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course Phy 201
q001: Lab investigation:You investigated the question of whether a fixed amount of energy imparted to a system has a consistent result.
The energy came from a rubber band chain.
The system was a rotating ramp with dominoes on its ends.
Give a description of the system and how it was used.
We taped a domino to each end of a rotating strap and balanced to strap on a domino. The rubber band chain was used to accelerate the strap. Time was recorded as well as degrees of rotation. The strap was accelerated at the end 30 cm from the point of balance, 15 cm from the center and 7.5 cm from the center.
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Give your data.
Data given in (deg, sec).
For 30 cm, (610,9.2), (900,10.9), (450,7.8), (610,8.3), (540,8.6).
For 15 cm, (1010,8.9), (440,6.2), (720,9.0), (900,11.8), (820,11.2).
For 7.5 cm, (635, 8.3), (900,9.2), (940,12.7), (720,10.2), (910,8.9).
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Show how you analyzed your data.
We took measurements from 30, 15, and 7.5 cm. The degrees ranged from 440 degrees to 1010 degrees and the time ranged from 6.2 seconds to 12.7 seconds.
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Give your conclusions.
We can conclude the angular V with the observations that we measured.
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
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Your conclusions would include calculation of velocities.
Your data would also include tension information for the chain. This could be a copy of the information you obtained previously for the rubber band chain.
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`q002. Your ramp has mass 200 grams and length 60 cm. By itself its moment of inertia is 1/12 M L^2, where M is its mass and L its length. What is its moment of inertia?
I = ½ ml^2
I = ½ (0.02kg)(6m)^2
I = 0.6 kgm
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The ramp was 60 cm long, not 6 meters. 6 meters would be close to 20 feet.
You'll want to correct your calculation of I, which will turn out to be .006 kg m^2.
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Your dominoes each had mass about 16 grams. How far was the center of each domino from the axis of rotation (your best estimate, recalling that the dimensions of a domino are 5 cm long by 2.5 cm wide by about .8 or .9 cm thick) and what was the moment of inertia of each?
I = ½ ml^2
I = ½(0.016kgm)(.3m)^2
I = 0.0012kgm.
I = ½(0.016kgm)(.15m)^2
I= 0.0003kgm.
I = ½(0.016kgm)(.075m)^2
I = 0.0000075kgm
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What therefore was the moment of inertia of your system?
Average I = 0.0005025 kgm
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You'll want to use the moment of inertia for each system, which will also include the .006 kg m^2 moment of inertia of the metal ramp itself.
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`q003. In the experiment your rubber band had some average tension as it snapped back from release. We want to find the work done on the system as it snapped back.
Using a reasonable rough estimate, how many dominoes do you think the rubber band would have supported at the stretched length of your chain?
5 dominos
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At .16 Newtons per supported domino, roughly how much force did the chain therefore exert at this length?
5*0.1666N = 0.833 N
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What approximate average force did the chain therefore exert between release and losing contact with the rotating system?
F = ma
F = (0.08kg)(9.8 m/s^2)
F = .784 J
Fave = (0J +.784 J)/2 = 0.392 J
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Through what distance do you think this average force was exerted?
The 15 cm that the rubber band was stretched. 40 cm- 25 cm = 15 cm.
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How much work did the chain therefore do on the system?
W = Fnet * ds
W = 0.392 J (1.5m)
W = 0.0588 Jm
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The unit J * m would follow correctly from your previous calculation.
However force is not measured in Joules, but in Newtons.
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1.5 meters is around the height of a typical adult female.
You'll need to recalculate this. Make sure your distances, etc., make sense in terms of the system you were observing.
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You have calculated the moment of inertia I for the system, and you have the definition KE = 1/2 I omega^2. If the work done by the chain on the system all went into kinetic energy, what then should omega have been at the instant the rubber band chain lost contact? Include units throughout your solution. You probably won't know how to interpret the units of your final solution, but don't worry about that.
KE = ½ mv^2 (or ½ Inertia* Omega^2)
Ke = ½ (0.06kgm)(2.94 rad/s)^2 = 0.259 J
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How much kinetic energy would the system therefore have lost as it coasted to rest?
It would lose all of the kenetic energy as it coasts to rest because there would be 0 KE at rest.
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How much work was therefore done by the net force acting on the system?
W = 0.784 J * .15m = .1176 J
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You previously calculated the approximate work done, based on the average force exerted.
Here you have used the maximum force, which occurred only at the instant of release. The force decreased to zero as the rubber band returned to its original length.
Note once more that force isn't measured in Joules.
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Through how many revolutions did the system you observed rotate, and how much kinetic energy would therefore have been lost per rotation?
.259 J / 1.47 rad/s = 0.18 Jrad/s were lost per rotation
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The system rotated at an average rate of 1.47 rad / s, but that's not the number of revolutions. Every 360 degrees (or 2 pi radians) would be one revolution.
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`q004. Suppose that another system gains 5 Joules of energy from an ideal (and hence conservative) rubber band chain, while losing 2 Joules of energy to friction. How many Joules of kinetic energy would you expect it to gain?
5J - 2J = 3J
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For this situation, identify each of the quantities `dW_noncons_ON, `dKE and `dW_cons_ON.
Dw noncons = 2 J of friction
dKE = vhange in KE = 3 J
dwcons = 5 J the rubberband chain
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Reconcile your conclusions with the equation `dW_noncons_ON = `dKE - `dW_cons_ON.
Delta KE = dw cons - dw noncons.
5J - 2J = 3J
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Identify `dPE for this situation.
dPE = -(dW cons)
dPE = - rubberband chain
dPE = -5J
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Write the above equation in terms of `dW_noncons_ON, `dKE and `dPE
dPE = -dWcons
dPE = -(dWcons + dWnoncons)
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Very good answers on this question.
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`q005. University Physics:
Consider a 60-cm ramp of mass 200 grams, rotating about an axis through its center. Give the mass, moment of inertia and uncertainty in the moment of inertia of each of the following:
A 1-mm segment which contains a point 12.4 cm from the axis.
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The part of the ramp whose distance from the axis lies between 10.0 cm and 12.0 cm.
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A segment of length `dr containing a point at distance r* from the axis.
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Suppose the ramp runs from x = 0 to x = 60 cm is partitioned, using the partition x_0 = 0, x_1, x_2, ..., x_(n-1), x_n = 60 cm, with the interval x_(i-1) <= x < x_i containing the point x_i *. Let `dx_i = x_i - x_(i-1).
In terms of these symbols what is your expression for each of the following:
The mass within the ith interval.
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The moment of inertia of the ith interval.
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The total moment of inertia of the ramp.
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The integral obtained by allowing the intervals widths to approach zero.
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The value of your integral.
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Compare your result to the value of the expression 1/12 M L^2.
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Good work, but you have some errors. Most should be easy and quick to correct., and it should greatly benefit you to do so.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
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