Physics 111. Tuesday, September 28, Circular Motion - centripetal acceleration Friction Tension - the massless string

Transcription

1 ics Tuesday, ember 28, 2004 Ch 6: Circular Motion - centripetal acceleration Friction Tension - the massless string

2 Announcements Help this week: Wednesday, 8-9 pm in NSC 128/119 Sunday, 6:30-8 pm in CCLIR 468

3 Announcements Don t forget to read over the lab write-up and be ready for the quiz.

4 Ch 4: Newton s Laws Note the following: How does your weight change while you re Ascending? Descending? Decelerating? Accelerating?

5 Applications of Newton s Laws Now that we have seen and applied Newton s Laws (at least a few times), let s introduce a couple of new forces. I will NOT cover everything in this chapter

6 v = constant But... v constant Uniform Circular Motion: the speed of the object does not change, but the object is accelerating throughout the motion.

7 For an object moving in a circle, the instantaneous velocity vectors are always tangent to the circle of motion! Uniform Circular Motion: the speed of the object does not change, but the object is accelerating throughout the motion.

8 Although our objects moves in a circle at constant speed, it still accelerates. WHY? Recall that acceleration and velocity are both vector quantities. Since acceleration is a change in velocity over a change in time, an object with a velocity that changes direction is accelerating!

9 v r a = c v r 2 ac v where r is the radius of the circle and v the speed of the object. Proof in the book and in class notes.

10 Recall our definition of acceleration: v i a = lim Δ v Δt 0 Δt = lim t f t i v f v i t f t i r i Δ r r f v f r i r f Δθ Color Code: velocity; position in yellow; acceleration

11 Velocity Vectors Again, we re using vector subtraction (remember, subtracting one vector from another is like adding its negative). By geometry, this angle is the same as that on the last slide between the position vectors! Δθ v i v f Δv Now, we ve got two similar, isosceles triangles.

12 r i Δ r r f Δθ Δ r r = Δ v v Δθ v i v f Δv Now, returning to our definition of acceleration a c = lim Δ v Δt 0 Δt = v Δr lim Δt 0 (Δt) r = v 2 r Δ v Notice that as Δt 0, the vector points towards the exact center of the circle.

13 For an object moving in a circle with a constant speed, we can define a period associated with the motion. The period of the rotation is simply the time required for the object to go around the circle exactly one time. T d = = v 2πr v

14 T = 2πr v [T ] = [2πr] [ v ] That s good. We knew period had to have units of time. [T ] = m m/s = s

15 A ball of mass m on the end of a string is moving in a circle. We know that the ball is experiencing an acceleration. T top view v a c m Newton s 2nd Law tells us that in order for an object to experience an acceleration, a net external force must be acting.

16 Worksheet Problem #1

17 Tues A rider in a barrel of fun finds herself stuck with her back to the wall.which diagram correctly shows the forces acting on her? Worksheet Problem #1 PI, Mazur (1997)

18 NOTE! Newton s 2nd Law takes on a slightly different form for problems involving circular motion. In these problems, Newton s 2nd Law says mv 2 F net = ma = c r (inward)

19 An automobile with mass 1500-kg moves at a constant speed over the crest of a hill. The driver moves in a vertical circle of radius 28.0 m. At the top of the hill, she notices that the car barely maintains contact with the road. Find the speed of the vehicle. Motion Diagram: Problem Solving Worksheet v a

20 An automobile with mass 1500-kg moves at a constant speed over the crest of a hill. The driver moves in a vertical circle of radius 28.0 m. At the top of the hill, she notices that the car barely maintains contact with the road. Find the speed of the vehicle. Pictorial Representation: W = weight of car N = normal force of road Unknown: v +y +x g r Knowns: m = 1500 kg g = m/s 2 r = 28.0 m N = 0 v ŷ

21 An automobile with mass 1500-kg moves at a constant speed over the crest of a hill. The driver moves in a vertical circle of radius 28.0 m. At the top of the hill, she notices that the car barely maintains contact with the road. Find the speed of the vehicle. Free-Body Diagram: W

