ME 242 Lecture 35 05.01

ME 242 Lecture 35 - Useful information KE = 1 2 mv G 2 1 2 I G ω 2 I G = mr 2 2 I G ≈ mr 2 Work-Energy Analysis KE 1 PE 1 W nc 1 − 2 = KE 2 PE

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1 ME 242 Dynamics May 1, 2009 Eric Wang Announcements • Final Exam: – One week from Friday – May 8 12-2pm – 2 pages of cheat sheets – Comprehensive – 3 “complete” problems – Review on Monday Course Evaluations • Complete ME242 course evaluations on WebCT Today’s Key Concepts Non-Intuitive Behavior of Rigid Bodies Rolling cylinders Gyroscopes In class problem • Steel vs. aluminum cylinder • Which one gets to the bottom first? Useful information KE = 1 2 mv G 2 + 1 2 I G ω 2 I G = mr 2 2
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2 Work-Energy analysis KE 1 +PE 1 + W nc 1 2 = KE 2 +PE 2 Another In class problem • Large aluminum cylinder vs. tiny aluminum tube • Which one gets to the bottom first?
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Unformatted text preview: Useful information KE = 1 2 mv G 2 + 1 2 I G ω 2 I G = mr 2 2 I G ≈ mr 2 Work-Energy Analysis KE 1 +PE 1 + W nc 1 − 2 = KE 2 +PE 2 Compare velocities Gyroscopes ( H G ) 2 = M G Δ t + ( H G ) 1 M G dt t 1 t 2 ∫ = ( H G ) 2 − ( H G ) 1 ( H G ) 1 ( H G ) 2 M G Δ t X Y Z Before After 3 What happens for Y-moment? ( H G ) 2 = M G Δ t + ( H G ) 1 M G dt t 1 t 2 ∫ = ( H G ) 2 − ( H G ) 1 ( H G ) 1 ( H G ) 2 M G Δ t X Y Z Before After Inertial Guidance Systems Put our gyroscope inside a 3-axis gimbal Impossible to apply a moment to the gyroscope...
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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 35 - Useful information KE = 1 2 mv G 2 1 2 I G ω 2 I G = mr 2 2 I G ≈ mr 2 Work-Energy Analysis KE 1 PE 1 W nc 1 − 2 = KE 2 PE

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