1. Physics 1501: Lecture 11, Pg 1 Physics 1501: Lecture 10 l Announcements çHomework #3 : due next Monday çBut HW 04 will be due the following Friday (same week). çMidterm 1: Monday Oct. 3 l Topics çReview: Forces and uniform circular motion çForces and non-uniform circular motion çAccelerated reference frames çWork & Energy çScalar Product

2. Physics 1501: Lecture 11, Pg 2 Nonuniform Circular Motion Earlier we saw that for an object moving in a circle with nonuniform speed then a = a r + a t. arar atat What are F r and F t ?

3. Physics 1501: Lecture 11, Pg 3 Lecture 10, ACT 1 l When a pilot executes a loop- the-loop (as in figure on the right) the aircraft moves in a vertical circle of radius R=2.70 km at a constant speed of v=225 m/s. Is the force exerted by the seat on the pilot: A) Larger B) Same C) Smaller then pilot’s weight (mg) at I) the bottom and II) at the top of the loop. 2.7 km

4. Physics 1501: Lecture 11, Pg 4 Lecture 10, ACT 1 Solution 2.7 km FcFc FcFc I) II) mg F c = N II + mg F c = m a r = mv 2 / R N II = mg (1- v 2 / Rg ) N II < mg ANSWER (C) NINI N II F c = N I - mg F c = m a r = mv 2 / R N I = mg (1+ v 2 / Rg ) N I > mg ANSWER (A) FcFc

5. Physics 1501: Lecture 11, Pg 5 My match box car is going to do a loop the loop. What must be its minimum speed at the top so that it can make the loop successfully ?? Example Exercise 1

6. Physics 1501: Lecture 11, Pg 6 Example Exercise 1 mgmg Radial : F r = N + mg cos  mv 2 /R Tangential : F t = mg sin  Solve the first for v. mgmg  N   

7. Physics 1501: Lecture 11, Pg 7 Example Exercise 1 mgmg N  To stay on the track there must be a non-zero normal force. Why ? The limiting condition is where N = 0. And at the top,  = 0.

8. Physics 1501: Lecture 11, Pg 8 My match box car is going to do a loop the loop. What must be its minimum speed at the bottom so that it can make the loop successfully ?? This is a difficult problem with just forces. We will deal with it with energy considerations. Example Exercise 2

9. Physics 1501: Lecture 11, Pg 9 My match box car is going to do a loop the loop. If the speed at the bottom is v B, what is the normal force at that point ?? Example Exercise 3

10. Physics 1501: Lecture 11, Pg 10 Example Exercise 3 mgmg Radial : F r = N + mg cos  mv 2 /R Tangential : F t = mg sin  Solve the first for N. N  At the bottom, v = v B, and .

11. Physics 1501: Lecture 11, Pg 11 I am building a roller coaster. I design it so that when I do a loop by myself I am weightless at the top, and I have speed v 1. Next my two friends get in with me so that the total weight of car and people doubles. How fast must the car go so we are still weightless at the top ?? Lecture 10, ACT 2 Nonuniform Circular Motion A) 1/2 v 1 B) v 1 C) 2 v 1 D) 4 v 1

12. Physics 1501: Lecture 11, Pg 12 Accelerated Reference Frames: The Accelerometer l Your first job is with General Motors. You are working on a project to design an accelerometer. The inner workings of this gadget consist of a weight of mass m that is hung inside a box that is attached to the ceiling of a car. You design the device with a very light string so that you can mathematically ignore it. The idea is that the angle the string makes with the vertical is determined by the car’s acceleration. Your preliminary task is to think about calibration of the accelerometer. First you calibrate the measurement for when the car travels on a flat road.

13. Physics 1501: Lecture 11, Pg 13 Accelerated Reference Frames: The Accelerometer a  i 1 We need to solve for the angle the plum bob makes with respect to vertical. First we will solve by using Newton’s Second Law and checking x and y components. Then we will consider other possible solution methods.

14. Physics 1501: Lecture 11, Pg 14 Accelerometer... 2. Draw a free-body diagram for the mass: We wish to solve for  in terms of the acceleration a. We will use  F = ma in two cartesian dimensions. m T g mg (gravitational force)  a x y  TXTX TYTY

15. Physics 1501: Lecture 11, Pg 15 Accelerometer... 3. Solving, i i: F X = T X = T sin  = ma j j: F Y = T Y  mg = T cos  mg = 0 l Eliminate T T gmggmg m amaama ji  TXTX TYTY  T sin  = ma T cos  = mg tan  = a / g

16. Physics 1501: Lecture 11, Pg 16 Accelerometer... 4. No Numbers involved 5. This answer has the right units (none) It does give 1 in terms of the acceleration, tan  = a / g

