#$&*
course Phy 241
RESUBMITTING THIS ASSIGNMENT, October 1, 2010 around 10am
ExperimentsQuestions related to experiments are denoted with `qx. Most of the questions for this assignment are related to experiments.
For the magnets on the rotating strap:
`qx001. How fast was each of the magnets moving, on the average, during the second 180 degree interval? All the magnets had the same angular velocity (deg / second), but what was the average speed of each?
****
3 cm from center (15.71 cm/s)
8 cm from center (41.9 cm/s)
14 cm from center (73.3 cm/s)
Our experiment WAS symmetrical on both sides. (same distances)
#$&*
`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?
****
3 cm magnet, end on the inside: (10.5 cm/s), outside: (20.9 cm/s)
8 cm magnet, end on the inside: (36.7 cm/s), outside: (47.1 cm/s)
14 cm magnet, end on the inside: (68.1 cm/s), outside: (78.5 cm/s)
The velocities for the center point of the magnets are in the 1st problem.
#$&*
`qx003. What was the KE of 1 gram of each magnet at its center, at the end closest to the axis of rotation, and at the end furthest from the axis of rotation?
****
3 cm: inside (55.13), center (123.40), outside (218.41)
8 cm: inside (673.45), center (877.81), outside (1109.21)
14 cm: inside (2318.81), center (2686.45), outside (3081.13)
#$&*
`qx004. Based on your results do you think the KE each magnet is greater or less than the KE of its center?
****
I’m not sure about this one. However, I DO know that velocity is linear with the distance from the center and on the other hand, KE is NOT linear with the velocity.
GOod. Since the KE vs. v function is concave up, a linear approximation based on initial and final KE values would be an underestimate (i.e., the trapezoidal approximation over an interval would be above the graph of the function, so the midpoint KE represented by the trapezoid would be greater than the average value of the function; the trapezoid has greater area than the region beneath the graph).
#$&*
`qx005. Give your best estimate of the KE of each magnet, assuming its mass is 50 grams.
****
3cm (6170 g*cm/s)
8 cm (43890 g*cm/s)
14 cm (134322 g*cm/s)
#$&*
`qx006. Assuming that the strap has a mass of 50 grams, estimate its average KE during this interval.
****
KE= ½ mv^2 (need to find velocity)
Arclength:
C=2pi(14)
C=28pi
C=28pi*(40/360)
C=9.77
?Where do I go from here?
You could start with an estimate based on midpoint velocity (the velocity of a point halfway between axis and end). That would be an overestimate (see your previous response and my comment). You could consider a graph of v^2 vs. v to estimate by how much the trapezoid overestimates the area beneath the curve (the integral of v^2 from v = 0 to v = v_max is 1/3 v_max^3, which can be compared with the corresponding trapezoidal area).
#$&*
`qx007. (univ, gen invited) Do you think the KE of 1 gram at the center of a magnet is equal to, greater or less than 1/2 m v_Ave^2, where v_Ave is the average velocity on the interval?
****
I’m thinking it would be equal to ½ m*v_Ave.
the squared proportionality of KE with v dictates that it's less; see also preceding notes
#$&*
For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
****
The period of the oscillation was very close in being uniform completely.
#$&*
`qx009. Was the period of the unbalanced vertical strap uniform?
****
No
#$&*
`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?
****
It wasn’t moving much up and down, therefore resulting in a decreased angular velocity with each cycle because the angle is small.
#$&*
For the experiment with toy cars and paperclips:
`qx011. Assume uniform acceleration for the trial with the greatest acceleration. Using your data find the final velocity for each (you probably already did this in the process of finding the acceleration for the 09/15 class). Assuming total mass 100 grams, find the change in KE from release to the end of the uniform-acceleration interval.
****
“Vf” for each trial:
4 big/1 little: 11.61 cm/s
5 big/1 little: 25.52 cm/s
6 big/1 little: 31.54 cm/s
“KE” for each trial:
6739.6 g*cm/s
32563.5 g*cm/s
49738.6 g*cm/s
I found the “vf” for each by plugging what I knew into the “vf=v0+a’dt” equation. I then used my “vf’s” and put them into the “KE” equation using 100 grams* ½ and then * vf^2.
???My numbers are really big. Did I do this right? Was I supposed to change grams into kg and change cm into m???
#$&*
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.
****
Approx. 90 cm/s
you should indicate how you got this
#$&*
`qx013. Based on the energy calculations you did in response to 09/15 question, what do you think should have been the maximum velocity of the car on each of your trials? You should be able to make a good first-order approximation, which assumes that the PE of the magnets converts totally into the KE of the car and magnet.
****
(Initial separation): 3, 5, 6, 7, 10, 15, 20
(Vf): 0.3, 0.27, 0.2, 0.19, 0.12, 0.08, 0.01
What I did was take my forces that I calculated in the 100915 lab and plugged them into the KE formula, used 100 grams as the mass and solved for “vf” for each different initial separation.
For example: (3 cm)
4.4 mJ= ½ (100 g) (vf)
Vf=0.3 m/s
#$&*
`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.
