Sig Figs: Understand 2 Sig Fig Solutions

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this part(a),

Why do they give 2 sig figs in the solution? $80 kg$ and $10 m/s$ both have 1 sig fig.

Thank you for your help!

 

[Lnewqban]

The answer is 4000 N-m.

 

[jbriggs444]

Why do they give 2 sig figs in the solution? $80 kg$ and $10 m/s$ both have 1 sig fig.

It depends on the convention being used in the textbook you are learning from. Because neither input is expressed in scientific notation, we cannot know whether the trailing "0" in "80" or in "10" are significant or not.

One can adopt the pessimistic attitude and judge them to have 1 significant figure as you have done.

Or one can adopt an optimistic attitude and judge them to have 2 significant figures as the text seems to have done.

There is no right or wrong here. If we do not know how reliable the inputs are, we cannot know how reliable the outputs will be. There is no certainty to be had about these inputs. Best advice is to use the conventions established in your text book and ask the teacher for clarification.

There are more reliable ways of doing an error analysis here. One (pessimistic) way would be to look at error bars. Figure the energy based on $\frac{1}{2}(75 \text{ kg})(9.5 \text{ m/s})^2 =$ you tell us and based on $\frac{1}{2}(85 \text{ kg})(10.5 \text{ m/s})^2 =$ you tell us.

The resulting error bar will be pretty bad. Less than one significant figure.

The rules of significant figures are crude estimates. They are important to get you used to thinking about the reliability of the numbers you manipulate. The rules are simple so that you do not need to get into the details of partial derivatives, statistical effects, adding in quadrature and other things that the first year student may not be well prepared for.

[Statistically independent errors tend to operate so that the variance of the result is the sum of the variances of the inputs. Standard deviation is (roughly) the square root of the variance. So squaring the relative errors, totalling and taking the square root yields a good estimate for the standard deviation of the distribution of possible results. More generally, if the result is a function of the inputs, the sensitivity of the result to each input is gauged by the partial derivative of the result with respect to that input. So you take the errors in the inputs, multiply each by their respective partial derivative, square, sum, take the square root and *voila*. But we do not want to hit a first year physics student with that load of stuff immediately. There is a bunch of 2nd and even 4th year math stuff buried in there]

 

[ChiralSuperfields]

The answer is 4000 N-m.

Thank you for your reply @Lnewqban!

 

[ChiralSuperfields]

It depends on the convention being used in the textbook you are learning from. Because neither input is expressed in scientific notation, we cannot know whether the trailing "0" in "80" or in "10" are significant or not.

One can adopt the pessimistic attitude and judge them to have 1 significant figure as you have done.

Or one can adopt an optimistic attitude and judge them to have 2 significant figures as the text seems to have done.

There is no right or wrong here. If we do not know how reliable the inputs are, we cannot know how reliable the outputs will be. There is no certainty to be had about these inputs. Best advice is to use the conventions established in your text book and ask the teacher for clarification.

There are more reliable ways of doing an error analysis here. One (pessimistic) way would be to look at error bars. Figure the energy based on $\frac{1}{2}(75 \text{ kg})(9.5 \text{ m/s})^2 =$ you tell us and based on $\frac{1}{2}(85 \text{ kg})(10.5 \text{ m/s})^2 =$ you tell us.

The resulting error bar will be pretty bad. Less than one significant figure.

The rules of significant figures are crude estimates. They are important to get you used to thinking about the reliability of the numbers you manipulate. The rules are simple so that you do not need to get into the details of partial derivatives, statistical effects, adding in quadrature and other things that the first year student may not be well prepared for.

[Statistically independent errors tend to operate so that the variance of the result is the sum of the variances of the inputs. Standard deviation is (roughly) the square root of the variance. So squaring the relative errors, totalling and taking the square root yields a good estimate for the standard deviation of the distribution of possible results. More generally, if the result is a function of the inputs, the sensitivity of the result to each input is gauged by the partial derivative of the result with respect to that input. So you take the errors in the inputs, multiply each by their respective partial derivative, square, sum, take the square root and *voila*. But we do not want to hit a first year physics student with that load of stuff immediately. There is a bunch of 2nd and even 4th year math stuff buried in there]

Thank you for you reply @jbriggs444! That is very helpful. The absolute uncertainty in the kinetic energy would be

$K = 4000 ± 1000 J$

Many thanks!

 

[jbriggs444]

$K = 4000 ± 1000 J$

Normally one takes the radius of uncertainty associated with significant digits to be half of the place value of the trailing significant digit.

 

[mfb]

Thank you for you reply @jbriggs444! That is very helpful. The absolute uncertainty in the kinetic energy would be

$K = 4000 ± 1000 J$

Maybe, but evaluating that is beyond the scope of the problem:
If you say a human has a mass of 80 kg then it's probably between 75 and 85.
If you say a human runs 10 m/s (great sprint speed), it won't be 5 m/s (long-distance running speed) and it won't be 15 m/s either (exceeding the world record). It's likely between 9 and 11 m/s.

Using the extremes in both cases leads to a kinetic energy between 3040 J and 5140 J.

Note how critical this range depends on the assumptions. If we say the velocity has to be between 9.5 and 10.5 m/s then the range narrows to 3380 J to 4690 J.

4 kJ, 4.0 kJ and 4000 J are all reasonable answers for this problem.

 

[jbriggs444]

Something that I have rarely shared is the notion of fractional significant figures. I've not seen the idea presented elsewhere.

One can think of "significant figures" as a way of expressing relative error. A crude way which is always rounded to the nearest integer. The obvious way to express this would be:$$\text{sig figs} = \text{round}(-\log(\frac{\text{error}}{\text{value}}))$$If you consider $9.99 \pm 0.005$ then you get $$-\log(\frac{0.005}{9.99}) = 3.3 \text{ significant figures}$$This rounds to 3, of course.

If you consider $1.00 \pm 0.005$ then you get $$-\log(\frac{0.005}{1.00}) = 2.3 \text{ significant figures}$$This rounds to 2. Which is not quite what we wanted.

So we need to tweak this simple model just a bit. We need for our formula applied to $9.99 \pm 0.005$ to return a result of just under $3.5$ prior to rounding. And for $1.00 \pm 0.005$ to return a result of exactly 2.5 prior to rounding. An offset of $0.5 - \log(2) \approx 0.199$ should do nicely. So we wind up with $$\text{sig figs} = \text{round}(-\log(\frac{\text{error}}{\text{value}})+0.199)$$

To some extent, this is idle musing. But it is concerning to consider that $9.99 \pm 0.005$ is about ten times more precise than $1.00 \pm 0.005$ even though those both have the same number of significant figures.

 

[mfb]

To some extent, this is idle musing. But it is concerning to consider that $9.99 \pm 0.005$ is about ten times more precise than $1.00 \pm 0.005$ even though those both have the same number of significant figures.

That's unavoidable if the classes differ by a factor 10. It's the reason significant figures can't be used in scientific work, they just give a very rough idea about the uncertainties.