(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.
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 482-mile trip in a typical midsize car produces about 3.38 x 109 J of energy. How fast would a 26.0-kg flywheel with a radius of 0.222 m have to rotate to store this much energy?
6 answers
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
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.
what problem difrence must an electron pass in order for its velovity to in crease from 10 to 30 Mm/s electron mass equal to 9.1*10kg
disk shaped flywheel with a mass m=50 kg and a radius of r=0.2m rotates at a frequency of 480 rpm left to it self under the influence of the friction force the flywheel stopped after t=50 find the moment of friction force considering it lonstant
5 answers