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course PHY 241
9/18 9 pmI realize this and the following assignments are late, but I think I will be caught up (or at least have made significant progress) by the 24th.
A ball rolls down a 60 cm incline and off the end. As it drops to the floor it travels an additional 16 cm in the horizontal direction, and it is known that the drop to the floor requires 0.4 seconds. From this we conclude that the ball was moving at 40 cm/s at the end of the ramp.
`q001. If the ball started from rest, and if the acceleration on the ramp was uniform, then what does the graph of velocity vs. clock time look like?
**** The graph is a rising straight line.
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What new quantity or quantities can we determine, based on our graph, and what is the value of each? (Quantities in which we might be interested include acceleration, average velocity, time interval, change in velocity and possibly others; which ones can be determined directly from the graph?).
**** We can determine the acceleration, average velocity, change in velocity, and distance travelled (though that is already given).
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`q002. From the given information we know that the average velocity of the ball is 20 cm/s. Having found this, how can we determine the time interval for the motion down the ramp?
**** The time interval is: 60 cm / ( 20 cm/s) = 3 s.
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`q003. If for an interval on which the graph of v vs. t is a straight line we know two of the three quantities initial velocity, final velocity and average velocity, then we can find the third.
Suppose for such an interval we know that the initial velocity is 10 cm / sec and the final velocity is 30 cm/s. What is the average velocity?
**** The average velocity is: (10 cm/s + 30 cm/s) / 2 = 20 cm/s.
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Suppose we know for a different interval that the initial velocity is 20 cm/s and the average velocity is 30 cm/s. What is the final velocity?
**** The final velocity is: 2 (30 cm/s) - 20 cm/s = 40 cm/s.
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On such an interval where the final velocity is 50 cm/s and the average velocity is 30 cm/s, what is the initial velocity?
**** The initial velocity is: 2 (30 cm/s) - 50 cm/s = 10 cm/s.
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Describe in words how you would get the final velocity for an interval on which the initial and average velocities are known.
**** When the velocity vs. clock time graph is linear, the average velocity is the average of the initial and the final velocities: (vi + vf) / 2 = va. This means the final velocity is: vf = 2 (va) - vi
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`q004. The average rate of change of A with respect to B is defined to be the change in A, divided by the change in B, where A and B represent specific quantities.
If the average velocity of an object on some interval is defined to be the average rate of change of position with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
**** The average velocity (v) is the change in position (ds) divided by the change in clock time (dt), so that v = ds / dt.
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If the average acceleration of an object on some interval is defined to be the average rate of change of velocity with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
**** If the average acceleration (a) is the change in velocity (dv) divided by the change in clock time, then dv = a * dt. We then take the integral of this equation.
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What quantity could you find for an interval on which the position of an object changes by 40 cm while the clock time changes by 8 seconds?
**** The average velocity, which is 40 cm / 8 seconds, or 5 cm/s.
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What quantity could you find for an interval on which the velocity of an object changes by 300 cm/sec while the clock time changes by 20 seconds?
**** The average acceleration, which is (300 cm/s) / 20 seconds, or 15 cm/s^2.
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If you know the average rate of change of A with respect to B, and you also know the change in B, how would you find the change in A?
**** The average rate of change of A with respect to B is the average rate of change of A per unit of B, so if I multiply the rate of change of A with the change in B, I will get how much A changed.
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If you know the average rate of change of A with respect to B, and you also know the change in A, how would you find the change in B?
**** In the last answer, I said the change in A is equal to the rate of change of A times the change in B. Therefore, if I want to find the change in B, I will divide the change in A by the rate of change in A.
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If you know the average velocity of an object on an interval, and know the change in position, how do you find the change in clock time?
**** v = ds / dt, so to find dt, I rewrite the equation as dt = ds / v.
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If you know the average acceleration of an object on an interval, and know the change in clock time, what other quantity could you find?
**** The change in velocity.
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`q005. Let's return to situation we started with, the ball on the ramp. We know its initial velocity to be 0 and its final velocity to be 40 cm/s, and the length of the ramp to be 60 cm.
Summarize everything we can reason out from this data, assuming our v vs. t graph to be a straight line and using the definitions of average velocity and average acceleration.
**** Change in position = 60 cm
Initial velocity = 0
Final velocity = 40 cm/s
Average velocity = 20 cm/s
Change in clock time = 3 s
Acceleration = (40 cm/s) / 3 s = 13.3 cm/s^2 approx.
Change in velocity = 40 cm/s
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`q006. For your first short rubber band, what was the color of this rubber band, and what lengths of this rubber band corresponded to what lengths of the rubber band chain?
**** Red
Rubber band length Chain length
8.5 38.1
13.2 49.0
13.8 50.2
19.2 64.3
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Give the same information for your second short rubber band.
