ME 242 Lecture 29 04.15

ME 242 Lecture 29 04.15 - M G = I G b 3 M A = I G b 3 + r G...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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 + r B / A + ( 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 + r B / A + ( 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...
View Full Document

Page1 / 4

ME 242 Lecture 29 04.15 - M G = I G b 3 M A = I G b 3 + r G...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online