Download presentation

Presentation is loading. Please wait.

Published byChristopher Brown Modified about 1 year ago

1
Chapter 6 – Circular Motion and Other applications on Newton’s Laws Uniform circular motion.Uniform circular motion. Drag forces and terminal speed.Drag forces and terminal speed. Other Applications of Newton’s LawsOther Applications of Newton’s Laws

2
Uniform Circular Motion A force, F r, is directed toward the center of the circleA force, F r, is directed toward the center of the circle This force is associated with an acceleration, a cThis force is associated with an acceleration, a c Applying Newton’s Second Law along the radial direction givesApplying Newton’s Second Law along the radial direction gives

3
Uniform Circular Motion A force causing a centripetal acceleration acts toward the center of the circleA force causing a centripetal acceleration acts toward the center of the circle It causes a change in the direction of the velocity vectorIt causes a change in the direction of the velocity vector If the force vanishes, the object would move in a straight-line path tangent to the circleIf the force vanishes, the object would move in a straight-line path tangent to the circle

4
Centripetal Force The force causing the centripetal acceleration is sometimes called the centripetal forceThe force causing the centripetal acceleration is sometimes called the centripetal force This is not a new force, it is a new role for a forceThis is not a new force, it is a new role for a force It is a force acting in the role of a force that causes a circular motionIt is a force acting in the role of a force that causes a circular motion

5
Conical Pendulum The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal directionThe object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction (1) Tcosθ = mg (2) Tsinθ = ma c = mv 2 /r

6
Conical Pendulum Dividing (2) by (1) and using sinθ/cosθ = tanθ we eliminate T and find:Dividing (2) by (1) and using sinθ/cosθ = tanθ we eliminate T and find: v is independent of mv is independent of m Using r = Lsinθ

7
Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00 with the vertical. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.

8
A 4.00-kg object is attached to a vertical rod by two strings, as in Figure. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

9
How fast can it spin? The speed at which the object moves depends on the mass of the object and the tension in the cord. The centripetal force is supplied by the tension:The speed at which the object moves depends on the mass of the object and the tension in the cord. The centripetal force is supplied by the tension: This shows that v increases with T and decreases with larger m.

10
Horizontal (Flat) Curve The force of static friction supplies the centripetal forceThe force of static friction supplies the centripetal force The maximum speed at which the car can negotiate the curve isThe maximum speed at which the car can negotiate the curve is Note, this does not depend on the mass of the carNote, this does not depend on the mass of the car FrFr

11
Banked Curve These are designed with friction equaling zeroThese are designed with friction equaling zero There is a component of the normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2).There is a component of the normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2). Dividing (1) by (2) gives:Dividing (1) by (2) gives:

12
Loop-the-Loop A pilot in a jet aircraft executes a loop-the loop. This is an example of a vertical circleA pilot in a jet aircraft executes a loop-the loop. This is an example of a vertical circle At the bottom of the loop, (b), the upward constant force, that keeps the pilot moving in a circular path at a constant speed, is greater than its weight, because the normal and gravitational forces act in opposite direction:At the bottom of the loop, (b), the upward constant force, that keeps the pilot moving in a circular path at a constant speed, is greater than its weight, because the normal and gravitational forces act in opposite direction:

13
Loop-the-Loop At the top of the circle, (c), the force exerted on the object is less than its weight, because both the gravitational force and the normal force, n top, exerted on the pilot by the seat act in same direction:At the top of the circle, (c), the force exerted on the object is less than its weight, because both the gravitational force and the normal force, n top, exerted on the pilot by the seat act in same direction:

14
A kg object is swung in a vertical circular path on a string m long. If its speed is 4.00 m/s at the top of the circle, what is the tension in the string there?

