2D Motion of Rigid Bodies: Falling Stick Example, Work-Energy Principle

Example: Falling Stick 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 3/12/2007 Lecture 10 2D Motion of Rigid Bodies: Falling Stick Example, Work-Energy Principle Example: Falling Stick Figure 1: Falling Stick. Stick has mass, M and Length, L. Figure by MIT OCW. What is the equation of motion if the stick slides without friction along flat a surface? Application: Joint slipping and sliding. Uniform Rod: Center of Mass C. Kinematics Constraint: yB = 0 ⇒ 3 - 1 generalized coordinates So we choose xC , φ to be the generalized coordinates because body is rigid. xB is related to xC . Forces not asked for in problem. r C = xC êx + L sin φêy 2 Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Example: Falling Stick 2 L φ̇ cos φêy 2 L L r̈ C = ẍC êx + φ¨ cos φ − φ̇2 sin φêy 2 2 ṙC = ẋC êx + Kinetics Free Body Diagram Figure 2: Free Body Diagram of Falling Stick. Figure by MIT OCW. Which principles to apply? Want to avoid using N (1) Angular momentum about B (2) Linear momentum in x (No forces in x-direction) Could use work-energy principle, because N does no work. Linear Momentum in x-direction Fx = d P dt x 0 = mẍc Angular Momentum about B τB = d H + vB × P dt B Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Example: Falling Stick vB = 3 d d L xB êx = (xC − cos φ)êx dt dt 2 � � L v B = ẋC + φ̇ sin φ êx 2 � H B = H C + rC × P � � � � L L L = IC φ̇êz + cos φêx + sin φêy ×m ẋc êx + φ̇ cos φêy 2 2 2 L2 mL = IC φ̇êz + m cos2 φφ̇êz − ẋc sin φêz 4 2 �� � � L2 mL 2 ˙ = IC + m cos φ φ − ẋc sin φ êz 4 2 � � mL2 ˙ z � = IC + cos2 φ φê 4 Note: We cannot write in this problem that IB ω = H B . Point B is moving, not stationary. Ḣ B = �� � � mL2 mL mL mL2 2 2 ¨ IC + cos φ φ − · 2 cos φ sin φφ̇ − ẍC sin φ − ẋC φ̇ cos φ êz 4 4 2 2 � � � � L L v B × P = ẋC + φ̇ sin φ êx × m ẋC êx + φ̇ cos φêy 2 2 � � mL2 L 2 sin φ cos φφ̇ êz v B × P = ẋc m φ̇ cos φ + 2 4 L cos φêz 2 Notice all terms are in the êz or −êz direction. τ B = −mg Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Example: Falling Stick 4 Figure 3: Free Body Diagram of Falling Stick. Figure by MIT OCW. Therefore: � � mL2 mL mL2 2 IC + cos φ φ¨ − cos φ sin φφ̇2 − ẋC φ̇ cos φ 4 2 2 mL mL2 + ẋC φ̇ cos φ + sin φ cos φφ̇2 2 4 � � mL2 mL2 cos2 φ φ¨ − cos φ sin φφ̇2 = IC + 4 4 L −mg cos φ = 2 L −mg cos φ = 2 � � mL2 mL2 2 IC + cos φ φ¨ − cos φ sin φφ̇2 4 4 0 = mẍC Alternative Approach: Apply Angular Momentum About Point C, The Center of Mass We should get the same answer by applying angular momentum about C. d d ¨ z H C = Iω = IC φê dt dt L ¨ z −N cos φêz = IC φê 2 τC = IC φ¨ = −N L cos φ 2 Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Work-Energy Principle for Rigid Bodies 5 Figure 4: Free Body Diagram of Falling Stick. The approach employed here is one that uses angular momentum about C. Figure by MIT OCW. Simpler expression but must find N . Use linear momentum in y-direction. � � L¨ L 2 N − mg = m φ cos φ − φ̇ sin φ 2 2 From above: −2IC φ¨ L cos φ � � L L¨ φ cos φ − φ̇2 sin φ 2IC φ¨ + mgL cos φ = −ml cos φ 2 2 � � mL2 mL2 L cos2 φ φ¨ − cos φφ̇2 −mg cos φ = IC + 2 4 4 N= Notice the equations are the same. Work-Energy Principle for Rigid Bodies We need a kinetic energy expression. If you can show that all non-conservative external forces do no work, V + T = Constant. V is potential energy defined based on center of mass. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Work-Energy Principle for Rigid Bodies Figure 5: Rigid Body. Figure by MIT OCW. 1� mi v i · v i 2 i 1� mi (v C + ω × ρi ) · (v C + ω × ρi ) = 2 i � � 1� mi v C · v C + 2v C · (ω × ρi ) + (ω × ρi ) · (ω × ρi ) = 2 i � 1 1� mi ρ2i ω · ω m i ρi + = M vc2 + v C · ω × 2 2 i i T = � i mi ρi = 0 because of center of mass. T = 1 1 2 M vC + IC ω 2 2 2 Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6