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Physics 207 – Lecture 16 Lecture 16 Goals: • Chapter 11 (Work) Employ conservative and non-conservative forces Relate force to potential energy Use the concept of power (i.e., energy per time) • Chapter 12 Define rotational inertia (“mass”) Define rotational kinetic energy Define center of mass Assignment: HW7 due Tuesday, Nov. 2nd For Wednesday: Read Chapter 12, Sections 1-3, 5 & 6 Exam 2 7:15 PM Thursday, Nov. 4th Physics 207: Lecture 16, Pg 3 Definition of Work, The basics Ingredients: Force ( F ), displacement ( ∆ r ) Work, W, of a constant force F acts through a displacement ∆ r : W = F ·∆ ∆r F θ ∆r (Work is a scalar) “Scalar or Dot Product” r r F ⋅ dr m ce a l p dis en If we know the angle the force makes with the path, the dot product gives us F cos θ and ∆r ∆W = r r v r F ⋅ ∆r = F ∆r cos θ Physics 207: Lecture 16, Pg 4 Page 1 t Physics 207 – Lecture 16 Exercise Work in the presence of friction and non-contact forces A box is pulled up a rough (µ > 0) incline by a rope-pulleyweight 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 v A. 2 B. 3 C. 4 D. 5 Physics 207: Lecture 16, Pg 5 Work with varying forces or curved paths If the path is curved or if the force varies with position then at each point dW = r r F ⋅ dr and W = r rf r r ∫r F ⋅ d r ri Physics 207: Lecture 16, Pg 7 Page 2 Physics 207 – Lecture 16 Work and Varying Forces (1D) Area = Fx ∆x F is increasing Here W = F ·∆ ∆r becomes dW = F dx Consider a varying force F(x) Fx x ∆x xf W = ∫ F ( x ) dx xi Physics 207: Lecture 16, Pg 8 Example: Spring-Mass energy transfer • How much will the spring compress (u = x – xequilibrium) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? uf Wbox to vo ui uf m Wbox spring at an equilibrium position ∆u = ∫ − ku du ui = - 12 ku 2 |uf ui 1 2 Wbox = - 2 k ∆u = ∆K Wbox V=0 t = ∫ F (u ) du F m spring compressed - 12 k ∆u 2 = 12 m0 2 − 12 mv 02 Physics 207: Lecture 16, Pg 10 Page 3 Physics 207 – Lecture 16 Relate work to potential energy Consider the ball moving up to height h (from time 1 to time 2) How does this relate to the potential energy? Work done by the Earth’s gravity on the ball) W = F • ∆x = -mg h mg ∆U = Uf – Ui = mg h - mg 0 = mg h h ∆U = -W This is a general result for all conservative forces (also path independent) mg Physics 207: Lecture 16, Pg 11 Conservative Forces & Potential Energy For any conservative force F we can define a potential energy function U in the following way: W = ∫ F ·dr = - ∆U The work done by a conservative force is equal and opposite to the change in the potential energy function. r U f ri f Ui Physics 207: Lecture 16, Pg 12 Page 4 Physics 207 – Lecture 16 Conservative Forces and Potential Energy So we can also describe work and changes in potential energy (for conservative forces) ∆U = - W but in 1D W = Fx ∆x Combining these two, ∆U = - Fx ∆x Letting small quantities go to infinitesimals, dU = - Fx dx Or, Fx = -dU / dx Physics 207: Lecture 16, Pg 13 A Non-Conservative Force, Friction Looking down on an air-hockey table with no air flowing (µ > 0). Now compare two paths in which the puck starts out with the same speed (Ki path 1 = Ki path 2) . Path 2 Path 1 Physics 207: Lecture 16, Pg 15 Page 5 Physics 207 – Lecture 16 A Non-Conservative Force Path 2 Path 1 Since path2 distance >path1 distance the puck will be traveling slower at the end of path 2. Work done by a non-conservative force irreversibly removes energy out of the “system”. Here W NC = Efinal - Einitial < 0 and reflects Ethermal Physics 207: Lecture 16, Pg 16 Work & Power: Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. Assuming identical friction, both engines do the same amount of work to get up the hill. Are the cars essentially the same ? NO. The Corvette can get up the hill quicker It has a more powerful engine. Physics 207: Lecture 16, Pg 22 Page 6 Physics 207 – Lecture 16 Work & Power: Power is the rate at which work is done. Average Power: Instantaneous Power: P = W∆t P = dW dt Units (SI) are Watts (W): 1 W = 1 J / 1s Example: A person of mass 200. kg walks up 10.0 m. If he/she climbs in 10.00 sec what is the average power used (g = 10 m/s2) Pavg = F h / t = mgh / t = 200. x 10.0 x 10.0 / 10.00 Watts P = 2000 W Physics 207: Lecture 16, Pg 23 Work & Power: Average Power: Instantaneous Power: Constant force P = Paverage = W∆t P = dW dt r r r r dW d ( F ⋅ x ) F ⋅ dx r r P= = = = F ⋅v dt dt dt Physics 207: Lecture 16, Pg 24 Page 7 Physics 207 – Lecture 16 Work & Power: Example 2 : Engine of a jet develops a trust of 15,000 N when plane is flying at 300 m/s. What is the horsepower of the engine ? r r P = F ⋅v P=Fv P = (15,000 N) (300 m/s) = 4.5 x 106 W = (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp ! Physics 207: Lecture 16, Pg 25 Exercise Work & Power Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following, Power mg∆h P = dW = dt ∆t A. Top Power C. Bottom Z3 time Power B. Middle time time r r P = F ⋅v Physics 207: Lecture 16, Pg 26 Page 8 Physics 207 – Lecture 16 Chap. 12: Rotational Dynamics Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt Rotation is common in the world around us. Many ideas developed for translational motion are transferable. Physics 207: Lecture 16, Pg 30 Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare A merry-go-round is spinning and we run and jump on it. What does it do? We are standing on the rim and our “friends” spin it faster. What happens to us? We are standing on the rim a walk towards the center. Does anything change? Physics 207: Lecture 16, Pg 32 Page 9 Physics 207 – Lecture 16 System of Particles (Distributed Mass): Until now, we have considered the behavior of very simple systems (one or two masses). But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. ω 1 2 Compare the velocities and kinetic energies at these two points. Physics 207: Lecture 16, Pg 37 System of Particles (Distributed Mass): 1 K= ½ m v 2 = ½ m (ω r)2 ω The rotation axis matters too! 2 K= ½ m (2v) 2 = ½ m (ω 2r)2 Twice the radius, four times the kinetic energy K = 12 mv 2 = 12 m(ωr ) 2 Physics 207: Lecture 16, Pg 38 Page 10 Physics 207 – Lecture 16 Rotation & Kinetic Energy Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). The kinetic energy of this system will be the sum of the kinetic energy of each piece: 4 K = 12 ∑ mi v i2 i =1 K = ½ m 1v12 + ½ m 2v22 + ½ m 3v32 + ½ m 4v42 m4 ω r4 m3 r1 m1 r2 r3 m2 Physics 207: Lecture 16, Pg 39 Rotation & Kinetic Energy Notice that v1 = ω r1 , v2 = ω r2 , v3 = ω r3 , v4 = ω r4 So we can rewrite the summation: 4 4 4 i =1 i =1 i =1 K = 12 ∑ mi v i2 = 12 ∑ miω 2 ri2 = 12 [ ∑ mi ri2 ]ω 2 We recognize the quantity, moment of inertia or I, and write: m 4 K = 12 I ω 2 ω r4 m3 r3 r1 r2 m1 m2 Physics 207: Lecture 16, Pg 40 Page 11 Physics 207 – Lecture 16 Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2 I1 = 2mL2 m m m m I2 = mL2 I = 2mL2 L Physics 207: Lecture 16, Pg 41 Calculating Moment of Inertia... For a discrete collection of point masses we found: N I = ∑ m i ri 2 i =1 For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm. dm An integral is required to find I : I = ∫ r 2 dm r Physics 207: Lecture 16, Pg 42 Page 12 Physics 207 – Lecture 16 Moments of Inertia Some examples of I for solid objects: dr r L R Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center. I = ½ M R2 Physics 207: Lecture 16, Pg 43 Lecture 16, Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density ρ = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the pull? W = F d = ½ I ω2 (A) ω2 ω1 Disk 1 (B) Disk 2 (C) Same start finish F F d Physics 207: Lecture 16, Pg 49 Page 13 Physics 207 – Lecture 16 Rotation & Kinetic Energy... The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System K = 12 I ω 2 K = 12 mv 2 v is “linear” velocity m is the mass. ω is angular velocity I is the moment of inertia about the rotation axis. I = ∑ m i ri 2 i Physics 207: Lecture 16, Pg 50 Connection with CM motion If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is 2 K T = 12 MVCM If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity ω then its kinetic energy is K R = 12 I CMω 2 What if the object is both moving linearly and rotating? 2 K = 12 I CMω 2 + 12 MVCM Physics 207: Lecture 16, Pg 51 Page 14 Physics 207 – Lecture 16 Connection with motion... So for a solid object which rotates about its center of mass and whose CM is moving: 2 K TOTAL = K Rotational + 12 MVCM VCM ω Physics 207: Lecture 16, Pg 52 System of Particles: Center of Mass (CM) If an object is not held then it will rotate about the center of mass. Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding centers of mass. m1 + m2 m1 + m2 mobile Physics 207: Lecture 16, Pg 53 Page 15 Physics 207 – Lecture 16 System of Particles: Center of Mass How do we describe the “position” of a system made up of many parts ? Define the Center of Mass (average position): For a collection of N individual point-like particles whose masses and positions we know: N r ∑ mi ri r RCM ≡ i =1 M RCM m2 m1 r1 r2 y x (In this case, N = 2) Physics 207: Lecture 16, Pg 54 Sample calculation: Consider the following mass distribution: r ∑ mi ri r RCM = i =1 = XCM î + YCM ĵ + ZCM k̂ M N XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters RCM = (12,6) YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters (12,12) 2m XCM = 12 meters YCM = 6 meters m (0,0) m (24,0) Physics 207: Lecture 16, Pg 55 Page 16 Physics 207 – Lecture 16 System of Particles: Center of Mass For a continuous solid, convert sums to an integral. dm y r r r ∫ r dm ∫ r dm = RCM = M ∫ dm r x where dm is an infinitesimal mass element. Physics 207: Lecture 16, Pg 56 Recap Assignment: HW7 due Nov. 2nd For Monday: Read Chapter 12, Sections 7-11 Physics 207: Lecture 16, Pg 57 Page 17