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course Phy 242
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University Physics students should do a quick read-through of Chapters 14, 17, 18, 19 and 20
You are expected to have made brief notes, at least recognizing things in the text that you think might be relevant to the systems we are working with in class.
`q001. Include a copy of your data for the domino measurements, including your estimated uncertainty:
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Measured from last year.
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Based on your data, calculate the volume of each domino in cm^3, and in (cm_reduced)^3, where cm_reduced is the measurement made with your paper ruler.
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`q002. Describe your experience in working with the buoyant balance.
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Our University Class didn't exactly make measurements of the system, just observed from the past class experements. But non theless, I could basically understand what was going on.
The plastic cup had a CD on it with two dominos clamping a paperclip together that acted as a fulctrum for the wooden stick on top of it. This wooden stick had a chain of paperclips which were suspended in a small cup of water, not touching the bottom.
Obviously there was a counter weight here, which they moved to balance the system. Then, they laid a piece of tape after sticking it to itself and ripping apart to create a static charge, and moved the system slightly by moving another piece of
tape close to the statically charged piece. The clost the counter weight moved to the center, the paper clips moved up, which in term, made the boyant force decrese. Therefore, the net force is acting down, as is the torque. Gravity is consistent.
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`q003. Report all data obtained during class, and include a description of what was measured and how:
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We made no measurements as the University Phy class, just very brief observations.
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`q004. Once the buoyant balance is constructed, a small mass is placed on the free end (i.e., the end on which the paperclips aren’t hanging into the water). The system is observed to rotate in such a way that this end moves lower.
Taking the direction of this rotation as the positive rotational direction, we conclude that the added mass results in an additional positive torque on the system.
The system oscillates for a short time and comes to rest, with the end containing the added mass a little lower than before.
Once the system is at rest, the net torque on it is zero.
Explain how we know that the torque in the positive direction is greater than before the mass was added.
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The system was basically at 180 degrees. (imagine looking sideways into a unit circle, the wooden strap being the point where i'm measuring at ""180"" degrees, or level.) the net torque is balance in both directions, by the buoyant force,
and the mass of the opposite force. After that mass is moved away from the paperclips, the moment arm created by that mass has increased, thus, increasing the amount of torque that is applied by this mass. This slight increase in torque
in your declared "" positive "" direction, is just enough that the buoyant force is less of a force, but still relevant. The system will oscillate in a slight SHM for a breif amount of time, until the net torque is ""positive"" this positive net torque
is in balance by the counter weight of gravity on the paperclips, and the lesser boyant force. Buoyant force decreases, Gravity obviously stays the same but more of the force of gravity is allowed to act on this system because the
buoyant force is less. Therefore is less net force acting on that side of the system. Therefore, the counter mass has to be the positive or greater net force or torque on this system.
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Is the torque in the negative direction now greater or less than before we added the mass?
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Even though the buoyant force has decresed, the force of gravity is ""increasing"" because the buoyant force is decreasing. ( I know gravity stays the same, but the gravity force in the system is more relevant now that there is less buoyant force,
I'm having a bit of trouble explaing this, but I think you understand what i'm trying to say.) Because this force of the mass has a greter moment arm, the torque on the paperclips has increased, to account for this moment arm increase. Therefore,
the torque in the negative direction is now greater. (equal and oppsite reaction)
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Good.
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What caused the torque in the negative direction to change?
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the buoyant force, which allowed the force of gravity to become more relevant, again hard to explain, but I understand completely.
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Suppose we add a square of paper having mass 0.6 gram to the ‘free’ side of the system, at a point 20 cm from the balancing point. How much additional torque will then be acting on the system?
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F=ma, = 0.6g * 981cm/s^2 = 588.6 g cm/s^2 * 20cm = 11,772 g cm^2/s^2 = additional torque applied
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0.6g * 981cm/s^2 isn't equal to 11,772 g cm^2/s^2 , but I know what you intend. Despite misusing = signs to indicate train of thought rather than equality, your intent and your final value of the torque are correct.
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Assuming that the paper clips are suspended from a point 20 cm from the balancing point, on the end opposite the ‘free end’, what must be the change in the torque on this side once the system has come to rest in its new position?
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If the paperclips are 20cm from the balancing point, the change must be equal and opposite to the additional torque applied earlier, -11,772 g cm^2/s^2. ( assuming that positive is rotating up with respect to the paperclips.)
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How is it that the torque on this end changes?
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the torqe changes because the buoyant force is less because the extra torque applied to the free end has increased the force of gravity has on the system. ( again hard to explain detailed) so if the system comes to rest,
the decreased buoyant force and the increased gravity of the paperclips, has to counter act against the extra torque on the free end, increasing the torque on the paperclips side.
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Does the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
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Yes, the weight changes because the force of gravity becomes more relevant, increases. It increases equal and opposite to the buoyant force.
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The weight of the clips doesn't change. Still the same amount of steel, acted on gravitationally by the same old Earth.
It's only the buoyant force that changes.
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Does the torque due to the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
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Yes, it has to increase to counter the new torque on the opposite side, it increases equal and opposite to the new torque applied on the opposite end. which is related to the decrease in buoyant force and increase in gravity on the clips.
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The buoyant force is unrelated to the weight of the clips.
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Does the buoyant force on the paper clips on this end change? If so does it increase or decrease, and by how much?
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Yes, decrease, equal and opposite to the torque applied opposite on this system.
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Does the buoyant force on the paper clips on this end change? If so does it increase or decrease, and by how much?
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same question as before.
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Suppose we add a square of paper having mass 0.6 gram 3 cm on each edge, to the ‘free’ side of the system, at a point 20 cm from the balancing point. A sheet of this paper, 8.5 inches x 11 inches, has a mass of 5 grams.
