A grinding wheel of 0.2 m in diameter, of mass 3kg, is rotating at 3600 revolutions per minute about an axis through its center.
What is its kinetic energy?
E=Iω22,E=\frac{I\omega^2}{2},E=2Iω2,
I=mr22=md28,I=\frac{mr^2}{2}=\frac{md^2}{8},I=2mr2=8md2,
ω=2πν60=πν30,\omega=\frac{2 \pi \nu}{60}=\frac{\pi \nu}{30},ω=602πν=30πν,
E=12⋅md28⋅π2ν2900=md2π2ν214400=1066 J.E=\frac 12 \cdot \frac{md^2}{8}\cdot \frac{\pi^2\nu^2}{900}=\frac{md^2\pi^2 \nu^2}{14400}=1066~J.E=21⋅8md2⋅900π2ν2=14400md2π2ν2=1066 J.
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