#$&*
PHY 231
Your 'cq_1_10.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed Q 10.1
A pendulum requires 2 seconds to complete a cycle, which consists of a complete back-and-forth oscillation (extreme point to equilibrium to opposite extreme point back to equilibrium and finally to the original extreme point). As long as the amplitude of the motion (the amplitude is the distance from the equilibrium position to the extreme point) is small compared to the length of the pendulum, the time required for a cycle is independent of the amplitude.
How long does it take to get from one extreme point to the other, how long from an extreme point to equilibrium, and how long to go from extreme point to equilibrium to opposite extreme point and back to equilibrium? #$&*
answer/question/discussion: ->->->->->->->->->->->-> :
Since it takes 2 seconds to complete a cycle, one would assume it only takes 1 second to go from one extreme to the other since that would be half way. This means that it should only take .5 seconds to go from extreme point to equilibrium. To go ¾ of the cycle would mean that it would be ¾ the time: (3/4 * 2 sec = 1.5sec)
What reasonable assumption did you make to arrive at your answers? #$&*
answer/question/discussion: ->->->->->->->->->->->-> :
I assumed that the pendulum was moving with the same average velocity through each amplitude. You also have to assume that it reaches the starting extreme point position on its way back; that the distance of every amplitude is exact.
@& Good. However read on.
Note that the speed of the pendulum actually decreases along with its amplitude. What you acxtually assumed was that the time for each 1/4 cycle was equal to the time for every other 1/4 cycle.
The specifics of the actual speed and amplitude get fairly complicated, as does the question of whether the 1/4 cycles really do take exactly the same time (they do, to a very high level of accuracy, take the same time, but the theory in the non-ideal real-world case gets way more complicated than we can deal with in this course. The theory for the ideal case, in which the bead doesn't lose any energy, is well within the scope of this course, as we'll see later in the semester).*@
@& You aren't expected to always be able to work these problems out completely.
You should always attempt, before looking at the given solution, to figure out as much as you can about what the problem means (as I expect you are doing).
Then if necessary you would move on to a self-critique.
Ideally this puts you on an upward learning curve as you work through subsequent problems.
Let's see how this works out on subsequent questions.
*@
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Question: `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
X motion:
Fnet = mass * a
5 N = 4 kg * a
A = 1.25
Acceleration of X comp. = 1.25 * cos(210) = -1.08 m/s/s
A = -1.08 m/s/s
Ds = 30 m
V0 = 10 m/s
Vf2 = v02 + 2a * ds
Vf2 = 102 + 2(-1.08) * 30 m = 35.2
Vf = 5.93 m/s
Y Motion:
V0 = 0 m/s
A = -.63 m/s/s
Dt = ds / vAve = 30 / 8 = 3.75 s
Dv = (v0 - vf) = aAve * dt
Dv = -.63 * 3.75
Because v0 = 0: vf = -2.36
Magnitude: a2 +b2 = c2
5.932 + -2.362 = c2
C = 6.38 m/s
Angle = arctan(-2.36 / 5.93) +360 = 338.29
confidence rating #$&*:
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Given Solution:
As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds.
During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second.
Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
STUDENT COMMENT
It makes sense when it explained, but I cannot seem to get the right answers. I guess it will take some time to figure everything out.
INSTRUCTOR RESPONSE
For the x direction you have acceleration, initial velocity and displacement, from which you can find time interval and final velocity.
Having found the time interval you can use the given information to find the final y velocity.
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Self-critique (if necessary):
I understand the first part but am having trouble comprehending the second part starting at the Y motion.
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Self-critique rating: 2
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@& You appear to have understood the second question better than the first.
You can use these questions for reference in the future, and I'll be glad to answer any additional questions related to this q_a_.*@