Page 1Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 1Lecture 15Goals:Goals:••Chapter 11Chapter 11 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 rotation about a fixed axis. Employ “conservation of angular momentum” conceptAssignment: HW7 due March 25th 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 2More Work: “2-D” Example (constant force) (Net) Work is Fx∆∆∆∆x= F cos(= F cos(-45°) ∆∆∆∆x= 50 x 0.71 Nm = 35 J= 50 x 0.71 Nm = 35 J Notice that work reflects energy transfer∆∆∆∆xF An angled force, F = 10 N, pushes a box across a frictionless floor for a distance ∆x = 5 m and ∆y = 0 m θ = -45°StartFinishFxPage 2Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 3Net Work: 1-D 2ndExample (constant force) Net Work is F ∆∆∆∆x= = --10 x 5 N m = 10 x 5 N m = --50 J50 J Work reflects energy transfer∆∆∆∆xF A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance ∆x = 5 m.θ = 180°StartFinishPhysics 207: Lecture 15, Pg 4Work in 3D….(assigning U to be external to the system)221221)(zizfzifzmvmvzFzzF −=∆=− x, y and z with constant F:221221)(yiyfyifymvmvyFyyF −=∆=−221221)(xixfxifxmvmvxFxxF −=∆=−2222221221 with zyxifzyxvvvvKmvmvzFyFxF++=∆=−=+∆+∆+∆Page 3Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 5 Useful for performing projections.A •••• î = Axî •••• î = 1î •••• j = 0îAAxAyθA tool: Scalar Product (or Dot Product)Calculation also in terms of magnitudes and relative angles.A • B = (Ax)(Bx) + (Ay)(By) + (Az)(Bz) Calculation can be made in terms of components.A • B | A | | B | cos θYou choose the way that works best for you!A · B |A| |B| cos(θ)Physics 207: Lecture 15, Pg 6Scalar 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 WorkPage 4Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 7Definition of Work, The basicsIngredients: Force ( F ), displacement ( ∆∆∆∆ r )“Scalar or Dot Product”θθθθ∆∆∆∆ rdisplacementFWork, W, of a constant force Facts through a displacement ∆∆∆∆ r :W = F ·∆∆∆∆ r(Work is a scalar)(Work is a scalar)If we know the angle the force makes with the path, the dot product gives us F cosθand ∆rIf the path is curved at each pointand rdFdWrr⋅=rdFrr⋅∫⋅=firrrdFWrrrrPhysics 207: Lecture 15, Pg 8Remember 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=+Page 5Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 9Definition of Work, The basicsIngredients: Force ( F ), displacement ( ∆∆∆∆ r )θθθθ∆∆∆∆rrdisplacementFFWork, W, of a constant force Facts through a displacement ∆∆∆∆ r :W = F ·∆∆∆∆ r(Work is a scalar)(Work is a scalar)Work tells you something about what happened on the path!Did something do work on you? Did you do work on something?If only one force acting: Did your speed change? Physics 207: Lecture 15, Pg 10ExerciseWork 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 6Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 11Exercise Work in the presence of friction and non-contact forces A box is pulled up a rough (µ > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces are doing work on the box ? And which are positive (T) and which are negative (f, mg)?(For mg only the component along the surface is relevant) Use a Free Body Diagram (A) 2 (B) 3 is correct(C) 4(D) 5vfmgNTPhysics 207: Lecture 15, Pg 12Work and Varying Forces (1D) Consider a varying force F(x)Fxx∆xArea = Fx∆xF is increasingHere W = F ·∆∆∆∆ rbecomes dW = F dxFθ = 0°StartFinishWork has units of energy and is a scalar!∫=fixxdxxFW )(F∆xPage 7Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 13• 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 and xiis the equilibrium position of the spring?∆xvimtiFspring compressedspring at an equilibrium positionV=0tm∫=fixxdxxFW )(box∫−=fiixxdxxxkW )(-boxfiixxxxkW |221box )( - −=K 0 )( -221221box ∆=+−= kxxkifW2i212 21221v0 - mmxk−=∆Example: Hooke’s Law Spring (xiequilibrium)Physics 207: Lecture 15, Pg 14• How much will the spring compress 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 and xeis the equilibrium position of the spring? (More difficult)∆xvimtiFspring compressedspring not at an equilibrium positionV=0tm∫=fixxdxxFW )(box∫−=fixxedxxxkW )(-boxfiexxxxkW |221box )( - −=K )( )( -221221box∆=−+−=eiefxxkxxkWExample: Hooke’s Law SpringPage 8Physics 207 – Lecture 12Physics 207: Lecture 15, Pg 15Work 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 16Conservative 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
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