Consider the pendulum shown in the figure. Calculate: y r₁" A A' 0.5h" B B' h" t" √2" C C' x r1 = 0.5 in r2 = 1.2 in h = 4 in t = 0.3 in The moment of inertia about the y axis. 3 tries remaining. 1 point(s) possible in 4 ENTER The moment of inertia about the x axis. 3 tries remaining. 1 point(s) possible in 4 ENTER The moment of inertia about the AA' axis. 3 tries remaining. 1 point(s) possible in 4 ENTER The moment of inertia about the BB' axis. 3 tries remaining. 1 point(s) possible in4 ENTER The moment of inertia about the CC' axis. in 4 ENTER 3 tries remaining. 1 point(s) possible The location of the x axis such that the moment of inertia around the x axis will be minimized. in ENTER 3 tries remaining. 1 point(s) possible Moment of Inertia IAA == = [(1) * ] + [12² + Ay³] + [(2)² + Ay² ] π(0.5)4 + 1+6+0.5)²] * [0.2(6)² + (0.2 × 6) (+ 0.5)³] + [(1.1)* + ˜ (1.1)²(1.1 + 6 12 0.04918.3 + 220.715 = 239.064 in.4 IBB = = = [π(r1)4 [(r₁)* + Ay³] + [th²] + [(2)² + Ay²] 4 [■(0.5)* 4 [π(1,1)* + π (0.5)² ({ + 0.5)³] + [02(6)³] + [™ (1.1)* + m (1.1)² ({ +1.1)*] 9.67 +3.6+65.05 12 = 78.32 in.4 Icc = [(1) + Ay³] + [+ Ay³] + [*] 4 + 1.1)²] + [0.2(6)² + (0.2 × 6) (+ 1.1)³] + [™(1,1)* ] + (0.5)2(0.5+ 6+1.1)2 45.414+23.772 + 1.15 == 12 70.336 in.4