15
Non-Uniform Circular Motion When the force acting on a particle moving in a circular path has a tangential component, the particles speed changes. The acceleration has a tangential components F rF r produces the centripetal acceleration F t produces the tangential acceleration. The total force is the vector sum or the radial force and tangential force:

16
Vertical Circle with Non-Uniform Speed The gravitational force exerts a tangential force on the objectThe gravitational force exerts a tangential force on the object –Look at the components of F g The gravitational force resolve into a tangential component mg sinθ and a radial component mg cosθ.The gravitational force resolve into a tangential component mg sinθ and a radial component mg cosθ.

17
Vertical Circle with Non-Uniform Speed The tension at any point can be found: Applying II NL to the tangential and radial directions:Applying II NL to the tangential and radial directions:

18
Tarzan (m = 85.0 kg) tries to cross a river by swinging from a vine. The vine is 10.0 m long, and his speed at the bottom of the swing (as he just clears the water) will be 8.00 m/s. Tarzan doesn't know that the vine has a breaking strength of N. Does he make it safely across the river?

19
Motion in accelerated Frames When Newton’s laws of motion was introduced in Chapter 5, we emphasized that they are valid only for inertial frames of reference. In this section, we will analyze the noninertial frames, that is, on that is accelerating. Example: Let’s consider a hockey puck on a table in a moving train. The train moving with a constant velocity represents an inertial frame. The puck at rest remains at rest, and Newton’s I low is obeyed.

20
Motion in accelerated Frames The accelerating train is not an inertial frame. For the observer on the train, there appears to be no visible force on the puck, but it will accelerate from rest toward the back of the train, as the train start to accelerate. The Newton’s I law is violated. The observer on the accelerating train, if he applied the N II law to the puck, might conclude that a force has acted on the puck to cause it to accelerate. We call an apparent force such as this a fictitious force.

21
Motion in Accelerated Frames A fictitious force results from an accelerated frame of referenceA fictitious force results from an accelerated frame of reference –A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force

22
“Centrifugal” Force From the frame of the passenger (b), a force appears to push her toward the doorFrom the frame of the passenger (b), a force appears to push her toward the door From the frame of the Earth, the car applies a leftward force on the passengerFrom the frame of the Earth, the car applies a leftward force on the passenger The outward force is often called a centrifugal forceThe outward force is often called a centrifugal force –It is a fictitious force due to the acceleration associated with the car’s change in direction

23
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μ s = 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?

24
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration of the elevator.

25
Coriolis Force” “ Coriolis Force” This is an apparent force caused by changing the radial position of an object in a rotating coordinate systemThis is an apparent force caused by changing the radial position of an object in a rotating coordinate system The result of the rotation is the curved path of the ballThe result of the rotation is the curved path of the ball

26
The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

27
Fictitious Forces, examples Although fictitious forces are not real forces, they can have real effectsAlthough fictitious forces are not real forces, they can have real effects Examples:Examples: –Objects in the car do slide –You feel pushed to the outside of a rotating platform –The Coriolis force is responsible for the rotation of weather systems and ocean currents

28
Fictitious Forces in Linear Systems The inertial observer (a) seesThe inertial observer (a) sees The noninertial observer (b) seesThe noninertial observer (b) sees

29
Fictitious Forces in a Rotating System According to the inertial observer (a), the tension is the centripetal forceAccording to the inertial observer (a), the tension is the centripetal force The noninertial observer (b) seesThe noninertial observer (b) sees

30
Motion with Resistive Forces Motion can be through a mediumMotion can be through a medium –Either a liquid or a gas The medium exerts a resistive force, R, on an object moving through the mediumThe medium exerts a resistive force, R, on an object moving through the medium The magnitude of R depends on the mediumThe magnitude of R depends on the medium The direction of R is opposite the direction of motion of the object relative to the mediumThe direction of R is opposite the direction of motion of the object relative to the medium R nearly always increases with increasing speedR nearly always increases with increasing speed

31
Motion with Resistive Forces The magnitude of R can depend on the speed in complex waysThe magnitude of R can depend on the speed in complex ways We will discuss only twoWe will discuss only two –R is proportional to v Good approximation for slow motions or small objectsGood approximation for slow motions or small objects –R is proportional to v 2 Good approximation for large objectsGood approximation for large objects