22 An automobile with mass 1500-kg moves at a constant speed over the crest of a hill. The driver moves in a vertical circle of radius 28.0 m. At the top of the hill, she notices that the car barely maintains contact with the road. Find the speed of the vehicle. Mathematical Representation: Write Newton s 2nd Law a = F net m = W m = mgŷ m = gŷ Use definition of centripetal acceleration a c = v2 r ( ŷ)

23 An automobile with mass 1500-kg moves at a constant speed over the crest of a hill. The driver moves in a vertical circle of radius 28.0 m. At the top of the hill, she notices that the car barely maintains contact with the road. Find the speed of the vehicle. Mathematical Representation: a c = v2 r ŷ = gŷ = a v = rg = (28.0m)(9.81 m ) = 13.3 s m 2 s

24 Let s look at a couple of demos involving blocks sliding across the floor. Again, use FBD s to help you understand the motion. Class Worksheet #2: Block Predictions Block across the floor at constant speed. Why is it not accelerating? Block across the floor after push. Why does it stop?

25 What is friction? A macroscopic (meaning, observable to the eye) force that operates at the interface between two surfaces in a direction to oppose the slipping of those surfaces across one another. As I push on the red block with some force, friction with the surface of the blue block acts to prevent the red block from sliding.

26 What is friction? First, make a pictorial representation and describe the forces present. N = normal force of bottom block on top block F ext = the external force pushing the block f s = frictional force of bottom block on top block W = Weight or the gravitational force of Earth on block

27 What is friction? Now let s draw the FBD for the red block F ext n W f s

28 f s F ext The frictional force at the boundary is described as the force of static friction, as it relates to the force observed on a pair of objects that do not slip relative to one another. We find that experimentally, the magnitude of the maximum force of static friction is proportional to the normal force exerted by the blue block on the red block.

29 f s F ext The frictional force at the boundary is described as the force of static friction, as it relates to the force observed on a pair of objects that do not slip relative to one another. The constant of proportionality is known as the coefficient of static friction (µ s ) and is a property of the materials at the interface.

30 a f k F ext If I push hard enough, the force of static friction will be broken and slipping will occur at the interface. The red block will begin to accelerate relative to the blue block. When the red block slips, we note that it still feels a frictional force. The magnitude of that force, however, has changed.

31 a f k F ext If I push hard enough, the force of static friction will be broken and slipping will occur at the interface. The red block will begin to accelerate relative to the blue block. We find experimentally that the the maximum value of new frictional force is still proportional to the normal force exerted by the blue block on the red block.

32 a f k F ext This force is described as the force of kinetic friction, as it relates to the force observed on two objects that slip relative to one another. The constant of proportionality is known as the coefficient of kinetic friction (µ k ) and is a property of the materials at the interface.

33 Worksheet Problem #3

34 Tues Worksheet Problem #3 You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about PI, Mazur (1997) 1. four times as great 2. twice as great 3. equally great 4. half as great 5. one-fourth as great as the force required before you changed the crate s orientation.

35 F ext v F ext f s f k Static Case: Object at rest f µ n s s Kinetic Case: Object in Motion f k = µ k n The frictional forces are always directed so as to oppose the slippage between the two surfaces.

36 Worksheet Problem #4

37 Tues Worksheet Problem #4 An object is held in place by friction on an inclined surface. The angle of inclination is increased until the object starts moving. If the surface is kept at this angle, the object 1. slows down. 2. moves at uniform speed. 3. speeds up. 4. none of the above PI, Mazur (1997)

38 Does friction always result in an object slowing down? Let s look at a couple of examples to find out... What s going to happen to the green object as I push the blue book across the table? Worksheet Problem #5a Consult with neighbors.

39 Does friction always result in an object slowing down? Let s look at a couple of examples to find out... What happened to the green object this time as I pushed the blue book across the table? Worksheet Problem #5b Consult with neighbors.