Lecture 11Problem 7.34 HintLocomotion: how fast can a biped walk?How fast can a biped walk?Slide 5Slide 6Orbiting satellites vT = (gr)½Geostationary orbitSlide 9Impulse & Linear MomentumCollisions are a fact of lifeForces vs time (and space, Ch. 10)Example 1Twice the massExample 1Slide 16Force curves are usually a bit different in the real worldExample 1 with Action-ReactionSlide 19Slide 20Applications of Momentum ConservationSlide 22Momentum ConservationSlide 24Exercise 1 Momentum is a Vector (!) quantitySlide 26Inelastic collision in 1-D: Example 2Slide 28Example 2 Inelastic Collision in 1-D with numbersExercise 2 Momentum ConservationPhysics 207: Lecture 11, Pg 1Lecture 11Goals:Goals:Assignment:Assignment:Read through Chapter 10MP HW5, due Wednesday 3/3• Chapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse Understand what momentum is and how it relates to forces Employ momentum conservation principles In problems with 1D and 2D Collisions In problems having an impulse (Force vs. time) Chapter 8: Use models with free fallPhysics 207: Lecture 11, Pg 2Problem 7.34 HintSuggested StepsTwo independent free body diagrams are necessaryDraw in the forces on the top and bottom blocksTop Block Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding)Bottom Block Forces: 1. normal to bottom surface 2. normal to top block interface3. rope tension (to the left)4. weight (2 kg) 5. friction with top block 6. friction with surface7. 20 NUse Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.Physics 207: Lecture 11, Pg 3Locomotion: how fast can a biped walk?Physics 207: Lecture 11, Pg 4How fast can a biped walk?What about weight?(a) A heavier person of equal height and proportions can walk faster than a lighter person(b) A lighter person of equal height and proportions can walk faster than a heavier person(c) To first order, size doesn’t matterPhysics 207: Lecture 11, Pg 5How fast can a biped walk?What about height?(a) A taller person of equal weight and proportions can walk faster than a shorter person(b) A shorter person of equal weight and proportions can walk faster than a taller person(c) To first order, height doesn’t matterPhysics 207: Lecture 11, Pg 6How fast can a biped walk?What can we say about the walker’s acceleration if there is UCM (a smooth walker) ?Acceleration is radial !So where does it, ar, come from?(i.e., what external forces are on the walker?)1. Weight of walker, downwards2. Friction with the ground, sidewaysPhysics 207: Lecture 11, Pg 7Orbiting satellites vT = (gr)½Physics 207: Lecture 11, Pg 8Geostationary orbitPhysics 207: Lecture 11, Pg 9Geostationary orbitThe radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth. Fgravity is proportional to r-2 and so little g is now ~10 m/s2 / 50 vT = (0.20 * 42000000)½ m/s = 3000 m/s At 3000 m/s, period T = 2 r / vT = 2 42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) Great for communication satellites (1st pointed out by Arthur C. Clarke)Physics 207: Lecture 11, Pg 10Impulse & Linear MomentumTransition from forces to conservation lawsNewton’s Laws Conservation LawsConservation Laws Newton’s Laws They are different faces of the same physics NOTE: We have studied “impulse” and “momentum” but we have not explicitly named them as suchConservation of momentum is far more general thanconservation of mechanical energyPhysics 207: Lecture 11, Pg 11Collisions are a fact of lifePhysics 207: Lecture 11, Pg 12Forces vs time (and space, Ch. 10)Underlying any “new” concept in Chapter 9 is (1) A net force changes velocity (either magnitude or direction) (2) For any action there is an equal and opposite reactionIf we emphasize Newton’s 3rd Law and look at the changes with time then this leads to the Conservation of Momentum PrinciplePhysics 207: Lecture 11, Pg 13Example 1A 2 kg block, initially at rest on frictionless horizontal surface, is acted on by a 10 N horizontal force for 2 seconds (in 1D).What is the final velocity?F is to the positive & F = ma thus a = F/m = 5 m/s2 v = v0 + a t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction)Notice: v - v0 = a t m (v - v0) = ma t m v = F t If the mass had been 4 kg … now what final velocity?F-+F (N)1002time (s)Physics 207: Lecture 11, Pg 14Twice the massSame forceSame timeHalf the acceleration (a = F / m’) Half the velocity ! ( 5 m/s )F (N)1002Time (sec)FBeforePhysics 207: Lecture 11, Pg 15Example 1 Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity). Here, mass times the velocity always gives the same value. (Always 20 kg m/s.)F (N)1002Time (sec)Area under curve is still the same !Force x change in time = mass x change in velocityPhysics 207: Lecture 11, Pg 16Example 1 There many situations in which the sum of the products “mass times velocity” is constant over time To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s)A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” F (N)1002Time (sec)Area under curve : “impulse”With: m v = Favg tPhysics 207: Lecture 11, Pg 17Force curves are usually a bit different in the real worldPhysics 207: Lecture 11, Pg 18Example 1 with Action-ReactionNow the 10 N force from before is applied by person A on person B while standing on a frictionless surfaceFor the force of A on B there is an equal and opposite force of B on AF (N)1002Time (sec)0-10A on BB on A MA x VA = Area of top curve MB x VB = Area of bottom curve Area (top) + Area (bottom) = 0Physics 207: Lecture 11, Pg 19Example 1 with Action-ReactionMA VA + MB VB = 0MA [VA(final) - VA(initial)] + MB [VB(final) - VB(initial)] = 0Rearranging terms MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial)which is constant regardless of M or V(Remember: frictionless surface)Physics 207: Lecture 11, Pg 20Example 1 with Action-Reaction MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial)which is
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