Answer to Question #186301 in Differential Geometry | Topology for Jethro

Question

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

Answer 1

Solution.

$A=4•4=16.$

The moment of inertia:

$I_y=\int_0^4 x^22\sqrt{x}dx=2\int_0^4 x^{\frac{5}{2}}dx=2•\frac{2}{7}x^{\frac{7}{2}}|_0^4=\frac{4}{7}•2^7=73\frac{1}{7}.$

The radius of gyration:

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

where m is mass of the area.