ME 242 Lecture 24 04.01

ME 242 Lecture 24 04.01 - Know the velocity vectors...

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1 ME 242 Dynamics April 1, 2009 Eric Wang Today’s Key Concepts Chapter 6: Kinematics Rigid Bodies Relative velocity example Instantaneous center of rotation ICR Example Example (prob. 6.1.26) Given: θ = 90°, = 5 rad/sec, β = 45° and cm, cm Find: both and angular velocity of triangle v B ˙ r B / E = 2 j r B / A = 5 i Start with easy part: Deal with the Unknowns… Using Link DA b 2 b 1 cos β cos β -sin β sin β b 2 j i b 1
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2 Using the Triangle b 2 b 1 Solve for the unknowns b 2 b 1 Instantaneous Center of Rotation (ICR) For 2D general motion: • There is always exists a point that has zero velocity • This point is called the ICR • The ICR is often not on the rigid body • The ICR can move over time Simple Case: • O is a fixed point and is the ICR • Velocity vectors are perpendicular to position vector v B = v O + ω × r B / O Less Obvious Case: v B = v ICR + × r B / ICR ICR
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Unformatted text preview: Know the velocity vectors Position vectors must be perpendicular to velocity vectors Find ICR at intersection r B / ICR ICR is not stationary ICR 3 Where is the ICR? ICR Where is the ICR? ICR Re-solve previous example ICR 5 5 = 45 Re-solve previous example ICR = 45 True or False ICR for link BC is shown in the correct position? Cant use the relative velocity ICR not useful for open links v C = v B + BC r C / B 4 True or False The ICR is shown in the correct position? True or False ICR The ICR is shown in the correct position? Take Home Points For 2D rigid body motion, there always exists and Instantaneous Center of Rotation (ICR) that has zero velocity The ICR may be outside of the rigid body The ICR will change location over time...
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This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.

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ME 242 Lecture 24 04.01 - Know the velocity vectors...

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