What is gravity? What is acceleration? What is force? Perhaps the Raspberry Pi and Sense-Hat with its Inertial Measurement Unit/sensors may be able to help us make sense of it all!

From the Sense-Hat documentation the Inertial Measurement Unit, IMU is an ST LSM9DS1 The manual can be found at: https://www.raspberrypi.org/blog/astro-pi-tech-specs/

Linear acceleration sensitivity = 0.061 mg/LSB (I don't quite understand this yet). Reading p12 of the manual:

To be honest I'm not sure about the direction of the axes and the pin 1 dot position in the manual. Not that it matters because we can find the axis directions using a simple python program.

pythonhosted.org/sense-hat/api/#imu-sensor

from sense_hat import Sense Hat sense = SenseHat() raw = sense.get_accelerometer_raw() print("x: {x}, y: {y}, z: {z}".format(**raw))and just steadily moving the Sense-Hat around and noting -1g when the axis is plumb (aligned with the centre of Planet Earth).

Other than a body in free fall {insert picture of Galileo dropping a Pi off the leaning Tower of Pisa} the pendulum is one of the next well understood and well studied mechanics problem. Galileo showed us the period of oscillation is the same irrespective of the amplitude of swing or the bob's mass. The period is determined by the length of the pendulum rod. Newton was born the same year Galileo died and he had a bit to say about motion and centripetal force. I've rigged up the Pi and Sense-Hat to form the pendulum bob mass. Using the Sense-Hat data logger software example I've managed to capture some acceleration data while the pendulum is swinging. Of course for a small swing where $sin \theta \approx \theta$ we are expecting to see simple harmonic motion, SHM.

The radial direction along the pendulum rod is the device x-axis and tangential direction is the device y-axis. The device z-axis is out of the plane of the pendulum. When the pendulum is at rest the x-axis records -1g. I forgot to record the other 2 axes. That was a school boy error.

The graph below shows the logged data from the IMU graphed using Mathematica. Please note when t=0 the pendulum was already swinging.

I would have thought the device z-axis acceleration would have been zero but it's approximately constant at 0.04g. That's 10 times greater than the sensitivity so it is significant. My only thoughts are:

The x and y accelerations are both oscillatory. The device x acceleration appears to have twice the frequency of the device y acceleration. (Hhmm not sure I was expecting that - maybe I missed that lecture.) The time period of the y data is approx. 1.2 seconds.

I think we should mathematically model the problem and see if these experimental results make any sense.

To get a handle on a dynamics problem I like to apply D'Alembert's Inertia force/torque to freeze the problem i.e. it's then an equation of statics rather than motion. I can then split the elements down into free body diagrams with equal and opposite forces/torques applied to all the elements. For plane statics problems the sum of the forces in any two directions and a moment about any point will equal zero.

From the above figure we can resolve the mg force on the pendulum bob into 2 components, and in this case it's convenient to use the radial and tangential force components because they're orientated with the device axes.

If you think those inertia forces look complicated - they are. The IMU is going along for the ride. Remember when you was young and liked the park swings you were the mass being impelled to move in circular arc. I found H. Stommel's book: 'An Introduction to the Corolis Force' shows how to derive these mysteries forces and in what situations you can apply them. They don't teach you that in school.

From the data captured we've got accelerations on both the x and y axes. School texts proceed on with the equating of torques because it's easier to understand.

Torque is equal to force x perpendicular distance i.e. tangential force component x the length of the pendulum: $mgsin \theta * l$. The radial force (cosine component) x 0(perp. dist) = 0 of course.

If we apply D'Alembert's inertia torque $I \ddot\theta$ to freeze the pendulum in space (hence it's now a problem of statics) we know the algebraic sum of the torques must equal zero.

$ I \ddot\theta - lmgsin \theta = 0$

The second moment of inertia (assuming the bob mass,m is much greater than the pendulum rod mass): $I = ml^2$

$\ddot\theta - {g \over l } sin \theta = 0$

$sin \theta = \theta$ for small angles then we've got a nice and easy 2nd order linear differential equation.

The natural frequency:

$\omega = \sqrt{g/l}$

I live on the surface of the Earth (unlike TP) where g = 9.81m/s^2. The Length of the pendulum rod was 0.4m.

$\omega = \sqrt{9.81/0.4} = 4.95$ radians/second

$f = \omega/2\pi = 0.79$ cycles per second(Hz)

and the Time period for 1 cycle. $T = 1/f = 1.26$ seconds {Not too bad}

Back to the funny forces:

If we equate the forces in the x direction and we assume the IMU is positioned at x=L and constant then I'm assuming $ \dot L = 0 $

$m L^2 \ddot \theta - mgsin \theta = 0 $ and that's the same equation we got when equateing the torques. Ah so that where the $I \ddot \theta $ comes from!

The device y-axis: Again assuming L is const and $\ddot L = 0$ and when $\theta$ is small then $cos \theta \approx 1$

$ T = mg - mL \dot \theta ^2$

The squared velocity term would make the periodic function appear to have twice the natural frequency. The wave would always be positive. http://www.wolframalpha.com/input/?i=sine+squared+x+%2C+sine+x

When the bob's angular velocity $\dot\theta = 0$ hence $\dot \theta ^2 = 0$ the centripetal force component disappears for split second at the pendulum swing extremes which happens twice per cycle which is what the IMU x axis data is showing.