acceleration of gravity

course Phy 231

2/23 6

007. Acceleration of Gravity

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Question: `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?

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Your solution:

50 cm/ 5 sec = 10 cm/sec is the average velocity. In order to get 10 cm/sec with an object having an initial velocity of 0 cm/sec, I have to plug it in to find the missing variable, vf. (0+vf)/2 = 10 cm/sec. The final velocity is 20 cm/sec. The change in velocity is also 20 cm/sec. The acceleration is then 20 cm/sec / 5 sec which is 4 cm/s^2. For the other one, since a lot of variables are given, an equation can be used:

ds = v0 * dt + ½ a* dt^2. V0 = 0 cm/sec, dt = 3 sec, ds = 50 cm.

50 cm = 0 cm/sec * 3 sec + ½ a * 3sec^2.

50 cm = ½ a* 9 sec^2

50 cm/9 sec = ½ a

5.56 cm/sec = ½ a

a = 11.11 cm/sec^2

For the last trial, the same process as the second one will be used.

50 cm = 0 cm/sec * 2 sec + ½ a * 2sec^2.

50 cm = ½ a * 4 sec^2

50 cm/4 sec = ½ a

a = 25 cm/s^2.

confidence rating #$&*:3

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Given Solution:

We can find the accelerations either using equations or direct reasoning.

To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s.

Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s.

This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2.

The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2.

Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2.

In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.

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Question: `q002. What are the ramp slopes associated with these accelerations?

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Your solution:

4 m/s^2, the slope is 0.5cm/50cm = 0.01

11 m/s^2, the slope is 1cm/50cm = 0.02

25 m/s^2, the slope is 1.5cm/50cm = 0.03

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Given Solution:

For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01.

For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02.

For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.

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Question: `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the best-fit straight line (i.e., the straight line that comes as close as possible, on the average, to the three points).

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Your solution:

(0.01, 4cm/s/s) is the point the graph starts on and (0.03, 25cm/s/s) is the point the graph ends on. (0.02, 11 cm/s/s) is the middle point. The slope of the graph is increasing at an increasing rate. It is not linear because the increase in values is not proportional, which can be shown in the fact that the graph is curving up.

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Given Solution:

The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis).

• The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2).

The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second.

The graph indicates that acceleration increases with increasing slope, which should be no surprise.

It is not clear from the graph whether a straight line is in fact the most appropriate model for the data.

• If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors.

• Alternatively the graph could that acceleration vs. ramp slope is increasing at an increasing rate.

STUDENT COMMENT

I am a little unclear about why the lines could be considered scattered from an actual linear behavior, unless it is because the line is not exactly straight and seems to have a curve.

INSTRUCTOR RESPONSE

The graph points don't line along a perfectly straight line. Based on these three points, you could possibly infer that the graph curves. A curved graph would be consistent with the data.

If experimental uncertainties are large enough, the graph could also be consistent with a best-fit straight line.

• The best-fit straight line for this data wouldn't go through any of the graph points; some points will lie above the line and some below.

• So the points would to an extent be scattered about the line.

• As long as the degree of scattering can be explained by the uncertainties in our measurements, then it is possible that the straight-line, or linear, model is consistent with the data.

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Question: `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05.

What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?

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Your solution:

The coordinates are (0, -5 cm/s^2) and (0.05, 45 cm/s^2). This then gives me the rise and run which amount to 0.05 for the run and 50 cm/s^2 for the rise. The slope is then 50 cm/s^2 / 0.05 which is 1000 cm/s^2.

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Given Solution:

A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way.

For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2.

Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.

STUDENT QUESTION

Once I graphed the points using the correct coordinates I was able to see the correct graph, and its slope of 48cm/s^2/ .05 =960cm/s^2.

??????????My question here is what if on my graph the points were at around 42cm/s^2 instead of the 48cm/s^2, and my answer

was 840 or so. Is there room for the variation in how you plot the graph, if I started with different number or something,

or my graph is not as precise.?????????????????????????

