Can someone please help me solve this? I previously entered 6.46 kg*m^2 and it w

Question

Can someone please help me solve this? I previously entered 6.46 kg*m^2 and it was incorrect. thank you!

Three small spherical masses are located in a plane at the positions shown below. -1 -2 -4 -5-4- -2 -1 0 1 23 4 5 x (m) The masses are Q-0.500 kg, R-0.300 kg, and S-0.600 kg. Calculate the moment of inertia (of the 3 masses) with respect to an axis perpendicular to the xy plane and passing through X-0 and y--1. [Since the masses are of small size, you can neglect the contribution due to moments of inertia about their centers of mass.]

Explanation / Answer

Q = 0.5 kg (x1 , y1) = ( 4, 2)

R = 0.3 kg (x2 , y2) = (-5 , 0)

S = 0.6 kg (x3 , y3) = (3 , -1)

location of axis P = (x , y) = (0 -1)

distance from Q to p , r1 = sqrt((x-x1)^2 + (y-y1)^2)

r1 = sqrt((0-4)^2+(-1-2)^2) = 5 m

distance from R to p , r2 = sqrt((x-x1)^2 + (y-y1)^2)

r2 = sqrt((0+5)^2+(-1-0)^2) = 5.1 m

distance from s to p , r3 = sqrt((x-x3)^2 + (y-y3)^2)

r3 = sqrt((0-3)^2+(-1+1)^2) = 3 m

moment of inertia

I = Q*r1^2 + R*r2^2 + S*r3^2

I = (0.5*5^2) + (0.3*5.1^2) + (0.6*3^2)

I = 25.703 kg m^2 <<<<<<___________ANSWER

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