1. Physics 207: Lecture 11, Pg 1 Lecture 11 Goals: Assignment: l Read through Chapter 10, 1 st four section l MP HW6, due Wednesday 3/3 Chapter 8: Employ rotational motion models with friction or in free fall Chapter 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)

2. Physics 207: Lecture 11, Pg 2 l One last reprisal of the Free Body Diagram l Remember 1 Normal Force is  to the surface 2 Friction is parallel to the contact surface 3 Radial (aka, centripetal) acceleration requires a net force Zero Gravity Ride

3. Physics 207: Lecture 11, Pg 3 Zero Gravity Ride A rider in a horizontal “0 gravity ride” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

4. Physics 207: Lecture 11, Pg 4 Banked Curves In the previous car scenario, we drew the following free body diagram for a race car going around a curve at constant speed on a flat track. Because the acceleration is radial (i.e., velocity changes in direction only) we need to modify our view of friction. n mgmg FfFf So, what differs on a banked curve?

5. Physics 207: Lecture 11, Pg 5 Banked Curves (high speed) 1 Draw a Free Body Diagram for a banked curve. 2 Use a rotated x-y coordinates 3 Resolve into components parallel and perpendicular to bank marmar xx xx y  N mgmg FfFf

6. Physics 207: Lecture 11, Pg 6 Banked Curves (constant high speed) 1 Draw a Free Body Diagram for a banked curve. 2 Use a rotated x-y coordinates 3 Resolve into components parallel and perpendicular to bank ( Note: For very small banking angles, one can approximate that F f is parallel to ma r. This is equivalent to the small angle approximation sin  = tan , but very effective at pushing the car toward the center of the curve!!) marmar xx xx y  N mgmg FfFf

7. Physics 207: Lecture 11, Pg 7 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)

8. Physics 207: Lecture 11, Pg 8 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!)

9. Physics 207: Lecture 11, Pg 9 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)

10. Physics 207: Lecture 11, Pg 10 Banked Curves, Testing your understanding Free Body Diagram for a banked curve. Use rotated x-y coordinates Resolve into components parallel and perpendicular to bank N mgmg FfFf At this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you begin to accelerate in that direction. How does the radial acceleration change? marmary x 

11. Physics 207: Lecture 11, Pg 11 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?

12. Physics 207: Lecture 11, Pg 12 Orbiting satellites v T = (gr) ½

13. Physics 207: Lecture 11, Pg 13 Locomotion: how fast can a biped walk?

14. Physics 207: Lecture 11, Pg 14 How 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 matter

15. Physics 207: Lecture 11, Pg 15 How 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 matter

16. Physics 207: Lecture 11, Pg 16 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

17. Physics 207: Lecture 11, Pg 17 Impulse & Linear Momentum l Transition from forces to conservation laws Newton’s Laws  Conservation Laws Conservation 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 such Conservation of momentum is far more general than conservation of mechanical energy

18. Physics 207: Lecture 11, Pg 18 Forces vs time (and space, Ch. 10) l 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 reaction l If we emphasize Newton’s 3 rd Law and emphasize changes with time then this leads to the Conservation of Momentum Principle

19. Physics 207: Lecture 11, Pg 19 Example 1 A 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? l F is to the positive & F = ma thus a = F/m = 5 m/s 2 v = v 0 + a  t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction) Notice: v - v 0 = a  t  m (v - v 0 ) = ma  t  m  v = F  t If the mass had been 4 kg …what is the final velocity? - + F (N) 10 0 2 time (s)

20. Physics 207: Lecture 11, Pg 20 Twice the mass l Same force l Same time Half the acceleration (a = F / m ’ ) l Half the velocity ! ( 5 m/s ) F (N) 10 0 2 Time (sec) Before

21. Physics 207: Lecture 11, Pg 21 Example 1 l Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity). l Here, mass times the velocity always gives the same value. (Always 20 kg m/s.) F (N) 10 0 2 Time (sec) Area under curve is still the same ! Force x change in time = mass x change in velocity

22. Physics 207: Lecture 11, Pg 22 Example l There many situations in which the sum of the product “mass times velocity” is constant over time l To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s) l A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” F (N) 10 0 2 Time (sec) Area under curve : “impulse” With: m  v = F avg  t

23. Physics 207: Lecture 11, Pg 23 Force curves are usually a bit different in the real world

24. Physics 207: Lecture 11, Pg 24 Example with Action-Reaction l Now the 10 N force from before is applied by person A on person B while standing on a frictionless surface l For the force of A on B there is an equal and opposite force of B on A F (N) 10 0 2 Time (sec) 0 -10 A on B B on A M A x  V A = Area of top curve M B x  V B = Area of bottom curve Area (top) + Area (bottom) = 0

25. Physics 207: Lecture 11, Pg 25 Example with Action-Reaction M A  V A + M B  V B = 0 M A [V A (final) - V A (initial) ] + M B [V B (final) - V B (initial) ] = 0 Rearranging terms M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V (Remember: frictionless surface)

26. Physics 207: Lecture 11, Pg 26 Example with Action-Reaction M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V Define MV to be the “momentum” and this is conserved in a system if and only if the system is not acted on by a net external force (choosing the system is key) Conservation of momentum is a special case of applying Newton’s Laws

27. Physics 207: Lecture 11, Pg 27 Applications of Momentum Conservation 234 Th 238 U Alpha Decay 4 He v1v1 v2v2 Explosions Radioactive decay: Collisions

28. Physics 207: Lecture 11, Pg 28 Impulse & Linear Momentum l Newton’s 2 nd Law: Fa F = ma l This is the most general statement of Newton’s 2 nd Law pv p ≡ mv pv (p is a vector since v is a vector) So p x = mv x and so on (y and z directions) l Definition: For a single particle, the momentum p is defined as:

29. Physics 207: Lecture 11, Pg 29 Momentum Conservation l Momentum conservation (recasts Newton’s 2 nd Law when net external F = 0) is an important principle l It is a vector expression (P x, P y and P z ).  And applies to any situation in which there is NO net external force applied (in terms of the x, y & z axes).

30. Physics 207: Lecture 11, Pg 30 Momentum Conservation l Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved l Momentum is a vector quantity and we can independently assess its conservation in the x, y and z directions (e.g., net forces in the z direction do not affect the momentum of the x & y directions)

31. Physics 207: Lecture 11, Pg 31 Exercise 2 Momentum Conservation A. Box 1 B. Box 2 C. same l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.  Which box ends up moving fastest ? 1 2

32. Physics 207: Lecture 11, Pg 32 Lecture 11 Assignment: l For Monday: Read through Chapter 10, 1 st four sections l MP HW6 due Wednesday 3/3