Work-Energy Principle

Dynamics of a Single Particle (Review) (continued) 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 2/12/2007 Lecture 2 Work-Energy Principle Dynamics of a Single Particle (Review) (continued) Reading: Williams 4–1, 4–2, 4–3 (Momentum Principles) - End of lecture last time Work-Energy Principle This material is covered in Williams 5–1, 5–2, and 5–3. Figure 1: Single particle moving. m moves along r(t) and experiences a force, F . Figure by MIT OCW. W12 =work done going from 1 → 2 Integral over sum = = = � � r2 F · dr = “Work done by force F(t)” (in going from r 1 to r 2 ) r1 t2 F · vdt (Integration over time) t1 � t2 t1 d (mv) · vdt where m is constant dt 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]. Dynamics of a Single Particle (Review) (continued) = = 2 t2 � � d 1 mv · v dt t1 dt 2 1 1 1 mv · v|tt21 = mv22 − mv12 2 2 2 � Define Kinetic Energy T = 12 m|v|2 ⇒ W12 = T2 − T1 = (Work done is the change in the kinetic energy) Consider the case: F = −∂V ∂r For example, F =( ⇒ W12 = � r2 F · dr = r1 � r2 r1 ∂V ∂V ı̂ − ĵ) ∂x ∂y −∂V · dr = V1 − V2 (Potential Energy) ∂r Thus, in this special case: T2 − T1 = V1 − V2 Or Conservation of Mechanical Energy T2 + V2 = T1 + V1 = Constant Forces which can be written F = called “Conservative Forces.” Examples: −∂V ∂r (i.e. as the gradient of a potential) are Gravity F = −mgĵ V = mgy ⇒ −∂V ∂V −∂V ı̂ − ĵ = −mgĵ = ∂r ∂x ∂y Linear Spring F = −kxı̂ V = −∂V ∂V 1 2 −∂V kx ⇒ ı̂ − ĵ = −kxı̂ = 2 ∂r ∂x ∂y Linear Dashpot: Frictional force that opposes the direciton of motion. De­ pends on velocity. EX: Slows motion down, door opener. 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]. Dynamics of a Single Particle (Review) (continued) 3 Figure 2: Free body diagram. The only force acting on the mass is the force due to gravity mg. Figure by MIT OCW. Figure 3: Mass attached to a spring. Figure by MIT OCW. Figure 4: Mass attached to linear dashpot. Figure by MIT OCW. 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]. Dynamics of a Single Particle (Review) (continued) 4 Figure 5: Vehicle falling on a curve. The vehicle with mass m starts at the top of the hill and falls along curve. Figure by MIT OCW. F = cẋı̂ �= −∂V ∂r Not conservative. m experiences same force as long as velocity is same. Position independent. The constant c meanst this is linear dashpot behavior. If all forces are potentiall or the ones that are not do work (e.g. always perpendicular to motion) we call the particle motion ”conservative” because mechanical energy (kinetic + potential) is conserved. Discussion: F = F conservative + F non−conservative Using conservation of mechanical energy can be useful 1. Scalar Equation 2. Need not account for forces that do no work Example 1: Vehicle Falling on a Curve A small vehicle is released from rest at the top of a circular path. Determine the angle θ0 to the position where the vehicle leaves the path and becomes a projectile. (Neglect friction and treat vehicle as a particle.) Problem relates to driving a car down a hill. Whenever you solve a problem, make sure that you draw a good diagram. 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]. Dynamics of a Single Particle (Review) (continued) 5 Figure 6: Free body diagram of vehicle falling along curve. The two forces on the vehicle are the normal force, N , and the force due to gravity mg. Figure by MIT OCW. When does this car become free? When N is zero, the vehicle has left the surface. How does the point of departure (θ) depend on R or m? Free Body Diagram (FBD) ên : Unit vector normal êt : Unit vector tangential r = R sin θı̂ + R cos θĵ ṙ = Rθ̇(cos θı̂ − sin θĵ) = Rθ̇êt ṗ = mv̇ = mRθ̈(cos θı̂ − sin θĵ) + mRθ̇2 (− sin θı̂ − cos θĵ) = mRθ̈êt − mRθ̇2 ên � F = −mgĵ + N ên = (N − mg cos θ)ên + mg cos θêt From Newton II: N − mg cos θ = −mRθ̇2 mg sin θ = mRθ¨ � F = ṗ Normal Tangential (1) (2) We have: −mg cos θ + N = −mRθ̇2 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]. Dynamics of a Single Particle (Review) (continued) 6 Vehicle will leave track when N = 0 −mg cos θ0 = −mRθ̇2 Rθ̇2 = g cos θ0 (This is true only when N = 0. We want to know θ̇2 = f (θ)?) Could use Equation (2) to get information about θ̇2 by integration, but instead we will use conservation of energy to get a relation θ̇2 = f (θ) Forces: Gravity - Conservative No Friction Normal Force: N is always ⊥ to motion. Does no work. We have a conservative system. Apply conservation of mechanical energy. TT OP + VT OP = Tθ + Vθ |v| = Rθ̇ 1 mR2 θ̇2 + mgR cos θ 2 2g θ̇2 = (1 − cos θ) R 0 + mgR = Putting into (1) and letting N = 0, Rθ̇2 = R 2Rg (1 − cos θ) = g cos θ0 . 2(1 − cos θ0 ) = cos θ0 2 3 θ0 = 48.2◦ cos θ0 = Note: This solution is independent of R and m. Linear momentum principle used to solve problem. Next time, use angular momentum principle to solve a problem. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. 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