?????????????Also I am still a little unclear, seeing how I had my point graphed wrong in the first place, about why we would

use the slope of the graph .5/ 50cm to get .01....I really cannot seem to rationalize this. ?????????????????????????

INSTRUCTOR RESPONSE

Your slope is based on two points; it can't be calculated based on a single point because you need two points to get the rise and run.

If the two points on your straight line were, say, (0, -10 cm/s^2) and (.05, 48 cm/s^2) then the rise would be 48 cm/s^2 - 10 cm/s^2 = 58 cm/s^2, your run would be .05 and your slope would be 58 cm/s^2 / .05 = 1060 cm/s^2.

If the rise between the two points on your straight line was 42 cm/s^2, with the run still .05, then your slope would be 42 cm/s^2 / .05 = 840 cm/s^2, as you say.

There is in fact only one best-fit line for the data, but since we're estimating the best-fit line it is expected that our estimates of its slope and location will differ somewhat. Certainly it would be possible for one person to get an estimate of 840 cm/s^2, while another might estimate 960 cm/s^2 and another 1060 cm/s^2.

If all three people used the same three data points, and actually found the unique line that best fits those three points, then all three would get identical results.

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Question: `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum.

Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical.

How long did it take, and how long did each cycle therefore last?

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Your solution:

It took me about 94 seconds to complete 100 complete back-and-forth cycles of a pendulum of length 30 cm. Each cycle therefore lasted 94 sec/ 100 cycles = 0.94 sec/cycle.

confidence rating #$&*:3

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Given Solution:

100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.

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Question: `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.

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Your solution:

T = 2*pi / sqrt(g) * sqrt(L)

T = 2*pi*sqrt(L) / sqrt (g)

T*sqrt(g) = 2*pi*sqrt(L)

sqrt (g) = 2 * pi * sqrt(L) / T

g = (2*pi*sqrt(L))^2 / T^2

g = (2*pi*sqrt(30))^2 / 1.08^2

g = 1015.39 cm/s/s

confidence rating #$&*:2 I was not sure if I was supposed to use 1.08 or 0.94 for time.

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Given Solution:

Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain

• T * `sqrt(g) = 2 `pi `sqrt(L).

Then dividing both sides by T we obtain

• `sqrt(g) = 2 `pi `sqrt(L) / T.

Squaring both sides we finally obtain

• g = 4 `pi^2 L / T^2.

Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain

• g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2.

You should check these calculations for accuracy, since they were mentally approximated.

STUDENT QUESTION

I do not understand how if you divide both sides by the sqrt (g), the you end up with t* sqrt(g) = 2' pi ' sqrt (L), when sqrt(L) was divided by 2 'pi.......why wouldnt it be t * sqrt (g) = 2'pi / sqrt(L)....what happens to the division sign there?????

INSTRUCTOR RESPONSE

The division sign is between pi and sqrt(g). It doesn't apply to sqrt(L).

Multiplications and divisions are done in order, as stated in the order of operations. This means that you multiply 2 by pi, then divide the result by sqrt(g), then multiply the result by sqrt(L).

If the sqrt(L) was to be part of the denominator, the expression would have to be written 2 pi / ( sqrt(g) * sqrt(L) ). In that case you would begin by multiplying sqrt(g) by sqrt(L), then you would do multiplications and divisions in order, multiplying 2 by pi then dividing by the product (sqrt(g) sqrt(L)).

The correct expression:

The expression 2 pi / sqrt(g) * sqrt(L) is represented in a computer algebra system as

This representation is consistent with the order of operations.

The common misinterpretation of the expression:

The expression 2 pi / sqrt(g) * sqrt(L) is often misinterpreted as

However the above represents 2 pi / (sqrt(g) * sqrt(L) ), not our original expression 2 pi / sqrt(g) * sqrt(L)

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acceleration of gravity

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