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element is a distance from the z axis, the moment of inertia about the z axis is However, we can relate the coordinates x, y of the mass element dm to the coordinates of CM , y CM in the original coordinate system centered on O, then from Figure 10.12a we see that the relationships between the CM and y " y1 ) y CM . Therefore, The first integral is, by definition, the moment of inertia about an axis that is parallel . The last integral is sim- ply MD 2 because and D 2 " x CM 2 ) y CM 2 . Therefore, we conclude that I " I CM ) MD
2 &dm " M &x1dm " &y1dm " 0 " &
[(x1) 2 ) (y1)
2 ]dm ) 2x CM
&
x1dm ) 2y CM
&
y1dm ) (x CM
2 ) y CM
2 ) & dm I " &
[(x1 ) x
CM ) 2 ) (y1 ) y CM ) 2 ]dm I " &
r
2 dm " &
(x
2 ) y 2 )dm r " √ x 2 ) y 2 SECTION 10.5 • Calculation of Moments of Inertia 305 Example 10.8 Applying the Parallel-Axis Theorem Consider once again the uniform rigid rod of mass M and Solution Intuitively, we expect the moment of inertia to be because there is mass up to a dis- tance of L away from the rotation axis, while the farthest dis- I
CM " 1 12
ML
2 between the center-of-mass axis and the y1 axis is D " L/2, So, it is four times more difficult to change the rotation of a 1 3
ML 2 I " I CM ) MD 2 " 1 12
ML 2 ) M
# L 2 $ 2 " (a) y x, y dm y ′ y CM O D r y x CM x x CM, y CM x ′ x CM (b) Axis through CM x y z Rotation axis O CM Figure 10.12 (a) The parallel-axis theorem: if the moment of inertia about an axis perpendicular to the figure through the center of mass is I CM , then the moment of inertia about the z axis is I z " I CM ) MD 2 . (b) Perspective drawing showing the z axis (the axis of rotation) and the parallel axis through the CM. |