****
I’m not sure what this question is asking. I do know that work does positive work against friction. Since friction does negative work on the traveling car, it reduces its KE. On the other hand, we saw that the closer the magnets were, the greater the KE was gained by the car. The initial magnet distances we took had PE, but the closer they were the more. The initial PE was then “transformed” into the initial KE done, which friction had a hold of and did work that was equal and opposite of its original KE. So depending on what you want to find, signs will be different in some cases. (positive and negative)
max KE will be less than PE because energy is being lost to friction at the same time the magnet is accelerating the car.
#$&*
`qx015. If frictional forces assume in the 9/15 document were in fact underestimated by a factor of 4, then how will this affect your results for the last two questions?
****
#$&*
`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?
****
1st direction: 32.7 cm/s^2, 23.5 cm/s^2, 20.5 cm/s^2, 19.4 cm/s^2, 18.3 cm/s^2
Opposite direction: 35.3 cm/s^2, 29.1 cm/s^2, 28.0 cm/s^2, 26.9 cm/s^2, 20.5 cm/s^2
#$&*
`qx017. Using the acceleration you obtained find the frictional force on the car, assuming mass 100 grams, and assuming also a constant frictional force.
****
(final velocities)
(1st direction: 33.3 cm/s, 40 cm/s, 45.2 cm/s, 46.2 cm/s, 46.7 cm/s)
(Opposite direction: 42.02 cm/s, 49.4 cm/s, 52.4 cm/s, 54.9 cm/s, 55.9 cm/s)
Frictional force on each:
1st direction: -1665 g*cm/s, -2000 g*cm/s, -2260 g*cm/s, -2310 g*cm/s, -2335 g*cm/s
Opposite direction: -2101 g*cm/s, -2470 g*cm/s, -2620 g*cm/s, -2745 g*cm/s, -2795 g*cm/s
What I did was first find the “vf” for each trial, in each direction, and then plug that into the KE formula, using 100 grams at my mass. Since frictional force does negative work on the car, all my answers are going to be negative.
#$&*
`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
****
1st direction: 1.02 sec, 1.7 sec, 2.21 sec, 2.38 sec, 2.55 sec
Opposite direction: 1.19 sec, 1.7 sec, 1.87 sec, 2.04 sec, 2.72 sec
I used the “’ds=((v0+vf)/2)*’dt and solved for ‘dt in each trial to give me my answers, which is measured in seconds.
#$&*
The following questions are for university physics students, though all but one are accessible to general college physics students, who are invited but by no means required to attempt them. Questions of this nature will be denoted by (univ; gen invited). Questions which actually require calculus are denoted (univ; calculus required). General College Physics students with a calculus background are invited to attempt these questions.
`qx019. (univ; gen invited) Looking at how v0 affects vf, with numbers: On a series of trials, a car begins motion on a 30 cm track with initial velocities 0, 5 cm/s, 10 cm/s, 15 cm/s and 20 cm/s. By analyzing the first trial in the standard way, the acceleration is found to be 8 cm/s^2. Using the equations of uniformly accelerated motion, find the symbolic form of the final velocity in terms of the symbols v0, a and `dx. Then plug the information common to all trials into this equation (i.e., plug in the values of `ds and a) to get an expression whose only unknown quantity is v0. Finally plug your values of v0 for the various trials into your expression, and obtain your values for vf. Sketch a graph of vf vs. v0 and explain as best you can, in terms of your direct experience with these systems, why the graph has the shape it does.
****
If “vf^2= (v0^2+2a’dx), then symbolically, solving for just “vf” will give me (+-) sq. rt of (v0+2a’dx). Using the equation “vf=(+-) sq. rt of (v0+2a’dx)”, plugging in what I found, I can solve it to get an expression whole only unknown is “v0” and can plug in my v0’s and obtain my vf’s. When I graphed my outcomes, the graph looked like a “wide” opened up half parabola. It was a curved line going up and out each time. The smaller the initial velocity, the less the final velocity is going to be. This is why the curve is increasing but the rate is not constant.
#$&*
`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.
****
The graph I did on my calculator also looked the same.
#$&*
`qx021. (univ, calculus required) Should the derivative of vf with respect to v0 be positive or negative? Don't answer in terms of your function, your graph or your results. There is a good common-sense answer based on the behavior of the system and the nature of uniform acceleration.
****
In this situation, the ball is rolling DOWN the ramp, therefore the acceleration is going to be positive instead of negative. From the equation “vf= + or – sq. rt of (v0+2a’dx), you would use the PLUS.
#$&*
`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?
****
I tried to derive this equation and had trouble with it. I don’t know if I’m using the wrong equation for “vf” or what. Would I have to use “u-substitution” on it? I don’t know if this would work or not.
Attempt:
[½ (v0+2a’dx)^(-1/2) ]*(2)
It's v0^2, not just v0.
The derivative is [½ (v0+2a’dx)^(-1/2) ]*(2 v0).
#$&*
`qx023. (univ; gen invited) Using the velocity and position functions for uniform acceleration, and the resulting equations of uniform acceleration, can you get an expression for the time required to achieve velocity v_mid_x in terms of v0, a and `dx?
****
#$&*
`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?
****
#$&*
`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?
"
Very good overall. See my notes for a few of the subtler points.