32
R Proportional To v The resistive force can be expressed as R = - b vThe resistive force can be expressed as R = - b v b depends on the property of the medium, and on the shape and dimensions of the objectb depends on the property of the medium, and on the shape and dimensions of the object The negative sign indicates R is in the opposite direction to vThe negative sign indicates R is in the opposite direction to v

33
R Proportional to v, Example Analyzing the motion results inAnalyzing the motion results in

34
R Proportional to v Initially, v = 0 and dv/dt = gInitially, v = 0 and dv/dt = g As t increases, R increases and a decreasesAs t increases, R increases and a decreases The acceleration approaches 0 when R → mgThe acceleration approaches 0 when R → mg At this point, v approaches the terminal speed of the objectAt this point, v approaches the terminal speed of the object

35
Terminal Speed To find the terminal speed, let a = 0 Solving the differential equation gives is the time constant and = m/b

36
For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speedFor objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed R = ½ D Av 2 –D is a dimensionless empirical quantity that called the drag coefficient is the density of air –A is the cross-sectional area of the object –v is the speed of the object R Proportional to v 2

37
Analysis of an object falling through air accounting for air resistanceAnalysis of an object falling through air accounting for air resistance

38
R Proportional to v 2, Terminal Speed The terminal speed will occur when the acceleration goes to zeroThe terminal speed will occur when the acceleration goes to zero Solving the equation givesSolving the equation gives

39
Some Terminal Speeds

40
Process for Problem-Solving Analytical MethodAnalytical Method –The process used so far involves the identification of well-behaved functional expressions generated from algebraic manipulation or techniques of calculus

41
Analytical Method Apply the method using this procedure:Apply the method using this procedure: –Sum all the forces acting on the particle to find the net force, F. –Use this net force to determine the acceleration from the relationship a = F/m –Use this acceleration to determine the velocity from the relationship dv/dt = a –Use this velocity to determine the position from the relationship dx/dt = v

42
Analytic Method, Example Applying the procedure:Applying the procedure: F g = ma y = - mg a y = -g and dv y /dt = -g v y (t) = v yi – gt y(t) = y i + v yi t – ½ gt 2

43
Numerical Modeling In many cases, the analytic method is not sufficient for solving “real” problemsIn many cases, the analytic method is not sufficient for solving “real” problems Numerical modeling can be used in place of the analytic method for these more complicated situationsNumerical modeling can be used in place of the analytic method for these more complicated situations The Euler method is one of the simplest numerical modeling techniquesThe Euler method is one of the simplest numerical modeling techniques

44
Euler Method In the Euler Method, derivatives are approximated as ratios of finite differences t is assumed to be very small, such that the change in acceleration during the time interval is also very small

45
Equations for the Euler Method

46
Euler Method It is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations into a tableIt is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations into a table Many small increments can be taken, and accurate results can be obtained by a computerMany small increments can be taken, and accurate results can be obtained by a computer

47
Euler Method Set Up

48
Euler Method Final One advantage of the method is that the dynamics are not obscured –The relationships among acceleration, force, velocity and position are clearly shown The time interval must be small –The method is completely reliable for infinitesimally small time increments –For practical reasons a finite increment must be chosen –A time increment can be chosen based on the initial conditions and used throughout the problem In certain cases, the time increment may need to be changed within the problem

49
Accuracy of the Euler Method The size of the time increment influences the accuracy of the resultsThe size of the time increment influences the accuracy of the results It is difficult to determine the accuracy of the result without knowing the analytical solutionIt is difficult to determine the accuracy of the result without knowing the analytical solution One method of determining the accuracy of the numerical solution is to repeat the solution with a smaller time increment and compare the resultsOne method of determining the accuracy of the numerical solution is to repeat the solution with a smaller time increment and compare the results –If the results agree, the results are correct to the precision of the number of significant figures of agreement

50
Euler Method, Numerical Example

51

52
II. Drag force and terminal speed -Fluid: anything that can flow. Example: gas, liquid. -Drag force: D - Appears when there is a relative velocity between a fluid and a body. Opposes the relative motion of a body in a fluid. Opposes the relative motion of a body in a fluid. - Points in the direction in which the fluid flows.

