The cat is moving in a circle, so it has a centripetal acceleration and a centripetal force. At the apex of the loop, that force is the sum of gravity, and resistance from the track. The track force is greater than or equal to zero, so acceleration due to gravity is less than or equal to the total centripetal acceleration.
g ≤ v²/r
So,
r ≤ v²/g
Taking top speed of a cat as 8.278m/s (from Wolfram Alpha), and g on earth as 9.81m/s², this gives us r ≤ 6.99m. So long as the cat can maintain its top speed all around the loop, it can successfully do a loop of up to 14 meters diameter. This is a lot bigger than I expected, to the extent that I suspect some flaw in my reasoning.
True. But if top speed allows for 14 meters, surely a 1.5 to 2 m loop should be possible (especially given a cat’s incredible reflexes and control given a lack of friction or even freefall). I’d guess that a cat, given enough motivation, could keep running under the little friction provided by the centripetal force for a few hundred milliseconds - likely long enough to complete a 1m loop, maybe even 2, given sufficient space for a top-speed run before entering…
I get the same math… Seems fucky but… This is assuming the sum of centripetal acceleration and gravity at the peak of the loop is zero. It may be physically possible for a cat to learn to manage a loop with such velocity but I imagine a cat wouldn’t be able to maintain a stride through a zero-g portion of the loop the first time it tried it.
So, instead let’s throw an assumption that the cat must maintain at minimum sum of -1g at the maxima of the loop. That may be badly phrased, assuming the cat must have at minimum a net force of at least one g between it’s paws and the surface of the loop it was currently using to accelerate…
3.5 meters = 11.5 feet
Radius, so still a freaking 7 meter diameter loop feels incredible…
The cat is moving in a circle, so it has a centripetal acceleration and a centripetal force. At the apex of the loop, that force is the sum of gravity, and resistance from the track. The track force is greater than or equal to zero, so acceleration due to gravity is less than or equal to the total centripetal acceleration.
g ≤ v²/r
So,
r ≤ v²/g
Taking top speed of a cat as 8.278m/s (from Wolfram Alpha), and g on earth as 9.81m/s², this gives us r ≤ 6.99m. So long as the cat can maintain its top speed all around the loop, it can successfully do a loop of up to 14 meters diameter. This is a lot bigger than I expected, to the extent that I suspect some flaw in my reasoning.
The cat won’t be able to maintain its top speed because of deceleration from lack of friction
Claws homie
Good point. It has zero contact force at the apex, so 14m is an upper bound on possible cat-loops.
True. But if top speed allows for 14 meters, surely a 1.5 to 2 m loop should be possible (especially given a cat’s incredible reflexes and control given a lack of friction or even freefall). I’d guess that a cat, given enough motivation, could keep running under the little friction provided by the centripetal force for a few hundred milliseconds - likely long enough to complete a 1m loop, maybe even 2, given sufficient space for a top-speed run before entering…
I get the same math… Seems fucky but… This is assuming the sum of centripetal acceleration and gravity at the peak of the loop is zero. It may be physically possible for a cat to learn to manage a loop with such velocity but I imagine a cat wouldn’t be able to maintain a stride through a zero-g portion of the loop the first time it tried it.
So, instead let’s throw an assumption that the cat must maintain at minimum sum of -1g at the maxima of the loop. That may be badly phrased, assuming the cat must have at minimum a net force of at least one g between it’s paws and the surface of the loop it was currently using to accelerate…
3.5 meters = 11.5 feet
Radius, so still a freaking 7 meter diameter loop feels incredible…
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