Answers to the Exercises - 3D Math Primer for Graphics and Game Development - page 790

12. (a) First we determine the total mass, which is 2200 kg. Then we take a weighted average

of the mass centers according to Equation (12.22).

0
@ 1000(0, 100, 225) + 600(0, 125, 75)

1
A

n

1
M

1
2200

+ 400(0, 100,−120) + 50(−100, 35, 230)
+ 50(100, 35, 230) + 50(−100, 35,−150)
+ 50(100, 35,−150)

r c = m i r I =

i

= (0, 101, 105)

(b) At the time of this writing, all the formulas are available on the Wikipedia article
List of moment of inertia tensors.
Body front:

2
4 (0.80 m) 2 + (1.50 m) 2

3
5

0 0

1000 kg
12

(2.00 m) 2 + (1.50 m) 2

J =

0 0

(2.00 m) 2 + (0.80 m) 2

0 0

2
4 241

3
5

0 0

(kg m 2 ) = 0 521 0

0 0 387

Body middle:

2
4 (1.30 m) 2 + (1.50 m) 2

3
5

0 0

600 kg
12

(2.00 m) 2 + (1.50 m) 2

J =

0 0

(2.00 m) 2 + (1.30 m) 2

0 0

2
4 197

3
5

0 0

(kg m 2 ) = 0 313 0

0 0 285

Body rear:

2
4 (0.80 m) 2 + (2.40 m) 2

3
5

0 0

400 kg
12

(2.00 m) 2 + (2.40 m) 2

J =

0 0

(2.00 m) 2 + (0.80 m) 2 0 0

2
4 213

3
5

0 0

(kg m 2 ) = 0 325 0

0 0 155

Each wheel:

2
4 2 (0.35 m) 2

3
5

0 0

12 (3(0.35 m) 2 + (0.20 m) 2 )

J = (50 kg) 0

1

0

1
12 (3(0.35 m) 2 + (0.20 m) 2 )

0 0

2
4 3.06

3
5

0 0

(kg m 2 ) = 0 1.70 0

0 0 1.70

(c) We apply the parallel axis theorem (Equation (12.31)) to each part. We must first
compute the position of each part relative to the center of mass of the truck, which
we denote as r ′ .