Question #186301

Find the moment of inertia and the radius of gyration for y = 2√x, y = 0, x = 4; about x = 0


1
Expert's answer
2021-05-07T11:28:09-0400

Solution.


A=44=16.A=4•4=16.

The moment of inertia:

Iy=04x22xdx=204x52dx=227x7204=4727=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}.

The radius of gyration:

R=Iym=5127m=5127m,R=\sqrt{\frac{I_y}{m}}=\sqrt{\frac{\frac{512}{7}}{m}}=\sqrt{\frac{512}{7m}},

where m is mass of the area.


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