How Do Significant Figures Affect Calculations in Physics Problems?

AI Thread Summary
Significant figures play a crucial role in calculations involving physics problems, particularly when dealing with subtraction of similar values. In the discussed scenario, subtracting the squares of two radii can lead to a loss of significant figures, especially when the result is close to zero. The initial approach of calculating the area using \(\pi (r_1^2 - r_2^2)\) produced a misleading number of significant figures due to the nature of the subtraction. It's important to note that the rules for significant figures are guidelines that can break down in specific cases, such as when subtracting nearly equal quantities. Understanding these nuances can help ensure more accurate results in physics calculations.
zed101
Messages
2
Reaction score
0
Hi! I'm confused, please I would love some help with this:

Homework Statement


if the inner radius of a ring is 3,56 cm and the outer radius is measured as 3,32 cm, compute the area of the ring

Homework Equations


When multiplying the number of significant figures stays the same, when adding or subtracting we keep same number of decimals.
Area of a circle: \pi r^2

The Attempt at a Solution



I'm basically subtracting \pi r_1^{2}-\pi r_2^2. When squaring the radii I get three significant figures and when subtrating I keep two decimals. The answer that way has two decimals for a total of three significant figures, my answer would be \pi 1.65 cm^2 (pi has infinite significant figures). However, if I rearrange the formula as \pi (r_1+r_2)(r_1-r_2) for r_1-r_2 I get 0, 24 cm which only has two significant figures, this crops the total number of significant figures down to two when multiplied with r_1+r_2. What am I doing wrong?
 
Last edited:
Physics news on Phys.org
Errors can become quite large, in proportion, when taking differences of similarly sized numbers. Suppose each of the original values has an error of +/- 0.005. The range of possible values for the difference of their squares is 1.58 to 1.72, so it's not even two significant digits.
This affected your calculation when you took the difference of the squares. Squaring produced 4 digits for each value, of which one disappeared when you took the difference. But after squaring each you should, technically, only have kept 3 sig figures in each, so when the difference lost the high order digit you should only have had 2 sig figures left.
 
Welcome to PF, zed101! :smile:

You're not doing anything wrong.
The rule is only a rule of thumb.

In particular, it breaks down when subtracting quantities with a result close to zero.
This is what happens in your case.
 
Hi! thanks for the replies! :) I see, so the rule of thumb is not always true, teachers never said that. I suppose I should go with the first method before I hand in my paper. I'll add that note about the problem hehe.
Thanks! I think I can work safely now :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top