**** Yellow
Rubber band length Chain length
10.7 40.9
12.1 51.0
12.5 58.1
13.4 65.8
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Sketch a graph of the length of the first short rubber band vs. the length of the chain. Note the convention that a graph of y vs. x has the y quantity on the vertical axis and the x quantity on the horizontal.
Do your points lie pretty close to a single straight line? Do you think there's a tendency of your data points to curve upward or downward?
**** Yes, the points lie pretty close to a straight line. There does not seem to be a tendency either an upward or downward curve.
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Sketch the straight line you think lies closest, on the average, to the points of your graph. What is the slope of this line?
**** The slope is about 11/15.
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Repeat for your second short rubber band.
**** The points lie pretty close to a straight line. The seems to be a slight tendency to curve downward. The slope of the line is about 3/15, or 1/5.
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If your two short rubber bands were allowed to oppose one another, you would be able to take data for the length of one vs. the length of the other. Based on your graphs, what do you think would be the slope of a graph of the length of your first rubber band vs. the length of the second?
**** The slope would be about 11/3.
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Based on your two graphs, construct a graph of the length of the first short rubber band vs. the length of the second. Describe your graph.
**** The graph is a straight line with a slope of about 10/3.
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`q007. Give your data for the the number of dominoes vs. the length of the rubber band chain.
****
Dominos Length
4 25.5
5 25.7
6 26.4
7 27.1
8 28.0
9 29.0
10 29.8
11 30.7
12 31.2
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For number vs. length, length would go in the first column, but OK.
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Sketch a graph of the number of dominoes vs. the length of the rubber band chain. Describe your graph.
**** The points form an increasing line that tends to have a slight upward curve to it.
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Sketch the straight line you think comes closest, on the average, to your data points. What is its slope?
**** The slope is about 6/8, or 3/4.
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For number vs. length it would be about 4/3.
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Relabel your graph using the following assumptions: Each cm marking on the ruler you used corresponds to an actual metric measurement of 0.006 meters, and each domino has a weight of 0.14 kilogram meters / second^2 (University Physics group) or 0.18 kilogram meters / second^2 (General College Physics group). In terms of your relabeling, what now is the slope of your straight line?
**** If I multiply the first and last length values by 0.006 meters and find their difference, I will get the rise. If I multiply 8 dominos by 0.14 kilogram meters / second^2, I will get the run. This makes the slope (0.0342 meters) / (1.12 kilogram meters / second^2), or 0.0305 seconds^2 / kilogram.
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`q008. Give your data for the number of cycles in 10 seconds vs. the number of dominoes.
****
Dominos Cycles
4 36
6 32
8 28
10 25
12 23
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Sketch a graph of your data. Is it plausible that your graph could be fit by a straight line?
**** The points could fit a straight line, but it is more likely that they form a curve.
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Figure out the time required for a single cycles, for each number of dominoes. Give your results.
****
Dominos Seconds per cycle
4 0.2778
6 0.3125
8 0.3571
10 0.4
12 0.4348
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Graph the square of the time required for a single cycle vs. the number of dominoes. Is it plausible that your graph could be well fit by a straight line? If so, give its slope.
**** The graph of the time for a single cycle squared vs. the number of dominoes would more likely fit a curve than a straight line, but since the curve is slight, a line could be used to describe this section of the graph. The slope of the line is 0.01398 seconds^2 / domino.
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University Physics Only:
Relabel your last graph, using the fact that the mass of a single domino is .018 kilograms. What is the slope of the best straight line you can fit to your data?
**** 0.7769 seconds^2 / kilogram.
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For each number of dominoes, figure out omega, in radians per second, for the corresponding oscillation. Each cycle corresponds to the 2 pi radians around the circle.
**** ω = dθ / dt
Dominos Seconds per cycle ω
4 0.2778 s 22.62 rad/s
6 0.3125 s 20.11 rad/s
8 0.3571 s 17.60 rad/s
10 0.4 s 15.71 rad/s
12 0.4348 s 14.45 rad/s
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For each number of dominoes, figure out sqrt( k / m ), where k is the slope of your graph of number of dominoes vs. length.
**** ω = sqrt( k / m )
Ω m
22.62 rad/s 0.00152
20.11 rad/s 0.00192
17.60 rad/s 0.00251
15.71 rad/s 0.00315
14.45 rad/s 0.00372
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For the 10-domino trial, write the equation
m * x '' = -k x
and give a solution of your equation.
**** 0.00315 * x '' = -0.7769 x
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You would get x '' = -0.78 / .0031 x, or about
x '' = 520 x
which has one solution
x = sin(sqrt(520) * t), or about
x = sin(23 t).
The 23 indicates 23 radians per second. Off by about a factor of 2 from the observation made for 12 dominoes, but still in the right ballpark.
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I am not sure what to do after this.
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Show that as long as you get omega right, it doesn't matter if you use the sine or the cosine function to express a solution.
**** Since I do not remember how sine and cosine apply to this problem, I have no idea how to answer.
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&#This looks good. See my notes. Let me know if you have any questions. &#