1. Lecture 15 – Drag Force & Circular Dynamics

2. Why is the drag force on a piece of paper so large?

3. Uniform Circular Motion α α v r

4. Circular Motion A particle is going around a center at (x,y) = (3m, -4m) in uniform circular motion with a radius of 3m and a period of 2 seconds. At t=0 it starts at a point on the negative y- axis on its anti-clockwise motion. Which statement is correct? a. The particle is at (1m, -2m) at t=2s. b. The particle’s velocity is constant at 37.7 m/s. c. The particle’s velocity points initially in positive y- direction. d. The particle’s acceleration points initially in positive x- direction.

5. Dynamics of circular motion Since a = v 2 /r, we can write down the force But: we don’t know what it is, i.e. what entity provides is Centripetal Force has constant strength, but changing direction Especially tricky for non-contact forces Examples –Car turning corner –Space shuttle orbiting Earth

6. Pre-lecture Exercise (6.5) How is the physics of an object in a vertical circular loop (Sample problem on page 125) different from the one in a Ferris wheel (as in problem 3 of Homework #5)? a. The normal force on the object at the bottom of the loop (but not the wheel) is upward. b. The normal force on the object at the top of the loop (but not the wheel) is downward. c. The centripetal acceleration at the bottom of the loop (but not the wheel) is upward. d. The centripetal acceleration at the top of the loop (but not the wheel) is downward. e. None of the above. Answer: b

7. Example: Simulating gravity in spaceship with R=10m

8. Lecture 16 – Energy & Work

9. Exam A > 80% B > 62% C > 56% D > 48% Average 60%

10. Post-lecture Exercise (6.5) A normal car is undergoing a banked circular turn, see the sample problem on page 128. The bank angle is 10 degrees and the radius of the curve 200m. What is the maximal velocity the car can go without sliding? v = 18.59 m/s –Normal force (tilted 10 degrees towards center) has to produce the necessary centripetal force, which limits the velocity, since N is fixed (mass cancels out!)

11. Work Work is “accomplished” by a force applied over some distance Angle matters! Work is scalar product between force vector and displacement vector Work is a number or scalar (with units Joules=J): it can be zero, even negative! Maximal work if force is parallel to displacement

12. You are holding a book in your hand, so it doesn’t fall down. Are you doing work on the book? Yes No Not enough information

13. Group Work: Page 1 of Work- energy theorem worksheet

14. Lecture 17 – Work-Energy Theorem

15. Reminder: Vectors Components vs polar coordinates 2D vs 3D vs nD Scalar product can be used to calculate angle in any dimension! Need to master vectors (ED is vectors)

16. Scalar Product

17. Varied Force

18. Double index notation: Quo vadis? There is a force involved, so there are two indices! Need to keep it straight: who is doing the work ON what? Still 2 indices necessary and same rule: swap indices, obtain minus sign!

19. Example: Work done by person on spring F= – kx is Hooke’s law It is the force of a spring, i.e. exerted by the spring on something else: F p,s = – kx = –F s,p Need to do integral over x from 0 until x??? Don’t confuse integration variable with value of variable: x’ = x Geometrical interpretation: area of triangle

20. Energy Useful label/name (like force, etc.) Rough definition of energy: ability to do work Different forms –Object in motion: kinetic energy (prop. to velocity) –Object in configuration where it could do work: potential energy –Chemical energy …

21. Kinetic Energy Object in motion (relative to another) can do work on contact  train on train in crash, etc. Quantitative definition: W= Fd = (ma)d ; use v f 2 = v i 2 +2ad = m{(v f 2 – v i 2 )/2d}d = ½ mv f 2 – ½ m v i 2 = K f - K i

22. Pre-lecture Exercise (7.1-7.4) Calculate the kinetic energy of a baseball (110g) going v = – 50 mph, i.e. traveling in the negative x- direction. K= ½ m v 2 –Needed to convert to kg and m/s –Sign of velocity does not matter! –In higher dimensions v 2 = v x 2 +v y 2 +v z 2

23. Two marbles, one twice as heavy as the other, are dropped to the ground from the top of a building. Just before hitting the ground, the heavier marble has… …as much kinetic energy as the lighter one. …twice as much kinetic energy as the lighter one. …half as much kinetic energy as the lighter one. …four times as much kinetic energy as the lighter one.

24. Work-energy theorem Net work done on an object is equal to the change of its kinetic energy Note that work and hence change of KE can be zero or even negative!

25. Pre-lecture Exercise (7.5-7.9) In the Work and Energy simulation, set the mass to 5 kg, the initial position to 0, the force to 5 N, and the initial velocity to - 3m/s. What is the ratio of the work being done on the block (from the start to the end of the simulation) and the difference in kinetic energy (kinetic energy at the end of the simulation minus the kinetic energy at the start of the simulation)? Answer: 1 (by Work-energy theorem or direct calculation)