The centripetal force $F_C$ of a uniform circular motion can be expressed as,


where $m$ is mass, $r$ is the radius, and $T$ is the time interval for one revolution (the period).



if we have the data for several $(T^2, r)$ pairs, by linear regression, the slope of the trendline $s=\frac{F_C}{4\pi^2m}\Leftrightarrow F_C=4\pi^2ms$.

So theoretically, $F_C=\frac{4\pi^2mr}{T^2}=4\pi^2ms$, but I get very different results for each method.

Here's some actual data from an experiment.

m = 0.1013 kg

T^2 (s^2)    r (m)        F_C = (4*π^2*m*r)/T^2 (N)
0.912        0.140        0.614
0.978        0.145        0.593
1.036        0.150        0.579

By linear regression, I get the trendline $r = 0.080T^2 + 0.066$, so $s=0.080$ and $F_C=4\pi^2ms=0.320$.

The difference of $F_C$ between the two methods is almost twice, and I really cannot explain what is causing the discrepancy.

  • $\begingroup$ Fitting this with a straight line is correct (do not attempt to force it through the origin), but then the size of your fitted intercept term is really big, signifying at least that this choice of parametrisation, for fitting, is not working. Maybe you need a way larger sample, or that you need to know what your individual data point's uncertainties are. Maybe a different fitting form would help. Do you have the goodness-of-fit? Also $F_C\neq4\pi^2ma$. $\endgroup$ May 11, 2023 at 13:37
  • $\begingroup$ @naturallyInconsistent I have only 5 pairs of data, and by using them all, I get the regression line y = 0.087x + 0.059 which is not really much different from what's shown here. The R^2 value is 0.998 for three and 0.992 for five. Why is it that $F_C \ne 4 \pi m a$? Is it because of the large intercept value? $\endgroup$
    – xiver77
    May 11, 2023 at 14:14
  • $\begingroup$ I would guess the problem is that you didn't do a nearly big enough range of $r$ values. So small errors in one measurement will drastically change both the slope and the intercept. As nautrallyInconsistent points out, the big intercept of your fit function shows immediately that your data doesn't agree with your model. Consider learning about fit parameter errors so that you can estimate the error on your $A$ parameter. I think you'll find it's big (although with only three points this too can be problematic if your data just happens to fall close to along one line) $\endgroup$
    – AXensen
    May 11, 2023 at 15:15
  • 1
    $\begingroup$ If you calculate slope with the term $\frac{F_C}{4\pi^2m}$, you should get 0.1535 (from the first data point). Your regressed equation has some type of data entry or calculation error in it. I also note that your derivation of $F_C=\frac{4\pi^2mr}{T^2}=4\pi^2ms$ was arrived at by a "circular" algebraic substitution. Very often, when a substitution is made back into an earlier equation, algebraic inconsistencies result, so such a practice should be avoided. $\endgroup$ May 11, 2023 at 19:48
  • 1
    $\begingroup$ @Amit, xiver77 can get a good answer from a linear regression, but he or she needs to be careful about "backwards substitution" into algebra derivations. $\endgroup$ May 12, 2023 at 0:11

1 Answer 1


You fit the model $r =slope \cdot T^2 + b + \epsilon$, where $slope = F_c/const$, $b$ is an offset, $const=4\pi m$, and $\epsilon$ is a random error. Therefore, if we solve for $F_c$, we get $$ F_c = slope \cdot const %\Rightarrow %slope = \frac{F_c}{const} \tag{1a}$$ and $$ F_c = const\;\frac{r-b-\epsilon}{T^2} %\Rightarrow %slope = \frac{r-b-\epsilon}{T^2} \tag{1b}$$ Thus, we should compare these two expressions. By omitting the random error, we find that these two terms pretty similar. The red line displays this fit enter image description here The fit shows that the intercept $b$ differs significantly from zero. This is why we are not allowed to omit $b$ in eq. (1b). However, if we adjust our fitting model and instead fit the model $r =slope \cdot T^2 + \epsilon$, we obtain $slope \approx 0.148$, which is approx. twice the value you reported. This fit is shown as green line in the plot. Again, plugging the numbers into the two equations yields "matching results".

As there are only three data points the "quality" of both fits are great, $R^2 > 99\%$. Using model selection methods (such as AIC or BIC) seams to be weird due to the fact that I used only three data points. Hence, I would recommend you get more data points and spread them further apart -- use larger differences in the radius. In addition, it would make sense to repeat some measurements multiple time. Also, when repeating the measurement make sure that you change the radius to some "initial value" and reset it to the value you wish to re-measure. Otherwise you do not take into account the error in measuring/adjusting the radius.

  • $\begingroup$ I'm having a hard time to understand some parts. When you put $F_C=const\frac{r-b}{T^2}$, there is a variable $r$, unlike in $F_C=const\times slope$, where you can calculate $F_C$ in one go. How will you put into account the $r$ values to get $F_C$? Also, how does forcing the intercept to be 0 work? When so, the $R^2$ becomes 0.278 with the given 3 data points. $\endgroup$
    – xiver77
    May 13, 2023 at 10:50
  • $\begingroup$ The model equation $r =slope \cdot T^2 + b + \epsilon$ is equivalent to $F_c = const\;\frac{r-b-\epsilon}{T^2}$. Therefore, if you have no trouble with the model equation, I don't understand why you have trouble with the second equation. $\endgroup$
    – Semoi
    May 13, 2023 at 19:55
  • $\begingroup$ The method of forcing the intercept to be zero depends on the program language. Therefore, there is no general answer. $\endgroup$
    – Semoi
    May 13, 2023 at 19:56
  • $\begingroup$ The programming language I use is called R. In the case of zero-intercept R changes the way it calculates $R^2$, see stats.stackexchange.com/questions/26176/… This is the reason I get 99%. $\endgroup$
    – Semoi
    May 13, 2023 at 20:17
  • $\begingroup$ In $const \times slope$, $const$ is given and $slope$ is given as the fitting line is determined. With, $const \times \frac{r-b}{T^2}$, $const$ is given and $b$ is given with the fitting line, but $r$ and $T^2$ is different at each data point. $\endgroup$
    – xiver77
    May 14, 2023 at 17:21

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