Presentation on theme: "Chapter 6 – Circular Motion and Other applications on Newton’s Laws"— Presentation transcript:
1 Chapter 6 – Circular Motion and Other applications on Newton’s Laws Uniform circular motion.Drag forces and terminal speed.Other Applications of Newton’s Laws
2 Uniform Circular Motion A force, Fr , is directed toward the center of the circleThis force is associated with an acceleration, acApplying 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 circleIt causes a change in the direction of the velocity vectorIf the force vanishes, the object would move in a straight-line path tangent to the circle
4 Centripetal ForceThe force causing the centripetal acceleration is sometimes called the centripetal forceThis is not a new force, it is a new role for a forceIt is a force acting in the role of a force that causes a circular motion
5 Conical PendulumThe object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction(1) Tcosθ = mg(2) Tsinθ = mac= mv2/r
6 Conical PendulumDividing (2) by (1) and using sinθ/cosθ = tanθ we eliminate T and find:v is independent of mUsing r = Lsinθ
7 Consider a conical pendulum with an 80. 0-kg bob on a 10 Consider a conical pendulum with an 80.0-kg bob on a 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: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 maximum speed at which the car can negotiate the curve isNote, this does not depend on the mass of the carFr
11 Banked Curve These 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).Dividing (1) by (2) gives:
12 Loop-the-LoopA pilot in a jet aircraft executes a loop-the loop. This is an example of a vertical circleAt 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-LoopAt 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, ntop, exerted on the pilot by the seat act in same direction:
14 A 0. 400-kg object is swung in a vertical circular path on a string 0 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 componentsFr produces the centripetal accelerationFt 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 objectLook at the components of FgThe gravitational force resolve into a tangential component mg sinθ and a radial component mg cosθ.
17 Vertical Circle with Non-Uniform Speed Applying II NL to the tangential and radial directions:The tension at any point can be found:
18 Tarzan (m = 85. 0 kg) tries to cross a river by swinging from a vine 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 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” ForceFrom the frame of the passenger (b), a force appears to push her toward the doorFrom the frame of the Earth, the car applies a leftward force on the passengerThe outward force is often called a centrifugal forceIt 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 μs = 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 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”This is an apparent force caused by changing the radial position of an object in a rotating coordinate systemThe 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 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 effectsExamples:Objects in the car do slideYou feel pushed to the outside of a rotating platformThe Coriolis force is responsible for the rotation of weather systems and ocean currents
28 Fictitious Forces in Linear Systems The inertial observer (a) seesThe noninertial observer (b) sees
29 Fictitious Forces in a Rotating System According to the inertial observer (a), the tension is the centripetal forceThe noninertial observer (b) sees
30 Motion with Resistive Forces Motion can be through a mediumEither a liquid or a gasThe medium exerts a resistive force, R, on an object moving through the mediumThe magnitude of R depends on the mediumThe direction of R is opposite the direction of motion of the object relative to the mediumR nearly always increases with increasing speed
31 Motion with Resistive Forces The magnitude of R can depend on the speed in complex waysWe will discuss only twoR is proportional to vGood approximation for slow motions or small objectsR is proportional to v2Good approximation for large objects
32 R Proportional To v The 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 objectThe negative sign indicates R is in the opposite direction to v
33 R Proportional to v, Example Analyzing the motion results in
34 R Proportional to v Initially, v = 0 and dv/dt = g As t increases, R increases and a decreasesThe acceleration approaches 0 when R → mgAt 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 givest is the time constant and t = m/b
36 R Proportional to v2For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speedR = ½ DrAv2D is a dimensionless empirical quantity that called the drag coefficient- r is the density of airA is the cross-sectional area of the objectv is the speed of the object
37 R Proportional to v2Analysis of an object falling through air accounting for air resistance
38 R Proportional to v2, Terminal Speed The terminal speed will occur when the acceleration goes to zeroSolving the equation gives
40 Process for Problem-Solving Analytical MethodThe 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: Sum all the forces acting on the particle to find the net force, SF.Use this net force to determine the acceleration from the relationship a =SF/mUse this acceleration to determine the velocity from the relationship dv/dt = aUse this velocity to determine the position from the relationship dx/dt = v
42 Analytic Method, Example Applying the procedure:Fg = may = - mgay = -g and dvy/dt = -gvy(t) = vyi – gty(t) = yi + vyi t – ½ gt2
43 Numerical ModelingIn many cases, the analytic method is not sufficient for solving “real” problemsNumerical modeling can be used in place of the analytic method for these more complicated situationsThe Euler method is one of the simplest numerical modeling techniques
44 Euler MethodIn the Euler Method, derivatives are approximated as ratios of finite differencesDt is assumed to be very small, such that the change in acceleration during the time interval is also very small
46 Euler MethodIt is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations into a tableMany small increments can be taken, and accurate results can be obtained by a computer
48 Euler Method FinalOne advantage of the method is that the dynamics are not obscuredThe relationships among acceleration, force, velocity and position are clearly shownThe time interval must be smallThe method is completely reliable for infinitesimally small time incrementsFor practical reasons a finite increment must be chosenA time increment can be chosen based on the initial conditions and used throughout the problemIn 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 resultsIt is difficult to determine the accuracy of the result without knowing the analytical solutionOne method of determining the accuracy of the numerical solution is to repeat the solution with a smaller time increment and compare the resultsIf the results agree, the results are correct to the precision of the number of significant figures of agreement
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.- Points in the direction in which the fluid flows.
53 * Body is blunt (baseball). * Fast relative motion turbulent air. 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-Terminal speed: vt- Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero Fg=D
54 III. Uniform circular motion Centripetal acceleration:v, a = const, but direction changes during motion.A centripetal force accelerates a body by changing the direction of the body’s velocity without changing its speed.Centripetal force:a, F are directed toward the centerof curvature of the particle’s path.
55 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?MgmgTN
56 33E. Calculate the drag force on a missile 53cm in diameter cruising with a speed of 250m/s at low altitude, where thedensity of air is 1.2kg/m3. Assume C=0.75
57 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 assume that the cord between B and Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25;assume that the cord between B andthe knot is horizontal. Find the maximumweight of block A for which the systemwill be stationary.NFgBT1fT1T2T3T3FgA
59 Block A Block B Light block A leads 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.Block ABlock BNANBABFgAFgBNANBTfk,Bfk,AMovementFgxATFgxBfkBfkATFgyAFgyBLight block A leads
60 determine whether the block moves, the direction of motion, 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 thefrictional force acting on the block:(a) 34N up, (b) 12N up, (c) 48N up,(d) 62N up, (e) 10N down, (f) 18N down.F=60NNPmg=22NN=F=60N22Nf
61 28. Blocks A and B have weights of 44N and 22N, respectively 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 μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15?NfTWcWA=44NTWB=22N
62 Fmin required to keep m from sliding down? 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: μs=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 slippingdown the larger block?NfF’F’mgFmin required to keep m from sliding down?MgTreat both blocks as a single system sliding across a frictionless floorMovement
63 The force of the boom on the car is capable of pointing any direction 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?The force of the boom on the car is capable of pointing any directionyFBW