Lecture 12 Goals: Chapter 8: Solve 2D motion problems with friction Chapter 9: Momentum & Impulse v Solve problems with 1D and 2D Collisions v Solve problems having an impulse (Force vs. time) Assignment: l HW5 due Tuesday 10/18 l For Monday: Read through Ch. 10.4 (Reading Quiz) Physics 207: Lecture 12, Pg 1 Eample, Circular Motion Forces with Friction (recall ma r = m v T 2 / r F f µ s ) l How fast can the race car go? (How fast can it round a corner with this radius of curvature?) r m car = 1600 kg µ S = 0.5 for tire/road r = 80 m g = 10 m/s 2 Physics 207: Lecture 12, Pg 2 Page 1
avigating a curve l Only one force is in the horizontal direction: static friction y -dir: F r = ma r = -m v T 2 / r = F s = -µ s (at maimum) y-dir: ma = 0 = mg = mg v T = (µ s m g r / m ) 1/2 F s mg v T = (µ s g r ) 1/2 = (0.5 10 80) 1/2 v T = 20 m/s (or 45 mph) m car = 1600 kg µ S = 0.5 for tire/road r = 80 m g = 10 m/s 2 Physics 207: Lecture 12, Pg 3 A slight variation l A horizontal disk is initially at rest and very slowly undergoes constant angular acceleration. A 2 kg puck is located a point 0.5 m away from the ais. At what angular velocity does it slip (assuming a T << a r at that time) if µ s =0.8? l Only one force is in the horizontal direction: static friction -dir: F r = ma r = -m v T 2 / r = F s = -µ s (at ω) y y-dir: ma = 0 = mg = mg v T = (µ s m g r / m ) 1/2 F s mg v T = (µ s g r ) 1/2 = (0.8 10 0.5) 1/2 v T = 2 m/s ω = v T / r = 4 rad/s m puck = 2 kg µ S = 0.8 r = 0.5 m g = 10 m/s 2 Physics 207: Lecture 12, Pg 4 Page 2
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 F f mg What differs on a banked curve? Physics 207: Lecture 12, Pg 6 Page 3
Banked Curves Free Body Diagram for a banked curve. (rotated -y coordinates) Resolve into components parallel and perpendicular to bank y F f o speed mg Physics 207: Lecture 12, Pg 7 Banked Curves Free Body Diagram for a banked curve. (rotated -y coordinates) Resolve into components parallel and perpendicular to bank Low speed ma r mg y F f ma r High speed F f mg Physics 207: Lecture 12, Pg 8 Page 4
Banked Curves, high speed 4 Apply ewton s 1 st and 2 nd Laws ma r sin ma r ma r cos Σ F = -ma r cos = - F f - mg sin F f y mg cos Σ F y = ma r sin = 0 - mg cos + mg sin Friction model F f µ (maimum speed when equal) Physics 207: Lecture 12, Pg 9 Banked Curves, low speed 4 Apply ewton s 1 st and 2 nd Laws ma r sin ma r ma r cos Σ F = -ma r cos = + F f - mg sin y F f mg cos Σ F y = ma r sin = 0 - mg cos + mg sin Friction model F f µ (minimum speed when equal but not less than zero!) Physics 207: Lecture 12, Pg 10 Page 5
Banked Curves, constant speed v ma = (gr) ½ [ (µ + tan ) / (1 - µ tan )] ½ v min = (gr) ½ [ (tan - µ) / (1 + µ tan )] ½ Dry pavement Typical values of r = 30 m, g = 9.8 m/s 2, µ = 0.8, = 20 v ma = 20 m/s (45 mph) v min = 0 m/s (as long as µ > 0.36 ) Wet Ice Typical values of r = 30 m, g = 9.8 m/s 2, µ = 0.1, =20 v ma = 12 m/s (25 mph) v min = 9 m/s (Ideal speed is when frictional force goes to zero) Physics 207: Lecture 12, Pg 11 avigating a hill Knight concept eercise: 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. F c = mg = m v 2 /r v = (gr) ½ (just like an object in orbit) ote this approach can also be used to estimate the maimum walking speed. Physics 207: Lecture 12, Pg 12 Page 6
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, a r, come from? (i.e., what eternal forces act on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways Physics 207: Lecture 12, Pg 14 Page 7
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, a r, come from? (i.e., what eternal 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! F r = m a r = m v 2 / r < mg Otherwise you will lose contact! a r = v 2 / r v ma = (gr) ½ v ma ~ 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 Page 8
