we vistited yesterday a fun park and saw a ride which was a big centrifuge only in vertical. this one had a speed meter but it was broken and i also wanted to know how much g-force is on there...

so my first thought was to calculate this with the centrifugal-accerleration. I have $\omega = \frac {0.5*2\pi} 1 [\frac {rad} s]$ and $r= 25[m]$, so the accerleration is $a_z = \omega^2r = \pi^2*25 = ~246$

then i can calculate the g force with $g_{force} = \frac {a_z} g$ which gives me about $24g$ that seems fairly too high!

what am i missing?

  • $\begingroup$ you have an orbital period of $T=2s$, that 'feels' a touch too fast over a $50$m diameter $\endgroup$ – Nic Aug 26 '11 at 10:00
  • $\begingroup$ mh we could only measure the time the thing need to turn by eye, so yes its a little bit sloppy... we tought that it needs from the highest to the lowest (=180° turn) arround one second. When i take half of that, i get a g-force of $6.2$ which seems also little bit too high or not? - edit: well then i have 0.5s 6.2g and then 0.5s the negative, seems fair :D $\endgroup$ – reox Aug 26 '11 at 10:06
  • $\begingroup$ better would be to count several rotations and take an average ;-) (I'm really fun at theme parks...) In calculating $\omega$ you have a factor of $2\pi$, that's for an entire revolution, 360$^{\circ}$, and $r$ is radius correct? Again that seems big. $\endgroup$ – Nic Aug 26 '11 at 10:12
  • $\begingroup$ yes, thats really hard to say... i will visit at day and see if i can determinate the height and the rotation speed. but how i calculated the force is correct right? $\endgroup$ – reox Aug 26 '11 at 10:18
  • $\begingroup$ Largely, though you also have to take into account the acceleration due to gravity of the Earth! $\endgroup$ – Nic Aug 26 '11 at 10:24

Ok, assuming a passenger on the ride to be traversing a vertical circle then at all times they are experiencing an acceleration $a_{circ}$, this is given by $a_{circ} = a_{track} + a_{gravity}$.

Therfore at the bottom of the circle: $a_{track} = a_{circ}+a_{gravity}$ (with gravity pointing down the two minus signs cancel)

At the top we get: $a_{track}=a_{circ}-a_{gravity}$.

The effects of which are we feel a greater push at the bottom than at the top by a difference of $2a_{circ}$'s. Which might seem counterintuitive at first but seems to fit reality quite well!

  • $\begingroup$ so $a_{circ}$ means $a_z = \omega^2r$? $\endgroup$ – reox Aug 26 '11 at 17:09
  • $\begingroup$ yup, its the net effect of the two accelerations. $\endgroup$ – Nic Aug 26 '11 at 18:36
  • $\begingroup$ The difference at the top and bottom is 2 a-circ, not 2 a-gravity,or 2 g's. Consider if the ride is moving very slowly, or not at all at the top and bottom points. $\endgroup$ – Jim Graber Aug 26 '11 at 21:28

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