# What is the acceleration magnitude of a simple pendulum?

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

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:

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

• Using $d$ for the length of the string is fairly confusing in a calculus context. I suggest using something else like $\ell$ or $r$. Sep 3 '21 at 13:24
• @BioPhysicist Thanks! I used $l$ originally in a related question on dsp.SE and was recommended to use $d$ instead :)
– Anna
Sep 3 '21 at 13:55
• 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. Sep 4 '21 at 21:53
• That makes sense, thank you @PhilipWood!
– Anna
Sep 5 '21 at 12:31
• My upvoted there is and I not know the reason. Sep 5 '21 at 12:33

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 – $$mg \cos \theta$$).

• 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?
– Anna
Sep 2 '21 at 18:39
• 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.
– Anna
Sep 2 '21 at 19:47
• "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. Sep 2 '21 at 19:50
• 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 :)
– Anna
Sep 2 '21 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 $$~a=\sqrt{\ddot x(t)+\ddot y(t)}$$

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

• 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?
– Anna
Sep 3 '21 at 14:12
• This is correct
– Eli
Sep 3 '21 at 14:17
• Why in this community are there few up votes? My it is +1 Sep 4 '21 at 21:41
• @Sebastiano I was also confused! I upvoted it and later it was back to zero. Someone also downvoted my question.
– Anna
Sep 5 '21 at 12:32
• 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!] Sep 5 '21 at 13:11