I'm trying to find the acceleration vector of a simple pendulum. The vector is labeled a in this image from Wikipedia:

enter image description here

Trying to solve it, I've split the acceleration into perpendicular components: radial (along the string) and tangential (along the velocity vector). I've found that:

$a_\textrm{tangential} = d \ddot\theta$

$a_\textrm{radial} = d \dot\theta^2 + g \cos\theta$

where d is the length of the rod and g the acceleration from gravity. However, combining these two causes an awful expression, which makes me think that I've got something wrong:

$|a| = \sqrt{ a_\textrm{tangential}^2 + a_\textrm{radial}^2 } = \sqrt{ d^2 \ddot\theta^2 + ( d \dot\theta^2 + g \cos\theta )^2 }$

What is the actual magnitude of the acceleration vector of a simple pendulum?

My goal is to find an expression for the acceleration magnitude ($\sqrt{a_x^2+a_y^2+a_z^2}$) as measured by a three-axis accelerometer inside the bob.

Edit: I just realised that since $\ddot\theta = -\frac{g}{d}\sin\theta$, we can simplify:

$|a| = \sqrt{ d^2 \ddot\theta^2 + ( d \dot\theta^2 + g \cos\theta )^2 } = \sqrt{ g^2 + d^2\dot\theta^4 + 2dg\dot\theta^2 \cos\theta }$

  • $\begingroup$ Using $d$ for the length of the string is fairly confusing in a calculus context. I suggest using something else like $\ell$ or $r$. $\endgroup$ Commented Sep 3, 2021 at 13:24
  • $\begingroup$ @BioPhysicist Thanks! I used $l$ originally in a related question on dsp.SE and was recommended to use $d$ instead :) $\endgroup$
    – Anna
    Commented Sep 3, 2021 at 13:55
  • $\begingroup$ I agree with BioPhysicist, but you do have to be careful to use a good font with $l$, or it can look like an $\text I$ or a 1. $\endgroup$ Commented Sep 4, 2021 at 21:53
  • $\begingroup$ That makes sense, thank you @PhilipWood! $\endgroup$
    – Anna
    Commented Sep 5, 2021 at 12:31
  • $\begingroup$ My upvoted there is and I not know the reason. $\endgroup$
    – Sebastiano
    Commented Sep 5, 2021 at 12:33

2 Answers 2


If $l$ is the pendulum length the radial acceleration is simply the centripetal acceleration $$a_\text{rad}=l\dot\theta^2.$$ The tangential acceleration is, as you say, $$a_\text{tan}=l\ddot\theta.$$ The expression for the resultant acceleration is not too bad if you are using the usual small angle approximations to treat the pendulum.

A nice little bit of trivia: for small angles: Maximum tangential acceleration = $g\left(\frac Al\right)$ But maximum radial acceleration = $g\left(\frac Al\right)^2$ In which $A$ is the amplitude, measured as an arc length.

You ought not to have included $g\cos\theta$ in your expression for $a_\text{rad}$. In this context, $g$ is not an acceleration, but the gravitational field strength. So the radial force component on the pendulum bob is (Tension in thread – $mg \cos \theta$).

  • $\begingroup$ Thanks a lot! I'll see how it goes with the small-angle approximation. The reason I include the gravitational field strength is that the accelerometer does measure it. When at rest on my table, it will read 9.81 m/s^2 in z-direction. Do you think it makes sense to include g as I have done then, or is it still a mistake? $\endgroup$
    – Anna
    Commented Sep 2, 2021 at 18:39
  • $\begingroup$ I just realised that it gets a bit simpler since $\ddot\theta = -\frac{g}{d}\sin\theta$, but still a hairy expression without the small-angle approximation. $\endgroup$
    – Anna
    Commented Sep 2, 2021 at 19:47
  • $\begingroup$ "When at rest on my table, it will read 9.81 m/s^2 in z-direction". That's because accelerometers measure the acceleration relative to a body in freefall. In applying Newton's laws $for most purposes$ we use the Earth's surface as our reference frame, and treat bodies on its surface as having zero acceleration. The gravitational force on a body at rest on the Earth's surface is balanced by an equal and opposite force (if we neglect the body's small acceleration due to the Earth's diurnal rotation. $\endgroup$ Commented Sep 2, 2021 at 19:50
  • $\begingroup$ Thank you @PhilipWood! Apologies for causing confusion with the wrong terms. Then I guess I'm trying to find the acceleration magnitude of a simple pendulum, relative to freefall :) $\endgroup$
    – Anna
    Commented Sep 2, 2021 at 20:13

the position vector to the mass point is

$$\mathbf R=\begin{bmatrix} x(t) \\ y(t) \\ \end{bmatrix}=\left[ \begin {array}{c} d\sin \left( \varphi \left( t \right) \right) \\ d\cos \left( \varphi \left( t \right) \right) \end {array} \right] $$

from here you obtain the acceleration $\mathbf{\ddot{R}}$ $$\mathbf{\ddot{R}}=\begin{bmatrix} \ddot x(t) \\ \ddot y(t) \\ \end{bmatrix}= \left[ \begin {array}{c} d\cos \left( \varphi \right) \ddot\varphi -d\sin \left( \varphi \right) {\dot\varphi }^{2} \\ -d\sin \left( \varphi \right) \ddot\varphi -d \cos \left( \varphi \right) {\dot\varphi }^{2}\end {array} \right] $$ substitute $\ddot\varphi=-\frac{g}{d}\,\sin(\varphi)$ from the equation of motion and obtain the magnitude $~a=\sqrt{\ddot x(t)+\ddot y(t)}$

$$a=\sqrt { \left( \sin \left( \varphi \right) \right) ^{2}{g}^{2}+{d}^{ 2}{\dot\varphi }^{4}} $$

  • $\begingroup$ Thank you, neat approach! In order to include the "acceleration" due to gravity (which is also measured by my accelerometer), I'll simply add find $\sqrt{\ddot{x}^2 + (\ddot{y} + g)^2}$, right? $\endgroup$
    – Anna
    Commented Sep 3, 2021 at 14:12
  • $\begingroup$ This is correct $\endgroup$
    – Eli
    Commented Sep 3, 2021 at 14:17
  • 1
    $\begingroup$ Why in this community are there few up votes? My it is +1 $\endgroup$
    – Sebastiano
    Commented Sep 4, 2021 at 21:41
  • 1
    $\begingroup$ @Sebastiano I was also confused! I upvoted it and later it was back to zero. Someone also downvoted my question. $\endgroup$
    – Anna
    Commented Sep 5, 2021 at 12:32
  • 1
    $\begingroup$ It's comforting to see that squaring and adding the radial and tangential accelerations from my answer (and using $\ddot \theta=-\frac gl \sin \theta$) gives the same answer as Eli's ! [That's if we use the usual convention for acceleration!] $\endgroup$ Commented Sep 5, 2021 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.