Lecture 12 Goals: • Chapter 8: • Solve 2D motion problems with friction Chapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time) Assignment: HW5 due Tuesday 10/18 For Monday: Read through Ch. 10.4 (Reading Quiz) Physics 207: Lecture 12, Pg 1 Example, Circular Motion Forces with Friction (recall mar = m |vT | 2 / r Ff ≤ ms N ) How fast can the race car go? (How fast can it round a corner with this radius of curvature?) mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2 r Physics 207: Lecture 12, Pg 2 Navigating a curve Only one force is in the horizontal direction: static friction x-dir: Fr = mar = -m |vT |2/ y-dir: ma = 0 = N – mg y r = Fs = -ms N (at maximum) N = mg vT = (ms m g r / m )1/2 vT = (ms g r )1/2 = (0.5 x 10 x 80)1/2 vT = 20 m/s (or 45 mph) N Fs mg mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2 Physics 207: Lecture 12, Pg 3 x Zero Gravity Ride A rider in a “0 gravity ride” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? A B C D E Physics 207: Lecture 12, Pg 5 Banked Curves In the previous car scenario, we drew the following free body diagram for a race car going around a curve on a flat track. n Ff mg What differs on a banked curve? Physics 207: Lecture 12, Pg 6 Banked Curves Free Body Diagram for a banked curve. (rotated x-y coordinates) Resolve into components parallel and perpendicular to bank y x N Ff No speed q mg Physics 207: Lecture 12, Pg 7 Banked Curves Free Body Diagram for a banked curve. (rotated x-y coordinates) Resolve into components parallel and perpendicular to bank y x N mar Ff Low speed q mg N mar High speed Ff q mg Physics 207: Lecture 12, Pg 8 Navigating a hill Knight concept exercise: A car is rolling over the top of a hill at speed v. At this instant, A. n > w. B. n = w. C. n < w. D. We can’t tell about n without knowing v. At what speed does the car lose contact? This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion. Fc = mg = m v2 /r v = (gr)½ (just like an object in orbit) Note this approach can also be used to estimate the maximum walking speed. Physics 207: Lecture 12, Pg 12 Locomotion of a biped: Top speed Physics 207: Lecture 12, Pg 13 How 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 act on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways Physics 207: Lecture 12, Pg 14 How 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, downwards 2. Friction with the ground, forwards & backwards! Physics 207: Lecture 12, Pg 15 How fast can a biped walk? Given a model then what does the physics say? Choose a position with the simplest constraints. If his radial acceleration is greater than g then he is on the wrong planet! Fr = m ar = m v2 / r < mg Otherwise you will lose contact! ar = v2 / r vmax = (gr)½ vmax ~ 3 m/s ! (So it pays to be tall and live on Jupiter) Olympic record pace over 20 km 4.2 m/s (Lateral motion about hips gives ~2 m/s more) Physics 207: Lecture 12, Pg 16 Impulse and Momentum: A new perspective Conservation Laws : Are there any relationships between mass and velocity that remain fixed in value? Physics 207: Lecture 12, Pg 17 Momentum Conservation FEXT dv d (mv) dP ma m dt dt dt and if FEXT 0 dP 0 implies that P constant dt P Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (usually when forces act over a short time) It is a vector expression so must consider Px, Py and Pz if Fx (external) = 0 then Px is constant if Fy (external) = 0 then Py is constant if Fz (external) = 0 then Pz is constant Physics 207: Lecture 12, Pg 18 Inelastic collision in 1-D: Example A 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 final speed V. In terms of m, M, and V : What is the momentum of the bullet with speed v ? x v V before after Physics 207: Lecture 12, Pg 19 Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ? mv Key question: Is x-momentum conserved ? P After P Before mv M 0 (m M )V aaaa v V after x before v (1 M / m)V Physics 207: Lecture 12, Pg 20 A perfectly inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). V v1 q m1 + m2 m1 m2 v2 before after If no external force momentum is conserved. Momentum is a vector so px, py and pz Physics 207: Lecture 12, Pg 21 A perfectly inelastic collision in 2-D If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved V v1 q m1 + m2 m1 m2 v2 before after x-dir px : m1 v1 = (m1 + m2 ) V cos q y-dir py : m2 v2 = (m1 + m2 ) V sin q Physics 207: Lecture 12, Pg 22 Exercise Momentum is a Vector (!) quantity A. B. C. D. A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction In regards to the block landing in the cart is momentum conserved? Yes No Yes & No Too little information given Physics 207: Lecture 12, Pg 23 Exercise Momentum is a Vector (!) quantity x-direction: No net force so Px is conserved. y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) Py is not conserved (system is block and cart only) 2 kg 5.0 m 30° Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m 7.5 m into the 10 kg cart 10 kg What is the final velocity of the cart? Physics 207: Lecture 12, Pg 24 Exercise Momentum is a Vector (!) quantity 1) ai = g sin 30° = 5 m/s2 x-direction: No net force so Px is conserved y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it 2) d = 5 m / sin 30° lands in the cart. j = ½ ai Dt2 N i 5.0 m 30° mg 30° Initial Final Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s V’x = 1.4 m/s 10 m = 2.5 m/s2 Dt2 2s = Dt v = ai Dt = 10 m/s vx= v cos 30° = 8.7 m/s 7.5 m y x Physics 207: Lecture 12, Pg 25 Elastic Collisions Elastic means that the objects do not stick. There are many more possible outcomes but, if no external force, then momentum will always be conserved Start with a 1-D problem. Before After Physics 207: Lecture 12, Pg 26 Billiards Consider the case where one ball is initially at rest. after before pa q pb vcm Pa f F The final direction of the red ball will depend on where the balls hit. Physics 207: Lecture 12, Pg 27 Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) Conservation of Momentum x-dir Px : m vbefore = m vafter cos q + m Vafter cos f y-dir Py : 0 = m vafter sin q + m Vafter sin f after before pafter q pb F Pafter f Physics 207: Lecture 12, Pg 28 Lecture 12 Assignment: HW5 due Tuesday 10/18 Read through 10.4 Physics 207: Lecture 12, Pg 29