ME 242 Lecture 29 04.15

# ME 242 Lecture 29 04.15 - ∑ M G = I G ˙ ˙ b 3 ∑ M A =...

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1 ME 242 Dynamics April 15, 2009 Eric Wang Today’s Key Concepts Chapter 7: Kinetics of Rigid Bodies Moment of inertia example 2D rotation example General 2D Motion Example of 2D general motion Use Inertia Tables When Possible Example (prob. 7.2.8) • Given: lineal density = ρ (kg/m) • Find: I G and I A Appendix B B C D E F Parallel Axis Theorem B C D E F

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2 Put it all together B C D E F True or False I A = I B + mr B / A 2 G A B I A = I G + mr G / A 2 I B = I G + mr G / B 2 I G = I B mr G / B 2 I A = I B mr G / B 2 + mr G / A 2 Types of Motion Translation Curvilinear Translation Rotation about General Motion fixed point Example (prob. 7.2.45) Given: uniform disk spinning on flat surface with friction μ . Initial angular velocity = ω o Find: length of time before stopping FBD=IRD mg ma r N ma z I G ˙ ˙ θ T f ma Differential Approach dr r R
3 Finish it up Equations of Motion • For General 2D Motion: F = ma G M G = I G ˙ ˙ θ b 3 F x = m ˙ ˙ x G F y = m ˙ ˙ y G Same as before (these equations always true for 2D motion) General Motion • 3 Forms of the Moment Equation:

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Unformatted text preview: ∑ M G = I G ˙ ˙ b 3 ∑ M A = I G ˙ ˙ b 3 + r G / A × ma G ∑ M A = I A ˙ ˙ b 3 + r G / A × ma A Rigid Body General 2D Motion Kinematics Kinetics a B = a A + α × r B / A + ω × ( × r B / A ) v B = v A + × r B / A ∑ F = ma G ∑ M G = I G ˙ ˙ b 3 ∑ M A = I G ˙ ˙ b 3 + r G / A × ma G ∑ M A = I A ˙ ˙ b 3 + r G / A × ma A Example (prob. 7.3.2) Given: k = 10 lb/ft r 1 = 0.5 ft weight = 15 lbs Spring is stretched 0.2 ft Find: angular acceleration at time of release FBD=IRD mg T ma x F sp ma y I ˙ ˙ 4 Force Equation: B Solve for unknowns: B Plug in givens: Take Home Points Rigid Body Kinematics Rigid Body Kinetics a B = a A + α × r B / A + ω × ( × r B / A ) v B = v A + × r B / A ∑ F = ma G ∑ M G = I G ˙ ˙ θ b 3 ∑ M A = I G ˙ ˙ b 3 + r G / A × ma G ∑ M A = I A ˙ ˙ b 3 + r G / A × ma A...
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ME 242 Lecture 29 04.15 - ∑ M G = I G ˙ ˙ b 3 ∑ M A =...

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