Transcript File
Ch 17 Notes – Part 1
Mass Moment of Inertia
• I = r^2 dm, over m Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Mass Moment of Inertia
• • • I = r^2 dV, over V I = r^2 dV, over V, constant rho Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Mass Moment of Inertia
To evaluate, use either the shell method or the disk method For shell: dV = (2 y)(z)dy Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Mass Moment of Inertia
To evaluate, use either the shell method or the disk method For disk: dV = ( y) 2 dz Here, we have to find (d Iz) first and then integrate that Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Example 17.1
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Example 17.1 (continued)
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Example 17.2
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Example 17.2 (continued)
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Ch 17 Notes – Part 1
• •
Parallel Axis Theorem
If the moment of inertia about an axis passing through the mass center of a body is known, we can find the moment of inertia of that body about any other parallel axis I z + I G + md 2 Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Radius of Gyration
The radius of gyration, k, is sometimes given. Its units are in length I = mk 2 k = sqrt(I/m) Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Composite Bodies
Sometimes you can break a complex body up into simple shapes. Find their moments of inertia and add them algebraically.
Use the parallel axis theorem to find their moments of inertia about a common axis.
Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Example 17.3
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Example 17.3 (continued)
Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Example 17.3 (continued)
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Example 17.4
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Example 17.4 (continued)
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Example 17.4 (continued)
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Ch 17 Notes – Part 1
Planar Kinetic Equations of Motion
F
F
x = m
a
G = m
a
Gx
F
y M G M P = m
a
Gy = I G = (
M
k ) P Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Equations of Motion: Translation
Rectilinear:
F
x
F
y M G = m( = 0
a
G ) = m(
a
G ) y x Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Ch 17 Notes – Part 1
Equations of Motion: Translation
Curvilinear:
F
n
Ft
M G = m( = 0
a
G )n = m(
a
G )t Engineering Mechanics: Dynamics, Thirteenth Edition R. C. Hibbeler Copyright ©2013 by Pearson Education, Inc.
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Example 17.5
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Example 17.5 (continued)
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Example 17.5 (continued)
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Example 17.6
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Example 17.6 (continued)
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Example 17.7
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Example 17.7 (continued)
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Example 17.7 (continued)
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Example 17.8
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Example 17.8 (continued)
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