Physics 1501: Lecture 10

1. Physics 1501: Lecture 10 • Announcements • Homework #3 : due next Monday • But HW 04 will be due the following Friday (same week). • Midterm 1: Monday Oct. 3 • Topics • Review: Forces and uniform circular motion • Forces and non-uniform circular motion • Accelerated reference frames • Work & Energy • Scalar Product

2. Nonuniform Circular Motion Earlier we saw that for an object moving in a circle with nonuniform speed then a = ar + at . at ar What are Fr and Ft ?

3. Lecture 10, ACT 1 • 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. NII mg Fc Fc 2.7 km Fc NI mg Lecture 10, ACT 1Solution Fc = NII + mg Fc = m ar = mv2 / R NII = mg (1- v2 / Rg ) NII < mg ANSWER (C) II) Fc = NI - mg Fc = m ar = mv2 / R NI = mg (1+ v2 / Rg ) NI > mg ANSWER (A) I)

5. Example Exercise 1 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 ??

6. q mg q N q q mg Example Exercise 1 Radial : Fr = N + mg cosq = mv2/R Tangential : Ft = mg sinq Solve the first for v.

7. N q mg Example Exercise 1 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, q = 0.

8. Example Exercise 2 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.

9. Example Exercise 3 My match box car is going to do a loop the loop. If the speed at the bottom is vB, what is the normal force at that point ??

10. mg Example Exercise 3 Radial : Fr = N + mg cosq = mv2/R Tangential : Ft = mg sinq Solve the first for N. q At the bottom, v = vB, and q = p. N

11. Lecture 10, ACT 2Nonuniform Circular Motion 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 v1. 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 ?? A) 1/2 v1 B) v1 C) 2 v1 D) 4 v1

12. Accelerated Reference Frames:The Accelerometer • 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. a  i Accelerated Reference Frames:The Accelerometer 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. TX  T TY y  m x mg (gravitational force) a Accelerometer... 2. Draw a free-body diagram for the mass: We wish to solve for  in terms of the acceleration a. We will use SF = ma in two cartesian dimensions.

15. TX TY  j i Accelerometer... 3. Solving, i: FX = TX = T sin = ma j: FY = TY - mg = T cos - mg = 0 • Eliminate T  T m ma T sin = ma tan q = a / g mg T cos = mg

16. Accelerometer... 4. No Numbers involved 5. This answer has the right units (none) It does give 1 in terms of the acceleration, tan q = a / g

17. Accelerometer – Other Thoughts 1 • Alternative solution using vectors (more elegant): • Find the total vector force FTOT: T (string tension) T  mg  m FTOT mg (gravitational force)

18. Accelerometer – Other Thoughts 1 • Alternative solution using vectors (more elegant): • Find the total vector force FTOT: • Recall that FTOT= ma: • So T (string tension)  T mg  m ma mg (gravitational force)

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

20. Accelerometer... • There must be some other force to balance the x component i: FX = TX + F ? = T sin + F ?= 0 j: FY = TY - mg = T cos- mg = 0 • And we must put F ? = -ma to get the same answer as before. • F? is known as a fictitious force.  T a mg

21. Lecture 10,ACT 3Accelerated 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. Chap.6: Work & Energy • One of the most important concepts in physics. • Alternative approach to mechanics. • Many applications beyond mechanics. • Thermodynamics (movement of heat). • Quantum mechanics... • Very useful tools. • You will learn new (sometimes much easier) ways to solve problems.

23. Forms of Energy • Kinetic: Energy of motion. • A car on the highway has kinetic energy. • We have to remove this energy to stop it. • The breaks of a car get HOT! • This is an example of turning one form of energy into another (thermal energy). • Kinetic energy is given by: K = (1/2) mv2

24. Energy Conservation • Energy cannot be destroyed or created. • Just changed from one form to another. • We say energy is conserved ! • True for any isolated system. • i.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. • The energy of the car “alone” is not conserved... • It is reduced by the braking. • Doing “work” on an otherwise isolated system will change its “energy”...

25. Definition of Work: Ingredients: Force (F), displacement ( r) Work, W, of a constant force F acting through a displacement r is: W = F·r = Frcos  = Frr F r  Fr displacement “Dot Product”

26. F r  Fcos  Definition of Work... • Only the component of Falong the displacement is doing work. • Example: Train on a track.

27. Review: Scalar Product ( or Dot Product) a Definition: a · b = ab cos  = a[b cos ] = aba = b[a cos ] = bab Some properties: a ·b =b ·a q(a ·b) = (qb) · a = b·(qa)(q is a scalar) a ·(b + c) = (a ·b)+ (a·c)(c is a vector) The dot product of perpendicular vectors is 0 !! ba  b a  b ab

28. y j x i k z Review: Examples of dot products i ·i = j ·j = k ·k = 1 i · j = j ·k = k ·i = 0 Suppose Then a = 1 i + 2 j + 3 k b = 4 i - 5 j + 6 k a·b = 1x4 + 2x(-5) + 3x6 = 12 a·a = 1x1 + 2x2 + 3x3 = 14 b·b = 4x4 + (-5)x(-5) + 6x6 = 77

29. a ay ax j i Review: Properties of dot products • Magnitude: a2 = |a|2 = a ·a =(axi + ay j ) ·(axi + ayj ) = ax2( i ·i ) + ay2( j ·j ) + 2axay( i ·j ) = ax2 + ay2 • Pythagorian Theorem !!

30. Review: Properties of dot products • Components: a = ax i + ay j + az k = (ax , ay , az ) = (a· i , a· j , a· k ) • Derivatives: • Apply to velocity • So if v is constant (like for UCM):

31. Back to the definition of Work: Work, W, of a force Facting through a displacement ris: W = F·r F r