The minimum required RPMs is a critical consideration. If the top is not spun fast enough it won't be able to spin freely like the Guinness Book of Records wants. Obviously, this is a case when it is better to have a theoretical estimate before building it. So I tried to calculate it using the equation I recently mentioned on this

post.

I assumed the top was a perfect cone made of expanded styrene. Adding the wooden and metal plates and the shaft only gave a small correction. Some of the assumptions and calculations are below for anybody interested in checking them, but the final result is that the critical spin at which wobble starts to increase rapidly is

**127 RPM**.

However, the goal is to be significantly above, so say about double:

**250 RPM**.

On the manual spun, only 100 RPMs were achieved.

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[Nerdy alert]

I modeled the top as a cone with height h = 3.88 m and radius r = 1.815 m.

The volume of the top is V = pi x r

^{2} x h / 3 = 3.1416 x (1.815 m)

^{2} x 3.88 m/ 3 = 13.4 m

^{3}The weight of expanded styrene is only about 15 to 45 kg/m

^{3} (it is mostly air). I assumed 20 kg/m

^{3} as this is compatible with a total weight of the top of 450 kg.

The mass of the Styrofoam cone would be M = 13.4 m

^{3} x 20 kg/m

^{3} = 268 kg.

The moment of inertia of a cone around its axis is available from tables like this

one, and it is:

I = 3/10 M r

^{2} = 3/10 x 268 kg x (1.85 m)

^{2} = 275 kg m

^{2}I did add the moments of inertia of the plywood boards and steel plates (the shaft and spool are close to the spin axis and can be neglected) and it only gave me a small correction (*)

The final value was I = 289 kg m

^{2}The center of mass of a cone is ¾ the height away from the apex, so in our case: L = ¾ x 3.88 m = 2.91 m

Plugging these values in the formula of the critical spin rate:

w = sqrt {4 M g L / I} = sqrt {4 x 450 kg x 9.8 m/s

^{2} x 2.91 m / 289 kg m

^{2} }= 13.3 rad/sec

or

**w = 127 RPM**========

* The moment of inertia of a cylinder or disk of radius r and mass m is: I = 1/2 m r

^{2}Four 4ft diameter 3/4 inch Plywood disk:

From web: 3/4 in thick plywood weighs 10.4 kg/m2, so the mass of one 4ft disk is: 1.17 m

^{2} x 10.4 kg/m

^{2} = 12 kg

I (four plywood disks) = 4 x {1/2 (12 kg) x (0.61 m)

^{2}} = 8.9 kg m

^{2}14ft long, 2 inch diameter, steel shaft

Weight: 7900 kg/m

^{3} x 0.0086 m

^{3} = 68 kg

I (shaft) = 1/2 (68 kg) x (0.0254 m)

^{2} = 0.022 kg m

^{2}Four 2ft diameter steel plates

Weight: 490 - 268 - 4x12 - 68 = 106 kg

I (four steel disks) = 1/2 x (106 kg) x (0.3 m)

^{2} = 4.8 kg m

^{2}I (total) = 275 + 8.9 +

~~0.022~~ + 4.8 = 289 kg m

^{2}