12.15. Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
Solve: (a) xcm
mi xi
mi
mA xA mB xB
mA mB
mC xC mD xD
mC mD
(100 g)(0 m) (200 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 m)
100 g 200 g 200 g 200 g
ycm
mA yA mB yB
mA mB
mC yC mD yD
mC mD
(100 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 cm) (200 g )(0 m)
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
I BD
where rA
rC
0.057 m
0.057 m
mArA2 mCrC2
7.07 cm and are the distances from the diagonal. Thus,
(0.10 m)cos 45
I BD
(0.100 kg)rA2 (0.200 kg)rC2
0.0015 kg m2
Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
Solve: (a) The hinges are at the edge of the door, so from Table 12.2,
1
2
I
25 kg 0.91 m
6.9 kg m2
3
(b) The distance from the axis through the center of mass along the height of the door is
0.91 m
d
0.15 m 0.305 m. Using the parallel–axis theorem,
2
1
2
2
25 kg 0.91 m
25 kg 0.305 cm
4.1 kg m 2
12
Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
I
I cm
Md 2
12.55. Model: The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve: (a) From Table 12.2, the moment of inertia of a disk about its center is
1
1
MR 2
(2.0 kg)(0.10 m) 2 0.010 kg m 2
2
2
(b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I
I
I center
Mh2
(0.010 kg m2 ) (2.0 kg)(0.10 m) 2
0.030 kg m2
Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the
edge than about the center.