Kr=1/2*(I*w^2)
I=1/2*(M*R^2)
w=(290*2pi)/60
Kr=1/2((290*2pi)/60)^2*(.5*8.0*10^4*1.8^2)
Kr=~~~(whatever that equals my question was different just be glad i did this at all ahah)
(this is all assuming you mean 10^4 and 290 is in rpm)
Find the energy stored in the solid flywheel of mass 8.0×104 and radius 1.8 when it's rotating at 290
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