How to correctly round in physics?

[Lotto]

Homework Statement

Round $v_0=\sqrt{5gl-\frac{1}{4\pi \epsilon_0}\frac{Q^2}{ml}}$ correctly.

$Q =1.0 \, \mathrm{\mu C}$, $l=10 \,\mathrm {cm}$, $m = 150 \,\mathrm g$, $\epsilon_0=8.85 \cdot 10^{-12} \, \mathrm {F\cdot m^{-1}}$, $g=9.81\,\mathrm{m\cdot s^{-2}}$.

Relevant Equations

$v_0=\sqrt{5gl-\frac{1}{4\pi \epsilon_0}\frac{Q^2}{ml}}$

I have never been sure how to round it according to the rules. My steps:

$\sqrt{5 \cdot 9.81 \cdot 0.10-\frac{1}{4\cdot \pi \cdot 8.85 \cdot 10^{-12}}\frac{(1.0\cdot 10^{-6})^2}{0.150\cdot 0.10}}$

In [ ] is number of significant digits using multiplication/adding etc. rules.

$\sqrt{5 \cdot 9.81 \cdot 0.10-\frac{1}{4\cdot \pi \cdot 8.85 \cdot 10^{-12}}\frac{(1.0\cdot 10^{-6})^2}{0.150\cdot 0.10}}$

$\sqrt{4.905[2]-8991804694[3]\cdot \frac{1.0[2]\cdot 10^{-12}}{0.0150[2]}}$

$\sqrt{4.905[2]-8991804694[3]\cdot 6.666667[2]\cdot 10^{-11}}$

$\sqrt{4.908[2]-0.59945[2]}$

$\sqrt{4.30855[2]}$

$v_0=2.1 \,\mathrm{m\cdot s^{-1}}.$

Is it correct according to the rules? But this is pain to calculate. I would normally put it all into a calculator and write the result on one decimal place, without using any rules.

How to round in calculations correctly and effieciently?

 

[jedishrfu]

Here's a reference:

https://www.dummies.com/article/aca...gnificant-figures-in-physics-problems-149123/

Basically in math, we round .5 to 1 and 3.5 to 4 but in physics we round up so the resulting number is even hence 3.5 rounds to 4 and 2.5 rounds to 2.0 ie we always round to an even number.

 

[kuruman]

But this is pain to calculate. I would normally put it all into a calculator and write the result on one decimal place, without using any rules.

Then stop using a calculator and start using a spreadsheet. You enter the input parameters only once to whatever accuracy you want. You set up the formula in the answer cell and format the cell to the number of decimal places that is appropriate according to the rules given your input parameters. The added advantage of the spreadsheet is that everything is before your eyes and easy to fix in case you make a mistake.

 

[Steve4Physics]

You should not round intermediate values. And you need not record them. Keep the precision as high as possible and only round as the final step.

But this is pain to calculate. I would normally put it all into a calculator

A pain-free approach should be used. Use your calculator or even better, as advised by @kuruman, use a spreadsheet where possible.

and write the result on one decimal place, without using any rules.

No. The correct level of rounding depends on the precision of the initial data and the type of calculation. Here, rounding to 2 significant figures is appropriate. By coincidence this gives the same answer as rounding to 1dp here.

Minor edit.

 

[Lotto]

For example, when I have to calculate $F_G=6mg$, where $m=150\,\mathrm{g}$ and $g=9.81\,\mathrm {m\cdot s^{-2}}$, how to round it? I mean, I have read that there is an unwritten rule in physics saying that numbers like $1,000\,\mathrm {kg \cdot m^{-3}}$ have four significant figures. So then $150$ should have three significant figures. So $F_G=8.83\,\mathrm{N}$.

What do you mean?

 

[berkeman]

Basically in math, we round .5 to 1 and 3.5 to 4 but in physics we round up so the resulting number is even hence 3.5 rounds to 4 and 2.5 rounds to 2.0 ie we always round to an even number.

That's interesting. I've never heard that before, even in my lower-division undergrad physics classes (which was long ago...). What is the reason for this departure from the math standard?

