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2. Strategy
Use Eq. (82) to find the rotational inertia. Use Eq. (79) for the center of mass.
Solution
(a)
22
2
(200 g)(5.0 cm)
(300 g)(0 cm)
(500 g)(4.0 cm)
13,000 g cm
C
xi
i
iA
Im
r
=
==
+
+
=
⋅
∑
(b)
2222
2
(200 g)(3.0 cm)
(300 g)(6.0 cm)
(500 g)(5.0 cm)
25,000 g cm
C
yi
i
r
=
+
+
=
⋅
∑
(c)
222
2
2
(200 g)[(3.0 cm)
(5.0 cm) ] (300 g)(6.0 cm)
(500 g)[(5.0 cm)
(4.0 cm) ]
38,000 g cm
C
zi
i
r
=
+
+
+
+
∑
=⋅
(d)
CM
(200 g)( 3.0 cm)
(300 g)(6.0 cm)
(500 g)( 5.0 cm)
1.3 cm
200 g
300 g
500 g
x
−+
+
−
−
++
CM
(200 g)(5.0 cm)
(300 g)(0 cm)
(500 g)( 4.0 cm)
1.0 cm
200 g
300 g
500 g
y
−
−
13. Strategy
Use Eq. (83) to compute the torque in each case.
Solution
(a)
The force is applied perpendicularly to the door, so
(1.26 m)(46.4 N)
58.5 N m .
rF
τ
=
⋅
(b)
The force is applied at 43.0
°
from the door’s surface, so
sin
(1.26 m)(46.4 N)sin 43.0
39.9 N m .
rF
rF
τθ
⊥
=
°
=
⋅
(c)
Since the force is applied such that its line of action passes through the axis of the door hinges—
the axis of rotation—there is no perpendicular component of the force and the torque is
0.
17.
Strategy
Use Eq. (83).
Solution
Find the magnitude of the torque.
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 Winter '08
 REHSE
 Physics, Center Of Mass, Inertia, Mass

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