By how much does the torque in the positive direction change as a result?
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3cm square has an area of 9 cm^2. the pressure is a force per unit area, which would be .6g * 981cm/s^2 / 9cm^2 = 65.4 g cm/s^2 with a moment arm of 18.5 cm to make a change in torque of 1209.9 g cm^2/s^2.
I chose 18.5cm for the moment arm because the centroid of a square is half way, half of 3 is 1.5. 20 - 1.5 is 18.5cm.
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If the hanging paper clips are 15 cm from the balancing point, then after the system has come to rest, by how much has the weight of the paperclips changed, and by how much has the buoyant force on the paper clips changed?
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I dont know exactly the weight of the paperclips because we did not take measurements on any of this. But by working in symbols, I can generalize this.
Weight of paperclips = W
Gravity = G
Buoyant force on clips = B
Using our change in torque to be 1210 gcm^2/s^2 as before, I will call this T
weight of clips has changed by, W - B = T therefore, W = T - B
Buyont force has change by, B = W - T which is consisten, because this will be a negative.
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`q005. BB’s are bouncing back and forth, without loss of energy, between two parallel walls, one on the left and one on the right. Assume that the walls are 36 cm apart, and that gravitational influences have negligible effect.
A BB shot from one wall, straight toward the other, will therefore keep bouncing back and forth between the walls forever. The speed of the BB, when it’s not in contact with the wall, will always be the same as the speed from which it was shot.
(This is of course an idealization; the coefficient of restitution for an actual BB colliding with a real wall is less than 1 so energy will in the real world be lost with each collision).
A BB is shot toward the right wall from the position of the left wall. A second BB is shot at the same velocity from the same position, just as the first BB hits the right wall, from which it rebounds toward the left wall without any loss of speed.
How far from the left wall will the two BB’s be when they meet up?
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3/4 of the way, 27cm from left wall.
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If that was the case the rebounding BB would move 9 cm while the new BB would move 27 cm.
However their speeds don't change, except very briefly while in contact with the walls.
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Assuming that they narrowly miss one another and continue bouncing back and forth between the walls, how far from the left wall will they be the next time they pass one another?
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1/4 of the way, 9cm from left wall.
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At what distances from the left wall will they pass for the third, fourth and fifth time?
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they will oscillate. even numbers will be 27cm, odd number pass times will be 9cm.
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As they continue bouncing back and forth, assuming their paths are just far enough apart to avoid collision, how many ‘passing points’ will there be between the two walls?
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2
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If the second BB was just a little slower than the first, how would the ‘passing point’ change as time goes on?
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The passing point would not be exactly 1/4 or 3/4 the total distance from the left wall, it will be more like 3/8 and 5/8 distance. ( a little closer to center)
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Suppose now that the second BB is fired at the instant the first BB is halfway to the right wall. How far from the left wall will they be when they meet?
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31.5cm from left wall. (7/8). half way of the half way point. which is 7/8.
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If we continue firing a new BB every time the last BB reaches the halfway point, assuming that the BB’s always manage always to narrowly miss one another, at what distances from the left wall will BB’s pass one another?
At the instant of firing, at what possible distances from the left wall will all the other BB’s be located?
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Either 7/8 or 1/8. 4.5cm or 31.5cm
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At the instant the first BB passes by the second, at what possible distances from the left wall will all the other BB’s be located?
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Eventually if you keep shooting BB's as the last BB hits the second wall, there will be a BB for each 8th interval of space inbetween the walls. So each 8th interval could be a passing point for at least two BB's.
so each 8th of 36cm. 9cm, 13.5cm, 18cm, 22.5cm, 27cm, 31.5cm. I'm not including 4.5cm and 36 because I'm not sure if you are counting the wall as a ""passing point""
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In what other ways might the firing be timed so that when two BB’s pass, every other BB is at the same distance from the left wall as at least one other BB?
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when the first BB has hit the other wall and come to the half way point coming back, fire a BB. then every other one will pass like you said.
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You'll want to rethink the BB situation. You started with an incorrect premise. Your subsequent thinking wasn't bad, so I don't think you'll have any trouble figuring it out.
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`q006. If an ideal gas is kept at constant volume, then the percent change in its pressure is the same as the percent change in its absolute temperature.
Every additional 10 cm of water supported in a thin vertical tube requires an increase in pressure equal to about a 1% of standard atmospheric pressure.
Based on this, how much additional pressure do you estimate corresponds to one unit on your ‘squeeze scale’?
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About a 2. Completely based on how hard I had to squeeze based on how much the height of the water changed, 2.
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`q007. If the initial absolute temperature of the gas is 300 Kelvin, then if it is heated enough to raise a column of water 40 cm high, then:
By how much will its temperature have changed?
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10cm is 1%, 40cm is 4%, which should be 4% of 300 kelvin, so changed by 12 degrees kelvin. now is 312 Kelvin.
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If the bottle contains 500 milliliters of air, how much energy would be required to achieve this change? Note that the volume of water in a thin tube can be regarded as negligible, so that the volume of the gas will remain unchanged.
(You should know how how to find approximately how many moles are contained in 500 milliliters of air at atmospheric pressure and typical ambient temperatures).
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I'm having trouble with this as far as energy change in air compared to water. Help?
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You can figure out how many moles; you could even solve PV = n R T for n, which isn't difficult, but there's another bit of information which also allows you to make a simple estimate.
The molar heat capacity is related to the gas constant R in a way I specified on the very first posting.
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How much would the temperature of the air in the bottle have to change to have the same effect as one unit on your ‘squeeze scale’?
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will be able to approximate after having some guidance for that last question. Other than that, I feel pretty good about this document.
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end document
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