 

[Steve4Physics]

For example, when I have to calculate $F_G=6mg$, where $m=150\,\mathrm{g}$ and $g=9.81\,\mathrm {m\cdot s^{-2}}$, how to round it?

Let's assume the '6' in your made-up equation $F_G=6mg$ is a precise integer (no uncertainty). Maybe you have six 150g massess hanging on a string.

$m=150g$ is the mass to 2 significant figures. (Because if there is no decimal point, trailing zeroes are not counted as 'significant figures'.) So the value is equivalent to $m= 0.15 ~kg$ (still 2 significant ifgures).

$g=9.81~m\cdot s^{-2}$ is 3 significant figures.

The basic rule is that when you multiply or divide numbers, you identify the number with the least number of s.f. The final answer should be rounded to this number of s.f. In your example it is $m$ with only 2 s.f.

$F_G = 6 \times 0.15 \times 9.81 = 8.829 N$
Rounded to 2 s.f. the final answer is $8.8~N$.

I mean, I have read that there is an unwritten rule in physics saying that numbers like $1,000\,\mathrm {kg \cdot m^{-3}}$ have four significant figures.

No. 1,000 (one thousand) written that way has only one significant figure. (True in physics and maths - and all other subjects!!!)

So then $150$ should have three significant figures. So $F_G=8.83\,\mathrm{N}$.

'$150$' has 2 significant figures as explained above. If you intended a precision of 3 s.f., you would not write $150$, but use $1.50\times10^2$ which removes the ambiguity.

 

[jedishrfu]

That's interesting. I've never heard that before, even in my lower-division undergrad physics classes (which was long ago...). What is the reason for this departure from the math standard?

It is likely an agreed-upon convention between the US NIST and the European ISO organizations. Here is the NIST reference on the rounding of numbers (Section B.7.1 page 43).

https://physics.nist.gov/cuu/pdf/sp811.pdf

and the ISO standard (page 20, B.3 Rule A)

https://www.iso.org/home.isoDocumen...O_bzO2QP14DxJAxwtmaRLmv-WwspRuGQ-iHB6VgCGRJnU

As an aside:

I ran into a similar change of convention issue some years ago regarding spherical coordinates where math uses $R \theta \phi$ but sciences use $R \phi \theta$ in conformance with European standards. It really tripped me up as I thought I had remembered things wrongly.

Apparently, it was adopted a few years after I had graduated and newer textbooks started to switch to the new convention.

https://en.wikipedia.org/wiki/Spherical_coordinate_system

Notice the first two diagrams, the first is the physics convention based on (ISO 80000-2:2019) whereas the second is the math convention extending 2D polar coordinates into 3D.

 

[apostolosdt]

That's the kind of rounding-off calculations you should be able to do mentally. Still,

1. Your final result--irrespective of how you found it--is acceptable, significant-figure-wise.

2. How come you came up with such intermediate results of so many digits, like those inside the radicands? Copied from the calculator's display?

3. To work that expression mentally (or otherwise), first change everything into the same system of units, like SI.

4. Advanced calculators like Hewlett-Packard or Texas Instruments easily accept numerical values along with whatever units you like and then, internally, transform them into the system of units of the user's choice. But that isn't working out mentally.

5. Rounding-off rules have only been revised once. When slide rules were replaced by electronic calculators, people started trusting the displayed rounding-off results. With a slide rule, after one had to reject a 5, one had to leave the last remaining digit IF IT WAS EVEN or increase it by 1 IF IT WAS ODD. But, with modern calculators, a rejected 5 will ALWAYS increase the last remaining digit by 1.

6. Finally, when mental calculation, after getting all units under the same system, we write or, better, mentally change the numbers in scientific format, do arithmetic only with the mantissa, and work out the exponents of 10. It's only a matter of practice, but it's worthwhile; especially, during a teaching session, it saves a lot of time.

7. With more practice, you should be able to carry out such numerical expressions without having to change the units into one system--just modify the exponents of 10 in the scientific format accordingly.