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course Phy 201
9/24 2This is a partial submission. I am just submitting the work I have completed, not the entire document.
For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
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Yes, using the clock to time the period of oscillation and using the midpoint as reference, it took approximately 9 seconds to go to the lowest point and then 7 seconds to get to the highest point. This may have been closer to 8 seconds each considering my judgment of the midpoint and trying to watch the clock at the same time. Of course, the balance slowed down or the distance traveled decreased as it continued to oscillate.
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`qx009. Was the period of the unbalanced vertical strap uniform?
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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 distance covered during the period decreased as the balance continued to oscillate. It may have taken the same amount of time to reach the low point and the high point, but the low point probably wasn’t as low nor the high point as high.
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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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greatest acceleration= 5.5cm/s^2
vAve=10.5cm/s, 9.8 cm/s, 10cm/s, 10.5cm/s
KE=1/2m(v^2) = ½(100g) (10.5cm/s) ^2=50g(110.3cm^2/s^2))=5515g cm^2/s^2
KE=1/2(100g) (9.8cm/s)^2= 50 g (96.04cm^2/s^2= 4802 g cm ^2/s^2
KE=1/2(100) (10 cm/s)^2= 50 g (100 cm^2/s^2)=5000 g cm^2/s^2
KE=1/2m(v^2) = ½(100g) (10.5cm/s) ^2=50g(110.3cm^2/s^2))= 5515g cm^2/s^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=1/2m(v^2)
KE/(1/2 m)=v^2
40 milliJ/ (1/2 (100g)=v^2
40 milliJ/(50g)=40000 J/50 g=800 J/g= v^2= sqrt (800J/g)= vf=28.9 cm/s
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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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Energy calculations are in units of g cm^2/s^2: 5515, 4802, 5000, 5515.
KE=1/2m*v^2
5515 g*cm^2/s^2= ½(100 cm) *v^2
5515 g*cm^2/s^2 /50 cm=v^2
110.3=v^2= sqrt(110.3)=10.5cm/s=vAve
vf=21 cm/s
KE=1/2m*v^2
4802 g*cm^2/s^2= ½(100 cm) *v^2
4802 g*cm^2/s^2 /50 cm=v^2
96.04=v^2= sqrt(96.04)=9.8cm/s=vAve
vf=19.6 cm/s
KE=1/2m*v^2
5000 g*cm^2/s^2= ½(100 cm) *v^2
5000 g*cm^2/s^2 /50 cm=v^2
100=v^2= sqrt(100)=10cm/s=vAve
vf=20 cm/s
KE=1/2m*v^2
5515 g*cm^2/s^2= ½(100 cm) *v^2
5515 g*cm^2/s^2 /50 cm=v^2
110.3=v^2= sqrt(110.3)=10.5cm/s=vAve
vf=21 cm/s
#### I get lost in the all units and bringing g*cm^2/s^2 back to cm/s
We'll sort that question out in class.
However your energy calculations on the 9/15 questions all involved multiplying a presumed frictional force by a displacement. I don't believe that any involved the velocity of the car. THe 5515 g cm^2/s^2 you used here appears to have been based on velocity assumptions.
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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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When you account for the work against friction, the KE is lower than the original because the Force of friction uses up part of the KE.
`dW=Ffrict*~dx
5515g *cm^2/s^2 or milliJ=Ffrict*12 cm
5515 milliJ/12 cm=Ffrict*12cm/12cm
5515milliJ/12 cm=Ffrict
459.6milliJ/cm=Ffrict
Do you subtract 459.6 milliJ from 5515 milliJ to determine how KE remains after accounting for the work against friction?
5515milliJ-459.6 milliJ=5055.4 milliJ
Very good, but:
If the magnets were originally in contact then 12 cm is a reasonable distance to use. However they weren't.
The work done against friction while the magnetic force was significant is not equal to the KE change of the system.
The work done by the magnetic force while the magnetic force was significant is not equal to the KE change of the system. You therefore appear to have calculated the average magnetic force, though your 12 cm interval should have been at least a few cm shorter.
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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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The resulting KE would be decreased by 4.
More frictional force implies more work against friction, not less.
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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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The average acceleration in one direction was 10 cm/s^2 and the average acceleration in the opposite direction was 9.3 cm/s^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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F=m*a
Direction 1—F=100g* -10cm/s^2= -1000 g*cm/s^2
Direction 2—F=100 g* - 9.3 cm/s^2= -930 g*cm/s^2
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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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Very good. See my notes.
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