Impulse and Momentum: A new perspective Conservation Laws : Are there any relationships between mass and velocity that remain fied in value? Physics 207: Lecture 12, Pg 17 r F EXT Momentum Conservation r r r r dv d( mv) dp = ma = m = = dt dt dt r dp r = 0 implies that P = constant dt and if F r EXT = 0 l Momentum conservation (recasts P ewton s 2 nd Law when net eternal F = 0) is an important principle (usually when forces act over a short time) l It is a vector epression so must consider P, P y and P z v if F (eternal) = 0 then P is constant v if F y (eternal) = 0 then P y is constant v if F z (eternal) = 0 then P z is constant Physics 207: Lecture 12, Pg 18 Page 9
Inelastic collision in 1-D: Eample l 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. l In terms of m, M, and V : What is the momentum of the bullet with speed v? v V before after Physics 207: Lecture 12, Pg 19 Inelastic collision in 1-D: Eample What is the momentum of the bullet with speed v? mv r v Key question: Is -momentum conserved? v P Before aaaa m v + M 0 = ( m + P After M )V V before after v= (1+M / m)v Physics 207: Lecture 12, Pg 20 Page 10
A perfectly inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). v 1 m 1 V m 1 + m 2 m 2 v 2 before after l If no eternal force momentum is conserved. l Momentum is a vector so p, p y and p z Physics 207: Lecture 12, Pg 21 A perfectly inelastic collision in 2-D l If no eternal force momentum is conserved. l Momentum is a vector so p, p y and p z are conseved V v 1 m 1 + m 2 m 1 m 2 v 2 before after l -dir p : m 1 v 1 = (m 1 + m 2 ) V cos l y-dir p y : m 2 v 2 = (m 1 + m 2 ) V sin Physics 207: Lecture 12, Pg 22 Page 11
Eercise Momentum is a Vector (!) quantity l A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction l In regards to the block landing in the cart is momentum conserved? A. Yes B. o C. Yes & o D. Too little information given Physics 207: Lecture 12, Pg 23 Eercise Momentum is a Vector (!) quantity l -direction: o net force so P is conserved. l y-direction: et force, interaction with the ground so depending on the system (i.e., do you include the Earth?) P y is not conserved (system is block and cart only) 2 kg 5.0 m 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 into the 10 kg cart 30 7.5 m 10 kg What is the final velocity of the cart? Physics 207: Lecture 12, Pg 24 Page 12
Eercise Momentum is a Vector (!) quantity l -direction: o net force so P is conserved l y-direction: v y of the cart + block will be zero and we can ignore v y of the block when it lands in the cart. j 5.0 m Initial Final P : MV + mv = (M+m) V M 0 + mv = (M+m) V V = m v / (M + m) = 2 (8.7)/ 12 m/s V = 1.4 m/s i 30 y mg 30 7.5 m 1) a i = g sin 30 = 5 m/s 2 2) d = 5 m / sin 30 = ½ a i t 2 10 m = 2.5 m/s 2 t 2 2s = t v = a i t = 10 m/s v = v cos 30 = 8.7 m/s Physics 207: Lecture 12, Pg 25 Elastic Collisions l Elastic means that the objects do not stick. l There are many more possible outcomes but, if no eternal force, then momentum will always be conserved l Start with a 1-D problem. Before After Physics 207: Lecture 12, Pg 26 Page 13
Billiards l Consider the case where one ball is initially at rest. before after p a p b v cm F P a φ The final direction of the red ball will depend on where the balls hit. Physics 207: Lecture 12, Pg 27 Billiards: Without eternal forces, conservation of momentum (and energy Ch. 10 & 11) l Conservation of Momentum l -dir P : m v before = m v after cos + m V after cos φ l y-dir P y : 0 = m v after sin + m V after sin φ before after p after p b F P after φ Physics 207: Lecture 12, Pg 28 Page 14
Lecture 12 l Assignment: v HW5 due Tuesday 10/18 v Read through 10.4 Physics 207: Lecture 12, Pg 29 Page 15