A hula hoop math owie
April 8, 2022 1:57 AM Subscribe
I am trying to figure out how to calculate the "spinning properties" of a hoop based on its diameter and weight.
I recently got into hula hooping and have a hoop that is 120 cm (~47 inches) in diameter and weighs about 800 g / 1.76 lbs. It's comfortable to spin around my waist, but a little too large for indoor hooping.
I know that a hoop is easier to spin the larger and the heavier it is. Therefore, I think I should be able to spin a smaller hoop (maybe 100 cm) if it weighs more than my current one. I would like to calculate the perfect weight for that smaller hoop. I assume centrifugal force is involved, but playing around with the formula didn't give me plausible results. Can you help?
I recently got into hula hooping and have a hoop that is 120 cm (~47 inches) in diameter and weighs about 800 g / 1.76 lbs. It's comfortable to spin around my waist, but a little too large for indoor hooping.
I know that a hoop is easier to spin the larger and the heavier it is. Therefore, I think I should be able to spin a smaller hoop (maybe 100 cm) if it weighs more than my current one. I would like to calculate the perfect weight for that smaller hoop. I assume centrifugal force is involved, but playing around with the formula didn't give me plausible results. Can you help?
I think the "spinning property" you are referring might be the moment of inertia.
You can find tables giving formulae for the moment of inertia of various shapes rotating around different axes. For a thin hoop/"thin-walled cylindrical tube" rotating about the central axis, the moment of inertia is given by I = M R^2.
(if the hoop is not thin, there's also a formula for I_z for the case of a "thick-walled cylindrical tube")
posted by James Scott-Brown at 4:17 AM on April 8, 2022 [2 favorites]
You can find tables giving formulae for the moment of inertia of various shapes rotating around different axes. For a thin hoop/"thin-walled cylindrical tube" rotating about the central axis, the moment of inertia is given by I = M R^2.
(if the hoop is not thin, there's also a formula for I_z for the case of a "thick-walled cylindrical tube")
posted by James Scott-Brown at 4:17 AM on April 8, 2022 [2 favorites]
Best answer: Going by James Scott-Brown's formula - if we want to solve for the same inertia experience:
M(1) * R(1)^2 = M(2) * R(2)^2
M(2) = (0.8*(0.6^2))/(0.5^2) = 1.152kg
Presuming the current hoop diameter is 120cm and the new one will be 100cm.
So you want a hoop weight of about 1.1 - 1.2kg to have the same inertia.
This of course presumes rotation about a central axis whereas the hoop axis moves about (your body is the axis) but we're comparing here so perhaps we can neglect that little issue!
posted by london explorer girl at 4:36 AM on April 8, 2022
M(1) * R(1)^2 = M(2) * R(2)^2
M(2) = (0.8*(0.6^2))/(0.5^2) = 1.152kg
Presuming the current hoop diameter is 120cm and the new one will be 100cm.
So you want a hoop weight of about 1.1 - 1.2kg to have the same inertia.
This of course presumes rotation about a central axis whereas the hoop axis moves about (your body is the axis) but we're comparing here so perhaps we can neglect that little issue!
posted by london explorer girl at 4:36 AM on April 8, 2022
Best answer: By the parallel axis theorem, if a thin hoop rotates about an axis a distance d away from the center,
I = M R^2 + M d^2.
If we assume an infinitely skinny person, d~R, so I = 2 M R^2, and the proportion remains the same.
If, instead, the person has a radius r, D=R-r, then
I = M R^2 + M (R-r)^2
If the person has a waist circumference of 90 cm, their radius is 90 cm /(2 pi) = 14 cm.
The desired mass is then 1.179 kg, only slightly higher than london explorer girl's answer.
posted by BrashTech at 5:54 AM on April 8, 2022 [1 favorite]
I = M R^2 + M d^2.
If we assume an infinitely skinny person, d~R, so I = 2 M R^2, and the proportion remains the same.
If, instead, the person has a radius r, D=R-r, then
I = M R^2 + M (R-r)^2
If the person has a waist circumference of 90 cm, their radius is 90 cm /(2 pi) = 14 cm.
The desired mass is then 1.179 kg, only slightly higher than london explorer girl's answer.
posted by BrashTech at 5:54 AM on April 8, 2022 [1 favorite]
Response by poster: Thank you! Alas, I am not an infinitely skinny person, but 1.2 kg sounds like a good place to start.
posted by Skybly at 8:03 AM on April 8, 2022
posted by Skybly at 8:03 AM on April 8, 2022
So, you're not only spinning a loop around and off-center axis, but due to the nature of hulahooping the loop is also spinning around it's central axis. Because the off-center axis is not fixed. Looking at it from the loop's point of view, your a circle rolling around on the inside, so it take a certain number of rotations of you to reach the point you started. So, at least two inertial rotations are going on.
(I thought there was some "hulahoop" formula that you had found "playing around with the formula didn't give me plausible results"...)
It would actually be a decent hard textbook question if phrased more specifically.
posted by zengargoyle at 4:59 PM on April 8, 2022
(I thought there was some "hulahoop" formula that you had found "playing around with the formula didn't give me plausible results"...)
It would actually be a decent hard textbook question if phrased more specifically.
posted by zengargoyle at 4:59 PM on April 8, 2022
HOW TO TWIRL A HULA-HOOP (PDF) - yeah, this is the physics/maths one.....
Demonstration: The Science of Hula Hooping - YouTube - google "hula hoop physics" and "videos" and or browse around.
Swiveling Science: Applying Physics to Hula-Hooping News and Research - Scientific American
It's going to depend on multiple variables creating a hyperspace but the best I could say is that there was an not too long ago proof that the plane slicing was within like within a reasonable percentage from the actual hyperplane so.... any reasonable aproximation isn't that far off from maximal.
You are better off just taking the common knowledge thing and experimenting and maybe learning to make your own hoops and not diving into the actual equations, it gets complicated pretty quickly. Luckily most of the terms are diminishing and can be ignored..... :)
posted by zengargoyle at 3:22 AM on April 9, 2022
Demonstration: The Science of Hula Hooping - YouTube - google "hula hoop physics" and "videos" and or browse around.
Swiveling Science: Applying Physics to Hula-Hooping News and Research - Scientific American
It's going to depend on multiple variables creating a hyperspace but the best I could say is that there was an not too long ago proof that the plane slicing was within like within a reasonable percentage from the actual hyperplane so.... any reasonable aproximation isn't that far off from maximal.
You are better off just taking the common knowledge thing and experimenting and maybe learning to make your own hoops and not diving into the actual equations, it gets complicated pretty quickly. Luckily most of the terms are diminishing and can be ignored..... :)
posted by zengargoyle at 3:22 AM on April 9, 2022
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posted by zengargoyle at 3:00 AM on April 8, 2022 [1 favorite]