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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 ˙ θ 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 + ω × 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? Can’t use the relative velocity ICR not useful for “open links” v C = v B + ω 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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