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I am trying to work out the mathematics behind the design of an old mechanical brake tester. The device sits in the passenger foot well of a vehicle and tests the brake efficiency. The device is basically a pendulum with a stylus attached, and as the brakes are applied the pendulum swings and draws up a card. The card is marked in acceleration, or deceleration in this case, (as a % of local gravity) in increments of 10 up to 100% (9.81 m/s2).

Brake Tester Diagram

I am trying to verify the mathematics behind how the instrument functions, by applying an angle to the device, essentially simulating an acceleration. When testing the equipment I can raise the stylus ever so slightly off the paper to remove the effects of friction, so applying an angle will result in a true acceleration reading.

I equated this to; Acceleration = g sin (34) Then I would need to rearrange this to calculate which angles I need to be applying to test each nominal acceleration reading (10,20 and so on).

Is friction a large enough factor that the card will be accounting for this? The distance between the scale divisions on the card get smaller and smaller as it progresses towards 100%. Should I be correcting for the initial friction coefficient. I have the measurements for the length of the pendulum 0.140m and the weight on the end of the pendulum 10 oz.

I found a similar thread of discussion but the results I am obtaining in my tests are slightly out from my calculations. However this error could be in the device itself.

Finding the acceleration at an angle

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    $\begingroup$ for what are you asking exactly? a relation between angle and acceleration? or the error created by friction? or you've made an experiment whose results deviated from calculations... $\endgroup$ – Tofi Apr 14 '16 at 12:17
  • $\begingroup$ Where is the center of mass of the pendulum and where is it pivot. It is hard to tell from the diagram. $\endgroup$ – John Alexiou Apr 14 '16 at 13:47
  • $\begingroup$ Thanks for replying. I am trying to relate angle to horizontal acceleration. $\endgroup$ – WelshJohn Jun 1 '16 at 18:42
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If the pivot is accelerating horizontally (together with the body) at a rate of $a_{pivot}$ then the angular acceleration of the pendulum is

$$ \ddot{\theta} = - \frac{m c (a_{pivot} \cos\theta + g \sin \theta)}{I_{zz} + m c^2} $$

where $c$ is the distance from the pivot to the center of mass, $m$ the total swinging mass and $I_{zz}$ the mass moment of inertia about the center of mass.

The equilibrium position is at

$$ \theta = - {\rm atan}\left(\frac{a_{pivot}}{g}\right) $$

The acceleration of the pendulum as a function of distance $\ell$ from the pivot is $$ a = a_{pivot} \left(1- \frac{m c \ell}{I_{zz} + m c^2} \right) $$

So if the stylus is located at the center of percussion $\ell = c + \frac{I_{zz}}{m c}$ the stylus point will not move in an inertial frame as $a = 0$ at $\theta=0$.

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  • $\begingroup$ Thanks for your help. However, I am still struggling to solve this. Maybe if I try a slightly different approach? If a linear acceleration equal to that of gravity was exerted on the pendulum in my example, how many degrees would the pendulum travel through? (linear acceleration = 9.81 gravity=9.81 mass=0.063kg length of pendulum = 0.140m centre of mass 0.1075m). $\endgroup$ – WelshJohn May 23 '16 at 22:46

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