt
[sqrt(v0^2 + 2a`dx) – v0]/ a = `dt
`dt/2 = [(sqrt(v0^2 + 2a`dx) – v0)/ a]/2 = v mid t
good
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`qx025. (univ; gen invited) Can you interpret the expressions for v_mid_x and v_mid_t to answer at least some of the open questions associated with the ordering of v0, vf, `dv, v_mid_x, v_mid_t and v_ave? Can you develop expressions that can be interpreted in order to answer the remaining questions?
?????Is this something to think about or do you want an answer, didn’t know because there was nowhere to put an answer?????????
Think about it.
Very good work, and good questions. See my notes.
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