Find the moment of inertia and the radius of gyration for y = 2√x, y = 0, x = 4; about x = 0
Solution.
A=4•4=16.A=4•4=16.A=4•4=16.
The moment of inertia:
Iy=∫04x22xdx=2∫04x52dx=2•27x72∣04=47•27=7317.I_y=\int_0^4 x^22\sqrt{x}dx=2\int_0^4 x^{\frac{5}{2}}dx=2•\frac{2}{7}x^{\frac{7}{2}}|_0^4=\frac{4}{7}•2^7=73\frac{1}{7}.Iy=∫04x22xdx=2∫04x25dx=2•72x27∣04=74•27=7371.
The radius of gyration:
R=Iym=5127m=5127m,R=\sqrt{\frac{I_y}{m}}=\sqrt{\frac{\frac{512}{7}}{m}}=\sqrt{\frac{512}{7m}},R=mIy=m7512=7m512,
where m is mass of the area.
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