17. Physics 1501: Lecture 11, Pg 17 Accelerometer – Other Thoughts 1 l Alternative solution using vectors (more elegant): F l Find the total vector force F TOT : T gmggmg F F TOT  m T T (string tension) g mg (gravitational force) 

18. Physics 1501: Lecture 11, Pg 18 l Alternative solution using vectors (more elegant): F l Find the total vector force F TOT : F l Recall that F TOT = ma: l So amaama T gmggmg  m T T (string tension) g mg (gravitational force)  Accelerometer – Other Thoughts 1

19. Physics 1501: Lecture 11, Pg 19 Accelerometer – Other Thoughts 2 l Think of this problem from the point of view of the person inside the car. l This person sees the plumb bob making the same angle with respect to the ground, but detects no acceleration. a 

20. Physics 1501: Lecture 11, Pg 20 Accelerometer... l There must be some other force to balance the x component i i: F X = T X + F ? = T sin  + F ? = 0 j j: F Y = T Y  mg = T cos  mg = 0 l And we must put F ? = -ma to get the same answer as before. l F? is known as a fictitious force.  a T mgmg

21. Physics 1501: Lecture 11, Pg 21 Lecture 10, ACT 3 Accelerated Reference Frames You are a passenger in a car and not wearing your seatbelt. Without increasing or decreasing speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is a correct description of the situation ? A) Before and after the collision there is a rightward force pushing you into the door. B) Starting at the time of the collision, the door exerts a leftward force on you. C) Both of the above. D) Neither of the above.

22. Physics 1501: Lecture 11, Pg 22 Chap.6: Work & Energy l One of the most important concepts in physics. çAlternative approach to mechanics. l Many applications beyond mechanics. çThermodynamics (movement of heat). çQuantum mechanics... l Very useful tools. çYou will learn new (sometimes much easier) ways to solve problems.

23. Physics 1501: Lecture 11, Pg 23 Forms of Energy l Kinetic l Kinetic: Energy of motion. çA car on the highway has kinetic energy. çWe have to remove this energy to stop it. çThe breaks of a car get HOT ! çThis is an example of turning one form of energy into another (thermal energy). çKinetic energy is given by: K = (1/2) mv 2

24. Physics 1501: Lecture 11, Pg 24 Energy Conservation l Energy cannot be destroyed or created. çJust changed from one form to another. energy is conserved l We say energy is conserved ! çTrue for any isolated system. çi.e when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same. çThe energy of the car “alone” is not conserved... »It is reduced by the braking. work energy l Doing “work” on an otherwise isolated system will change its “energy”...

25. Physics 1501: Lecture 11, Pg 25 Definition of Work: Ingredients: Fr Ingredients: Force ( F ), displacement (  r ) F Work, W, of a constant force F r acting through a displacement  r is: F rrr W = F ·  r = F  r cos  = F r  r  F r rr r displacement FrFr “Dot Product”

26. Physics 1501: Lecture 11, Pg 26 Definition of Work... l Only the component of F along the displacement is doing work. çExample: Train on a track. r rr r F  F cos 

27. Physics 1501: Lecture 11, Pg 27 Review: Scalar Product ( or Dot Product) Definition: a b a · b= ab cos  = a[b cos  ] = ab a = b[a cos  ] = ba b Some properties: a bb a a ·b= b ·a a bba b a q(a ·b) = (qb) · a = b · (qa) (q is a scalar) a b ca b a c c a · (b + c) = (a ·b) + (a ·c) (c is a vector) The dot product of perpendicular vectors is 0 !!  a abab b  a b baba

28. Physics 1501: Lecture 11, Pg 28 Review: Examples of dot products SupposeThen a i j k a = 1 i + 2 j + 3 k b i j k b = 4 i - 5 j + 6 k ab a · b = 1x4 + 2x(-5) + 3x6 = 12 aa a · a = 1x1 + 2x2 + 3x3 = 14 bb b · b = 4x4 + (-5)x(-5) + 6x6 = 77 i i j j k k i · i = j · j = k · k = 1 i j j k k i i · j = j · k = k · i = 0 x y z i j k

29. Physics 1501: Lecture 11, Pg 29 Review: Properties of dot products l Magnitude: a 2 = |a| 2 = a · a i j i j = (a x i + a y j ) · (a x i + a y j ) i i j j i j = a x 2 ( i · i ) + a y 2 ( j · j ) + 2a x a y ( i · j ) = a x 2 + a y 2 çPythagorian Theorem !! a axax ayay i j

30. Physics 1501: Lecture 11, Pg 30 Review: Properties of dot products l Components: a i j ka ia ja k a = a x i + a y j + a z k = (a x, a y, a z ) = (a · i, a · j, a · k ) l Derivatives: çApply to velocity çSo if v is constant (like for UCM):

31. Physics 1501: Lecture 11, Pg 31 Back to the definition of Work: F Work, W, of a force F acting r through a displacement  r is: F r W = F ·  r F r rr r