1. Physics 207: Lecture 12, Pg 1 Lecture 12 Goals: Assignment: l HW5 due Tuesday 10/18 l For Monday: Read through Ch. 10.4 (Reading Quiz) Chapter 8: Chapter 8: Solve 2D motion problems with friction Solve 2D motion problems with friction Chapter 9: Momentum & Impulse Chapter 9: Momentum & Impulse  Solve problems with 1D and 2D Collisions  Solve problems having an impulse (Force vs. time)

2. Physics 207: Lecture 12, Pg 2 Example, Circular Motion Forces with Friction (recall ma r = m |v T | 2 / r F f ≤  s N ) l How fast can the race car go? (How fast can it round a corner with this radius of curvature?) m car = 1600 kg  S = 0.5 for tire/road r = 80 m g = 10 m/s 2 r

3. Physics 207: Lecture 12, Pg 3 l Only one force is in the horizontal direction: static friction x-dir: F r = ma r = -m |v T | 2 / r = F s = -  s N (at maximum) y-dir: ma = 0 = N – mg N = mg v T = (  s m g r / m ) 1/2 v T = (  s g r ) 1/2 = (  x 10 x 80) 1/2 v T = 20 m/s (or 45 mph) Navigating a curve N mgmg FsFs m car = 1600 kg  S = 0.5 for tire/road r = 80 m g = 10 m/s 2 y x

4. Physics 207: Lecture 12, Pg 4 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 axis. At what angular velocity does it slip (assuming a T << a r at that time) if  s =0.8 ? l Only one force is in the horizontal direction: static friction x-dir: F r = ma r = -m |v T | 2 / r = F s = -  s N (at  ) y-dir: ma = 0 = N – mg N = mg v T = (  s m g r / m ) 1/2 v T = (  s g r ) 1/2 = (  x 10 x 0.5) 1/2 v T = 2 m/s   = v T / r = 4 rad/s A slight variation… m puck = 2 kg  S = 0.8 r = 0.5 m g = 10 m/s 2 y x N mgmg FsFs

5. Physics 207: Lecture 12, Pg 5 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

6. Physics 207: Lecture 12, Pg 6 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 mgmg FfFf What differs on a banked curve?

7. 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 xx xx y N mgmg FfFf  No speed

8. Physics 207: Lecture 12, Pg 8 Banked Curves Free Body Diagram for a banked curve. (rotated x-y coordinates) Resolve into components parallel and perpendicular to bank xx xx y N mgmg FfFf marmar  Low speed N mgmg FfFf marmar  High speed

9. Physics 207: Lecture 12, Pg 9 Banked Curves, high speed 4 Apply Newton’s 1 st and 2 nd Laws marmar xx xx y  N mg cos  FfFf mg sin   ma r cos  ma r sin   F x = -ma r cos  = - F f - mg sin   F y = ma r sin  = 0 - mg cos  + N Friction model  F f ≤  N (maximum speed when equal)

10. Physics 207: Lecture 12, Pg 10 Banked Curves, low speed 4 Apply Newton’s 1 st and 2 nd Laws marmar xx xx y  N mg cos  FfFf mg sin   ma r cos  ma r sin   F x = -ma r cos  = + F f - mg sin   F y = ma r sin  = 0 - mg cos  + N Friction model  F f ≤  N (minimum speed when equal but not less than zero!)

11. Physics 207: Lecture 12, Pg 11 Banked Curves, constant speed v max = (gr) ½ [ (  + tan  ) / (1 -  tan  )] ½ v min = (gr) ½ [  (tan  -  ) / (1 +  tan  )] ½ Dry pavement “Typical” values of r = 30 m, g = 9.8 m/s 2,  = 0.8,  = 20 ° v max = 20 m/s (45 mph) v min = 0 m/s (as long as  > 0.36 ) Wet Ice “Typical values” of r = 30 m, g = 9.8 m/s 2,  = 0.1,  =20 ° v max = 12 m/s (25 mph) v min = 9 m/s (Ideal speed is when frictional force goes to zero)

12. Physics 207: Lecture 12, Pg 12 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. 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) Note this approach can also be used to estimate the maximum walking speed. At what speed does the car lose contact?

13. Physics 207: Lecture 12, Pg 13 Locomotion of a biped: Top speed

14. 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, a r, come from? (i.e., what external forces act on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways

15. Physics 207: Lecture 12, Pg 15 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 external forces are on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, forwards & backwards!

16. Physics 207: Lecture 12, Pg 16 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 max = (gr) ½ v max ~ 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)

17. Physics 207: Lecture 12, Pg 17 Impulse and Momentum: A new perspective Conservation Laws : Are there any relationships between mass and velocity that remain fixed in value?

18. Physics 207: Lecture 12, Pg 18 Momentum Conservation l Momentum conservation (recasts Newton’s 2 nd Law when net external F = 0) is an important principle (usually when forces act over a short time) l It is a vector expression so must consider P x, P y and P z  if F x (external) = 0 then P x is constant  if F y (external) = 0 then P y is constant  if F z (external) = 0 then P z is constant

19. Physics 207: Lecture 12, Pg 19 Inelastic collision in 1-D: Example 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 beforeafter x

20. Physics 207: Lecture 12, Pg 20 Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ?  Key question: Is x-momentum conserved ? v V beforeafter x P Before P After

21. Physics 207: Lecture 12, Pg 21 A perfectly inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z 

22. Physics 207: Lecture 12, Pg 22 A perfectly inelastic collision in 2-D vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 x-dir p x : m 1 v 1 = (m 1 + m 2 ) V cos  y-dir p y : m 2 v 2 = (m 1 + m 2 ) V sin  l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z are conseved 

23. Physics 207: Lecture 12, Pg 23 Exercise Momentum is a Vector (!) quantity A. Yes B. No C. Yes & No D. Too little information given 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?

24. Physics 207: Lecture 12, Pg 24 Exercise Momentum is a Vector (!) quantity 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 What is the final velocity of the cart? l x-direction: No net force so P x is conserved. l y-direction: Net 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) 5.0 m 30 ° 7.5 m 10 kg 2 kg

25. Physics 207: Lecture 12, Pg 25 Exercise Momentum is a Vector (!) quantity Initial Final P x : MV x + mv x = (M+m) V ’ x M 0 + mv x = (M+m) V ’ x V ’ x = m v x / (M + m) = 2 (8.7)/ 12 m/s V ’ x = 1.4 m/s l x-direction: No net force so P x 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. 5.0 m 30 ° 7.5 m N mg 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 x = v cos 30° = 8.7 m/s i j x y 30 °

26. Physics 207: Lecture 12, Pg 26 Elastic Collisions l Elastic means that the objects do not stick. l There are many more possible outcomes but, if no external force, then momentum will always be conserved l Start with a 1-D problem. BeforeAfter

27. Physics 207: Lecture 12, Pg 27 Billiards l Consider the case where one ball is initially at rest. p p a  ppbppb F PPa PPa  before after The final direction of the red ball will depend on where the balls hit. v v cm

28. Physics 207: Lecture 12, Pg 28 Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) l Conservation of Momentum x-dir P x : m v before = m v after cos  + m V after cos  y-dir P y : 0 = m v after sin  + m V after sin  p p after  ppbppb F P P after  before after

29. Physics 207: Lecture 12, Pg 29 Lecture 12 l Assignment:  HW5 due Tuesday 10/18  Read through 10.4