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I o car = ((17,500 N / 9.81 m/s 2 ) / 12 ) ((1.80 m) 2 + (4.70 m) 2 ) =
3765.5 N − s 2 − m
I o driver = (m/12) (w 2 + L 2 )
I o driver = ((850 N / 9.81 m/s 2 ) / 12) ((0.50 m) 2 + (0.90 m) 2 ) = 7.7
N − s 2 − m
I o fuel = (m/12) (w 2 + L 2 )
I o fuel = ((993 N / 9.81 m/s 2 ) / 12) ((0.90 m) 2 + (0.50 m) 2 ) = 8.9 N
− s 2 − m
Since these are the moments of inertia of each component about its own neutral axis,
we now need to use the parallel axis theorem to transfer these moments to the neutral
axis of the body, which is located at the body center of gravity that we recently calculated.
To do this, we must find the distance from the body center of gravity to each component's
center of gravity. The distances squared from each component to the body center of
gravity are:
d 2 car = (x cg car − X cg ) 2 + (y cg car − Y cg ) 2
d 2 car = (30.50 m − 30.42 m) 2 + (30.50 m − 30.53 m) 2 = 0.01 m 2
d 2 driver = (x cg driver − X cg ) 2 + (y cg driver − Y cg ) 2
d 2 driver = (31.50 m − 30.42 m) 2 + (31.25 m − 30.53 m) 2 = 1.68
m 2
d 2 fuel = (x cg fuel − X cg ) 2 + (y cg fuel − Y cg ) 2
d 2 fuel = (28.00 m − 30.42 m) 2 + (30.50 m − 30.53 m) 2 = 5.86 m 2
Now we can apply the parallel axis theorem as follows:
I cg car = I o + md 2
I cg car = 3765.5 N − s 2 − m + (17,500 N / 9.81 m/s 2 ) (0.01 m 2 ) =
3783.34 N − s 2 − m
I cg driver = I o + md 2
I cg driver = 7.7 N − s 2 − m + (850 N / 9.81 m/s 2 ) (1.68 m 2 ) =
153.27 N − s 2 − m
I cg fuel = I o + md 2
I cg fuel = 8.9 N − s 2 − m + (993 N / 9.81 m/s 2 ) (5.86 m 2 ) =
602.07 N − s 2 − m
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