Page 1Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 1Lecture 15Goals:Goals:••Chapter 11Chapter 11 Employ the dot product Employ conservative and non-conservative forces Use the concept of power (i.e., energy per time)••Chapter 12Chapter 12 Extend the particle model to rigid-bodies Understand the equilibrium of an extended object. Understand rigid object rotation about a fixed axis. Employ “conservation of angular momentum” conceptAssignment: HW7 due March 10th For Thursday: Read Chapter 12, Sections 7-11do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia””Physics 207: Lecture 15, Pg 2 Useful for finding parallel componentsA •••• î = Axî •••• î = 1î •••• = 0A • B = (Ax)(Bx) + (Ay)(By) + (Az)(Bz) Calculation can be made in terms of components.Calculation also in terms of magnitudes and relative angles.Scalar Product (or Dot Product)A • B | A | | B | cos θYou choose the way that works best for you!θcosBABArrrr≡•îAAxAyθPage 2Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 3Scalar Product (or Dot Product)Compare:A • B = (Ax)(Bx) + (Ay)(By) + (Az)(Bz)Redefine A F (force), B ∆r (displacement)Notice:F • ∆r = (Fx)(∆x) + (Fy)(∆z ) + (Fz)(∆z)So hereF • ∆r = WMore generally a Force acting over a Distance does Work Physics 207: Lecture 15, Pg 4Work in terms of the dot productIngredients: Force ( F ), displacement ( ∆∆∆∆ r )Looks just like a Dot Product!θθθθ∆∆∆∆ rdisplacementFWork, W, of a constant force Facts through a displacement ∆∆∆∆ r :If the path is curved at each pointand rdFdWrr⋅=rdFrr⋅∫⋅=firrrdFWrrrrrFrFWrrrr∆=∆= •θcosPage 3Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 5Remember that a real trajectory implies forces acting on an object Only tangential forces yield work! The distance over which FTangis applied: Workavpathand timeaaa= 0Two possible options:Change in the magnitude of vChange in the direction of va= 0a= 0aaa=+aradialatanga=+FradiallFtangF=+Physics 207: Lecture 15, Pg 6Energy and WorkWork, W, is the process of energy transfer in which a force component parallel to the path acts over a distance; individually it effects a change in energy of the “system”.1.K or Kinetic Energy 2.U or Potential Energy (Conservative)and if there are losses (e.g., friction, non-conservative)3. EThThermal EnergyPositive W if energy transferred to a systemPage 4Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 7A. U KB. U EThC. K UD. K EThE. There is no transformation because energy is conserved.A child slides down a playground slide at constant speed. The energy transformation isPhysics 207: Lecture 15, Pg 8ExerciseWork in the presence of friction and non-contact forcesA. 2B. 3C. 4D. 5 A box is pulled up a rough (µ > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces (including non-contact ones) are doing work on the box ? Of these which are positive and which are negative? State the system (here, just the box) Use a Free Body Diagram Compare force and path vPage 5Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 9Work and Varying Forces (1D) Consider a varying force F(x)Fxx∆xArea = Fx∆xF is increasingHere W = F ·∆∆∆∆ rbecomes dW = FxdxFθ = 0°StartFinishWork has units of energy and is a scalar!∫=fixxxdxxFW )(F∆xPhysics 207: Lecture 15, Pg 10• How much will the spring compress (i.e. ∆x = xf- xi) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vi) on frictionless surface as shown below with xi= xeq, the equilibrium position of the spring?∆xvimtiFspring compressedspring at an equilibrium positionV=0tm∫=fixxxdxxFW )(box∫−=fixxeqdxxxkW )(-boxfiixxxxkW |221box )( - −=K 0 )( -221221box ∆=+−= kxxkifW2i212 21221v0 - mmxk−=∆Example: Hooke’s Law Spring (xiequilibrium)Page 6Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 11Work signs∆xvimtiFspring compressedspring at an equilibrium positionV=0tmNotice that the spring force is opposite the displacementFor the mass m, work is negativeFor the spring, work is positive They are opposite, and equal (spring is conservative) Physics 207: Lecture 15, Pg 12Conservative Forces & Potential Energy For any conservative force F we can define a potential energy function U in the following way:The work done by a conservative force is equal and opposite to the change in the potential energy function.W = F ·dr- ∆U ∫rirfUfUiPage 7Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 13Conservative Forces and Potential Energy So we can also describe work and changes in potential energy (for conservative forces)∆U = - W Recalling (if 1D)W = Fx∆x Combining these two,∆U = - Fx∆x Letting small quantities go to infinitesimals,dU = - Fxdx Or,Fx= -dU / dxPhysics 207: Lecture 15, Pg 14ExerciseWork Done by Gravity An frictionless track is at an angle of 30°with respect to the horizontal. A cart (mass 1 kg) is released from rest. It slides 1 meter downwards along the track bounces and then slides upwards to its original position. How much total work is done by gravity on the cart when it reaches its original position? (g = 10 m/s2)1 meter30°(A) 5 J (B) 10 J (C) 20 J (D) 0 J h = 1 m sin 30°= 0.5 mPage 8Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 15Home Exercise: Work & Friction Two blocks having mass m1and m2where m1> m2. They are sliding on a frictionless floor and have the same kinetic energywhen they encounter a long rough stretch (i.e. µ > 0) which slows them down to a stop. Which one will go farther before stopping? Hint: How much work does friction do on each block ?(A)(A) m1(B)(B) m2(C)(C) They will go the same distancem1m2v2v1Physics 207: Lecture 15, Pg 16Exercise: Work & Friction W = F d = - µ N d = - µ mg d = ∆K = 0 – ½ mv2 - µ m1g d1= - µ m2g d2 d1 / d2 = m2 / m1 (A)(A) m1(B) (B) m2(C)(C) They will go the same distancem1m2v2v1Page 9Physics 207 – Lecture 15Physics 207: Lecture 15, Pg 17Home Exercise Work/Energy for Non-Conservative Forces The air track is once again at an angle of 30°with resp ect to horizontal. The cart (with mass 1 kg) is released 1 meterfrom the bottom and hits the bumper at a speed, v1. This time the vacuum/ air generator breaks half-way through
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