53
Terminal speed: v t - Terminal speed: v t Assumptions: * Fluid = air. * Body is blunt (baseball). * Fast relative motion turbulent air. C = drag coefficient (0.4-1). ρ = air density (mass/volume). A = effective body’s cross sectional area area perpendicular to v - Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero F g =D

54
III. Uniform circular motion -Centripetal acceleration: A centripetal force accelerates a body by changing the direction of the body’s velocity without changing its speed. v, a = const, but direction changes during motion v, a = const, but direction changes during motion. -Centripetal force: a, F are directed toward the center of curvature of the particle’s path.

55
A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed keeps the cylinder at rest? 49. A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed keeps the cylinder at rest? Mg mg T T N

56
33E. Calculate the drag force on a missile 53cm in diameter cruising with a speed of 250m/s at low altitude, where the density of air is 1.2kg/m 3. Assume C=0.75

57
The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-dive position. Assuming that the diver’s drag coefficient C does not change from one point to another, find the ratio of the effective cross sectional area A in the slower position to that of the faster position. 32. The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-dive position. Assuming that the diver’s drag coefficient C does not change from one point to another, find the ratio of the effective cross sectional area A in the slower position to that of the faster position.

58
Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. F gA N F gB T1T1 f T3T3 T1T1 T2T2 T3T3

59
Two blocks of weights 3.6N and 7.2N, are connected by a massless string and slide down a 30º inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string. (c) Describe the motion if, instead, the heavier block leads. A B F gA F gB NANA NBNB T T f k,B f k,A Movement Block A Block B NANA NBNB f kA f kB T F gxA F gyA F gxB F gyB T Light block A leads

60
A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N. The coefficient of static friction between the wall and the block is 0.55 and the coefficient of kinetic friction between them is A second force P acting parallel to the wall is applied to the block. For the following magnitudes and directions of P, 74. A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N. The coefficient of static friction between the wall and the block is 0.55 and the coefficient of kinetic friction between them is A second force P acting parallel to the wall is applied to the block. For the following magnitudes and directions of P, determine whether the block moves, the direction of motion, and the magnitude and direction of the frictional force acting on the block: (a) 34N up, (b) 12N up, (c) 48N up, (d) 62N up, (e) 10N down, (f) 18N down. F=60N mg=22N N N=F=60N P 22N f

61
Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ s between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μ k between A and the table is 0.15? 28. Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ s between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μ k between A and the table is 0.15? T T W B =22N W A= 44N f WcWc N

62
The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficient of static friction between the blocks is: μ s =0.38 but the surface beneath the larger block is frictionless. 29. The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficient of static friction between the blocks is: μ s =0.38 but the surface beneath the larger block is frictionless. What is the minimum value of the horizontal force F required to keep the smaller block from slipping What is the minimum value of the horizontal force F required to keep the smaller block from slipping down the larger block? down the larger block? Mg N F’ mg f F’ Movement F min required to keep m from sliding down? Treat both blocks as a single system sliding across a frictionless floor

63
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weigh of the car and riders is 5kN, and the radius of the circle is 10m. 44. An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weigh of the car and riders is 5kN, and the radius of the circle is 10m. What are the magnitude and the direction of the force of the boom on the car at the top of the circle if the car’s speed is (a) 5m/s (b) 12m/s? What are the magnitude and the direction of the force of the boom on the car at the top of the circle if the car’s speed is (a) 5m/s (b) 12m/s? The force of the boom on the car is capable of pointing any direction FBFB W y

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google