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course phy 201
9/25 1:30 am
`q000. General College Physics: Give your data for today's experiment, along with a brief description of what you did.****
A long metal plate was balanced on top of a die. We then took a rubber band and stretched it back a set distance. It was released and snapped against the metal plate, from which point we measured the rotations.
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`q001. If you balance the metal ramp on a domino, then place one domino at one end of the ramp, the ramp will not remain balanced. In order to maintain balance we will add a stack of two dominoes to the other side. Where should the two-domino stack be placed?
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It would be placed halfway between the middle and the end opposite the first domino
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If you rotate the ramp, which will be moving faster, the domino at the end or the two-domino stack?
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The two domino stack. The speed of rotation increases closer to the center
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The angular velocity of the motion (i.e., number of degrees per second) is the same everywhere.
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The stack closer to the center travels around a smaller circle than the stack at the end. It takes the same amount of time to complete a circle, so the closer stack isn't traveling as fast.
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`q002. In the experiment where you dropped the domino at the same time as you released the washer, you obtained an upper and a lower bound on the acceleration of gravity. What were these bounds?
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70cm and 50 cm
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These are upper and lower bounds on the distance traveled during a quarter-cycle of the pendulum, not on the corresponding acceleration of the falling object.
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`q003. If a system has zero acceleration and is given a velocity of 1 millimeter per second, how fast will it be moving 10 seconds later?
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1 millimeter per second
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If a system is given a velocity of 10 cm/s toward the north, and has an acceleration of 1 cm/s toward the south, what will be its velocity 15 seconds later?
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5 cm/s towards the south
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What will be its position at that time?
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37.5 cm north
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`q004. If we had five dominoes on each side of an Atwood system, then assuming the dominoes to all have the same mass the system would have zero acceleration.
Suppose we had 15 dominoes on one side and 5 on the other. Clearly the system would accelerate pretty rapidly in the direction in which the 15-domino side descends. Now suppose we release this system and at the same time drop a domino, which falls freely to the floor. Which will have the greater acceleration, the Atwood system or the freely falling domino?
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The free falling domino
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`q005. Our first model of the Atwood machine will be as follows: The machine consists of two equal masses suspended over a pulley of negligible mass and with negligible friction, plus an additional mass added to one side. We hypothesize that the acceleration of the Atwood system will be in the same proportion with the acceleration of gravity as the additional mass to the total mass of the system. As a first approximation we will use 1000 cm/s^2 as the acceleration of gravity.
Suppose the system consists of five dominoes on each side, all of equal mass, and an additional unspecified mass.
If the system accelerates at 20 cm / s^2, then what is the unspecified mass, as a percent of the mass of the system?
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2%
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What is the mass of the unspecified mass as a percent of the mass of a single domino?
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each domino is 20% of gravity. Each domino equivilant to roughly 200 cm/s^2 of acceleration. Percentile: 10%
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Right, if the system consists of five dominoes.
However there are five dominoes on each side, so there are 10 dominoes in the system.
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If the unspecified mass was removed and a single domino added to one side, what would be the acceleration of the system?
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200 cm/s^2
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This would be correct if there were two dominoes on one side and three on the others.
However there are ten dominoes in the original system. When another is added there are now 11.
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If an additional four dominoes were added to the same side, what would be the acceleration of the system?
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1000 cm/s^2
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This isn't consistent with your answer to the very first question.
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If five more dominoes were added to the same side, what would be the acceleration of the system?
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2000 cm/s^2
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You can't have more dominoes on one side than you have in the whole system, so you can't have more than 100% of the mass on one side.
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`q006. If the slope of an incline is less than 0.1, then if friction has only negligible effects an object moving down the incline will accelerate at a rate equal to slope * acceleration of gravity. If a domino has thickness 0.9 cm and a ramp has length 60 cm:
What will be the acceleration of such an object on this ramp when it is supported by one domino?
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slope: 0.9/59.98 = 0.015 acceleration = 0.015 * 1000 cm/s^2 = 15 cm/s^2
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What will be the acceleration of the object if the ramp is supported by four dominoes?
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slope: 3.6/59.89 = 0.06; acceleration = 0.06 * 1000 cm/s^2 = 60 cm/s^2
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At what rate does the acceleration of the object change with respect to ramp slope?
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a one-to-one ratio
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You need to go back to the definition of average rate and reason out the correct answer to this question.
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Suppose the table on which the ramp is set up isn't quite level, so that the end of the table on which the lower end rests is 0.3 cm lower than the end on which the ramp is supported by dominoes. Answer the same questions as above.
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Impossible to determine. Without a right angle, the length on the bottom cannot be determined, and slope cannot be figured.
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If one end is supported by a domino on the table, and the other end lies on the table which is 0.3 cm lower than the end supporting the domino, then you can figure out the rise of the ramp.
Knowing its length you could also figure out the run of the ramp. However the angle is so small that the run will not differ significantly from the length of the ramp.
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Does the slope of the table affect the answer to the third question?
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I would assume not
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You need to work it out and carefully answer the questions, especially the third question about rate of change.
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`q007. Suppose the y axis of an xy coordinate plane points straight upward, and that a rubber band chain pulls straight down, in the negative y direction, on a paperclip at the origin. If no other force acts on the paperclip, it's going to accelerate in the direction of this force.
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In what direction would a second rubber band have to pull in order to balance this force and keep the paperclip stationary?
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It would have to pull strait up
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Suppose the only force available is directed at 10 degrees to the left of the positive y axis. Why is it impossible for this force alone to keep the paperclip stationary?
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because some of the force is directed away from directly up. This means that not all the force is pulling against the opposing force.
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Let's bring in a third rubber band, and let it act on the paperclip in the direction of the positive x axis. Is it possible to balance the system using this rubber band in combination with the previous force at 10 degrees to the left of the positive y axis? If so, what force do you think would be necessary, and what would be the force at 10 degrees to the left of the positive y axis, as percents of the original downward force?
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It is possible. The upward force of both rubber bands must equal the downward force of the paperclip. I assume the positive force from the rubber band would be roughly 90%.
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Assuming that the third force can, in some direction, balance the other two, in what direction do you think the amount of required force would be the least? In what direction do you think it would be the greatest?
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Likely, it would be around 80 degrees off to the right.
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In what directions would it be possible for a third force to balance the other two?
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Strait down still. Altering the angle in any way will redistribute the forces.
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Lots of good answers, and some more that are on the right track but not quite right.
This should be easy and not too time consuming to revise.
&#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).
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