What is the acceleration magnitude of a simple pendulum?

Question

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:

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

Answer 1

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 – $mg \cos \theta$).

Answer 2

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 $~a=\sqrt{\ddot x(t)+\ddot y(t)}$

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