Lecture 13Newton’s Laws rearrangedEnergySlide 4Inelastic collision in 1-D: Example 1Slide 6Kinetic & Potential energiesSlide 8Example of a conservative system: The simple pendulum.Example: The simple pendulum.Slide 11Slide 12Example The Loop-the-Loop … againSlide 14Example SkateboardSlide 16Potential Energy, Energy Transfer and PathSlide 18Slide 19Elastic vs. Inelastic CollisionsSlide 21Variable force devices: Hooke’s Law SpringsExercise 2 Hooke’s LawSlide 26Force vs. Energy for a Hooke’s Law springSlide 28Physics 207: Lecture 13, Pg 1Lecture 13Goals:Goals:Assignments: Assignments: HW6, due tomorrow, HW7 due Wednesday, Mar. 10HW6, due tomorrow, HW7 due Wednesday, Mar. 10For Thursday, Read all of Chapter 11For Thursday, Read all of Chapter 11•Chapter 10Chapter 10 Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Develop and exploit conservation of energy principlein problem solving•Chapter 11Chapter 11 Understand the relationship between force, displacement and workPhysics 207: Lecture 13, Pg 2Newton’s Laws rearrangedFrom motion in the y-dir, Fy = m ay and let accel. be constant y(t) = y0 + vy0 t + ½ ay t2 y = y(t)-y0= vy0 t + ½ ay t2 vy (t) = vy0 + ay t t = (vy - vy0) / ay and eliminating t yields ay y= ½ (vy2 - vy02 ) / ay -mg y= ½ m (vy2 - vy02 )Physics 207: Lecture 13, Pg 3Energy -mg y= ½ m (vy2 - vy02 ) -mg yf – yi) = ½ m ( vyf2 -vyi2 )Rearranging to give initial on the left and final on the right ½ m vyi2 + mgyi = ½ m vyf2 + mgyf We now define mgy as the “gravitational potential energy”A relationship between y- displacement and change in the y-speed squaredPhysics 207: Lecture 13, Pg 4EnergyNotice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyfwhere vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 terms are defined to be kinetic energies(A scalar quantity of motion)Physics 207: Lecture 13, Pg 5Inelastic collision in 1-D: Example 1A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved?vVbefore afterxPhysics 207: Lecture 13, Pg 6Inelastic collision in 1-D: Example 1What is the momentum of the bullet with speed v ? What is the initial energy of the system ? What is the final energy of the system ? Is momentum conserved (yes)? Is energy conserved? Examine Ebefore-EaftervVbefore afterxvm v21 vv212mm V)(212Mm V)( 0 M v Mmm )(1v21 vv)(21 v21 V]V)[(21 v21222MmmmMmmmmMmmNo!Physics 207: Lecture 13, Pg 7Kinetic & Potential energiesKinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of massesPotential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energyMechanical energy, EMech, is defined to be the sum of U and K.Physics 207: Lecture 13, Pg 8EnergyIf only “conservative” forces are present, the total mechanical If only “conservative” forces are present, the total mechanical energy energy ((sum of potential, U, and kinetic energies, K) of a system) of a system is is conservedconservedFor an object in a gravitational “field” Emech = K + UK and U may change, but Emech = K + U remains a fixed value.Emech = K + U = constant Emech is called “mechanical energy”K ≡ ½ mv2U ≡ mgy ½ m vyi2 + mgyi = ½ m vyf2 + mgyfPhysics 207: Lecture 13, Pg 9Example of a conservative system: The simple pendulum.Suppose we release a mass m from rest a distance h1 above its lowest possible point.What is the maximum speed of the mass and where does this happen ?To what height h2 does it rise on the other side ?vh1h2mPhysics 207: Lecture 13, Pg 10Example: The simple pendulum.yy=0y=h1 What is the maximum speed of the mass and where does this happen ?E = K + U = constant and so K is maximum when U is a minimum.Physics 207: Lecture 13, Pg 11Example: The simple pendulum.vh1yy=h1y=0 What is the maximum speed of the mass and where does this happen ?E = K + U = constant and so K is maximum when U is a minimumE = mgh1 at topE = mgh1 = ½ mv2 at bottom of the swingPhysics 207: Lecture 13, Pg 12Example: The simple pendulum.yy=h1=h2y=0 To what height h2 does it rise on the other side?E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.E = mgh1 = mgh2 or h1 = h2Physics 207: Lecture 13, Pg 13ExampleThe Loop-the-Loop … againTo complete the loop the loop, how high do we have to let the release the car?Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) Exploit the fact that E = U + K = constant ! (frictionless)(A) 2R (B) 3R (C) 5/2 R (D) 23/2 Rh ?RCar has mass mRecall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top isgRvUb=mghU=mg2Ry=0U=0Physics 207: Lecture 13, Pg 14ExampleThe Loop-the-Loop … againUse E = K + U = constantmgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgRh = 5/2 RRgRvh ?Physics 207: Lecture 13, Pg 15What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as a “point”Assume zero of gravitational U is at bottom of the hillR=10 m..m = 25 kgExampleSkateboard..R=10 m30°y=0Physics 207: Lecture 13, Pg 16What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as “point”Assume zero of gravitational U is at bottom of the hillR=10 my= 5 m..m = 25 kgExampleSkateboard..R=10 m30°Use E = K + U = constant Ebefore = Eafter0 + m g R = ½ mv2 + mgR / 2 mgR/2 = ½ mv2 gR = v2 v= (gR)½ v = (10 x 10)½ = 10 m/sPhysics 207:
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