Posted by StressedStudent on Saturday, November 3, 2012 at 11:17pm.
(1/2) I w^2 = 3.38*10^9
I = (1/2) m r^2 = (1/2)(26)(.222)^2
= .641 kg m^2
(1/2) I = .320 kg m^2
so
.320 w^2 = 3.38 * 10^9
w^2 = 10.55 *10^9 = 1.055 * 10^10
w = 1.03 * 10^5 radians/s
= .163 * 10^5 revolutions/sec
= 9.81 * 10^5 rpm
981,000 rpm
forget about it. If the axis were level, you could not turn it without flipping over but no matter how strong the material it would have flown apart by then.
I suppose you could have contra rotating flywheels pointed at each other to prevent the gyroscopic flip over problem, but you still can not get that sort of rpm practically even in a vacuum with magnetic suspension bearings without the material flying apart from centripetal acceleration stress. Something has o hold the material at the outside in toward the center and that is the radial tensile stress in the material.
Set the rotational kinetic energy equal to 3.39*10^9 Joules, and solve for the angular velocity, w.
The rotational KE is (1/2)*I*w^2
The moment of inertia of a solid disc about the center is
I = (1/2)MR^2
By the way, we tried this.
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