Answer:
Ix = Iy = [tex]\frac{ρπR^{4} }{16}[/tex]
Radius of gyration x = y = [tex]\frac{R}{4}[/tex]
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is [tex]\frac{MR^{2} }{2}[/tex]. Moment of inertia of quarter disk about its perpendicular is [tex]\frac{MR^{2} }{8}[/tex].
Now using perpendicular axis theorem, Ix = Iy = [tex]\frac{MR^{2} }{16}[/tex] = [tex]\frac{ρπR^{4} }{16}[/tex].
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.
