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course phy 201
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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180/4counts=45d/c
360 degree/15counts=24d/c
Good. Those are the angular velocities.
What velocities do these correspond to for each magnet? That is, for example, at how many cm/s was each moving? The answer will be different for each magnet.
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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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Well the ends were moving faster than the center point.
knowing how far the ends were from the axis of rotation, you can find the speed of each
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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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KE=½ mv^2
45d/c=½ (1g) v^2
45d/c=(.5g) v^2
Divide both sides by .5g
45 d/c would not be an energy; that is an angular velocity
you have to find the speed of the magnet, then plug that in for v
90=v^2
Square both sides
v=9.48
what you have done is the square root, not the square.
Had 45 d/c been an energy, your result would have been a good solution for v.
KE=½ mv^2
24d/c=½ 1(g) v^2
24d/c=(.5g)v^2
Divide both sides by .5g
48=v^2
Square both sides
v=6.92
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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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Less
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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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KE=½ mv^2
KE=½(50g)v2
45d/c=½(50g)V^2
45d/c=25g V^2
Divide both sided by 25g
1.8=v^2
Square both sides
v=1.34
KE=1/2mv2
KE=½(50g)v2
24d/c=½(50g)v^2
24d/c=25g V^2
Divide both sided by 25g
.96=v^2
Square both sided
V=.97
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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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KE=½ mv^2
KE=½(50g)v2
45d/c=½(50g)V^2
45d/c=25g V^2
Divide both sided by 25g
1.8=v^2
Square both sides
v=1.34
KE=1/2mv2
KE=½(50g)v2
24d/c=½(50g)v^2
24d/c=25g V^2
Divide both sided by 25g
.96=v^2
Square both sided
V=.97
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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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Less because the acceleration would be less as well at the center
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For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
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5 seconds
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`qx009. Was the period of the unbalanced vertical strap uniform?
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4seconds
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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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The time was changing also you could notice that the Equaliburm was getting shorter.
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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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Final Velocity
1.66
3
3.33
4.28
Those are plausible final velocities, depending of course on their units.
What are the units of each?
KE=½ MV^2
KE=½(100g)v^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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Is the energy it possesses because of its motion
KE=½ mv^2
40millijoules=½(100g)v^2
40millijoules=(50g) v^2
Divide both sides by 50g
40/50=v^2
.8millijoules/g=v^2
Sq both sides
V=.89 millijoules/g
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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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The highest speed an object can reach.
2.75 seconds
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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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Because it takes friction into account, the magnetic force would be stronger .
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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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It would be less in all of them, the KE and frictional force.
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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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Up
2.55
Down
1.63
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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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KE=½ mv^2
2.55=½ 100g v^2
2.55=50g v^2
Divide by 50g
.051g=V^2
Square both sides
V=.22
acceleration is not KE
the KE definition does not include any variable which is equal to the force
you need to use F = m a
KE=½ mv^2
1.63=½ 100g v^2
1.62=50g v^2
Divide by 50g
.032g=v^2
Square both sides
V=.18
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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?
.22 and .18
... 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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&#Please let me know if you have questions. &#
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