Question #185644

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?


1
Expert's answer
2021-05-03T10:33:16-0400

E=Iω22,E=\frac{I\omega^2}{2},

I=mr22=md28,I=\frac{mr^2}{2}=\frac{md^2}{8},

ω=2πν60=πν30,\omega=\frac{2 \pi \nu}{60}=\frac{\pi \nu}{30},

E=12md28π2ν2900=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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS