Chapter 7 Section 1 Circular Motion Preview

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Chapter 7Section 1 Circular MotionPreview Objectives Tangential Speed Centripetal Acceleration Centripetal Force Describing a Rotating System Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionObjectives Solve problems involving centripetal acceleration. Solve problems involving centripetal force. Explain how the apparent existence of an outwardforce in circular motion can be explained as inertiaresisting the centripetal force. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionTangential Speed The tangential speed (vt) of an object in circularmotion is the object’s speed along an imaginary linedrawn tangent to the circular path. Tangential speed depends on the distance from theobject to the center of the circular path. When the tangential speed is constant, the motion isdescribed as uniform circular motion. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal AccelerationClick below to watch the Visual Concept.Visual Concept Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Acceleration The acceleration of an object moving in a circularpath and at constant speed is due to a change indirection. An acceleration of this nature is called a centripetalacceleration.CENTRIPETAL ACCELERATIONvt 2ac r(tangential speed)2centripetal acceleration radius of circular path Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Acceleration, continued (a) As the particle movesfrom A to B, the direction ofthe particle’s velocity vectorchanges. (b) For short time intervals, v is directed toward thecenter of the circle. Centripetal acceleration isalways directed toward thecenter of a circle. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Acceleration, continued You have seen that centripetal accelerationresults from a change in direction. In circular motion, an acceleration due to achange in speed is called tangentialacceleration. To understand the difference between centripetaland tangential acceleration, consider a cartraveling in a circular track.– Because the car is moving in a circle, the car has acentripetal component of acceleration.– If the car’s speed changes, the car also has a tangentialcomponent of acceleration. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Force Consider a ball of mass m that is being whirled in ahorizontal circular path of radius r with constant speed. The force exerted by the string has horizontal and verticalcomponents. The vertical component is equal andopposite to the gravitational force. Thus, the horizontalcomponent is the net force. This net force, which is is directed toward the center of thecircle, is a centripetal force. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Force, continuedNewton’s second law can be combined with theequation for centripetal acceleration to derive anequation for centripetal force:vt 2ac rmvt 2Fc mac rmass (tangential speed)2centripetal force radius of circular path Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Force, continued Centripetal force is simply the name given to thenet force on an object in uniform circular motion. Any type of force or combination of forces canprovide this net force.– For example, friction between a race car’s tiresand a circular track is a centripetal force thatkeeps the car in a circular path.– As another example, gravitational force is acentripetal force that keeps the moon in itsorbit. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionCentripetal Force, continued If the centripetal force vanishes, the object stopsmoving in a circular path. A ball that is on the end of astring is whirled in a verticalcircular path.– If the string breaks at the positionshown in (a), the ball will movevertically upward in free fall.– If the string breaks at the top of theball’s path, as in (b), the ball willmove along a parabolic path. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionDescribing a Rotating System To better understand the motion of a rotatingsystem, consider a car traveling at high speed andapproaching an exit ramp that curves to the left. As the driver makes the sharp left turn, thepassenger slides to the right and hits the door. What causes the passenger to move toward thedoor? Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 1 Circular MotionDescribing a Rotating System, continued As the car enters the ramp and travels along acurved path, the passenger, because of inertia,tends to move along the original straight path. If a sufficiently large centripetal force acts on thepassenger, the person will move along the samecurved path that the car does. The origin of thecentripetal force is the force of friction between thepassenger and the car seat. If this frictional force is not sufficient, the passengerslides across the seat as the car turns underneath. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationPreview Objectives Gravitational Force Applying the Law of Gravitation Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationObjectives Explain how Newton’s law of universal gravitationaccounts for various phenomena, including satelliteand planetary orbits, falling objects, and the tides. Apply Newton’s law of universal gravitation to solveproblems. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationGravitational Force Orbiting objects are in free fall. To see how this idea is true, we can use a thoughtexperiment that Newton developed. Consider acannon sitting on a high mountaintop.Each successive cannonballhas a greater initial speed, sothe horizontal distance thatthe ball travels increases. Ifthe initial speed is greatenough, the curvature ofEarth will cause thecannonball to continue fallingwithout ever landing. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationGravitational Force, continued The centripetal force that holds the planets in orbitis the same force that pulls an apple toward theground—gravitational force. Gravitational force is the mutual force of attractionbetween particles of matter. Gravitational force depends on the masses and onthe distance between them. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationGravitational Force, continued Newton developed the following equation to describequantitatively the magnitude of the gravitational forceif distance r separates masses m1 and m2:Newton's Law of Universal Gravitationm1m2Fg G 2rmass 1 mass 2gravitational force constant (distance between masses) 2 The constant G, called the constant of universalgravitation, equals 6.673 10–11 N m2/kg. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationNewton’s Law of Universal GravitationClick below to watch the Visual Concept.Visual Concept Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationGravitational Force, continued The gravitational forces that two masses exert oneach other are always equal in magnitude andopposite in direction. This is an example of Newton’s third law of motion. One example is the Earth-moon system, shown onthe next slide. As a result of these forces, the moon and Earth eachorbit the center of mass of the Earth-moon system.Because Earth has a much greater mass than themoon, this center of mass lies within Earth. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationNewton’s Law of Universal Gravitation Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationApplying the Law of Gravitation Newton’s law of gravitation accounts for ocean tides. High and low tides are partly due to the gravitationalforce exerted on Earth by its moon. The tides result from the difference between thegravitational force at Earth’s surface and at Earth’scenter. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationApplying the Law of Gravitation, continued Cavendish applied Newton’s law of universalgravitation to find the value of G and Earth’s mass. When two masses, the distance between them, andthe gravitational force are known, Newton’s law ofuniversal gravitation can be used to find G. Once the value of G is known, the law can be usedagain to find Earth’s mass. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 2 Newton’s Law ofUniversal GravitationApplying the Law of Gravitation, continued Gravity is a field force. Gravitational field strength,g, equals Fg/m. The gravitational field, g,is a vector with magnitudeg that points in the directionof Fg. Gravitational fieldThe gravitational field vectorsstrength equals free-fall represent Earth’s gravitationalfield at each point.acceleration. Houghton Mifflin Harcourt Publishing Company

Section 2 Newton’s Law ofUniversal GravitationChapter 7Applying the Law of Gravitation, continued weight mass gravitational field strength Because it depends on gravitational fieldstrength, weight changes with location:weight mgFgGmmE GmEg 22mmrr On the surface of any planet, the value of g, aswell as your weight, will depend on the planet’smass and radius. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpacePreview Objectives Kepler’s Laws Sample Problem Weight and Weightlessness Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceObjectives Describe Kepler’s laws of planetary motion. Relate Newton’s mathematical analysis ofgravitational force to the elliptical planetary orbitsproposed by Kepler. Solve problems involving orbital speed and period. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceKepler’s LawsKepler’s laws describe the motion of the planets. First Law: Each planet travels in an elliptical orbitaround the sun, and the sun is at one of the focalpoints. Second Law: An imaginary line drawn from the sunto any planet sweeps out equal areas in equal timeintervals. Third Law: The square of a planet’s orbital period(T2) is proportional to the cube of the averagedistance (r3) between the planet and the sun. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceKepler’s Laws, continued Kepler’s laws were developed a generation beforeNewton’s law of universal gravitation. Newton demonstrated that Kepler’s laws areconsistent with the law of universal gravitation. The fact that Kepler’s laws closely matchedobservations gave additional support for Newton’stheory of gravitation. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceKepler’s Laws, continuedAccording to Kepler’s second law, if the time aplanet takes to travel the arc on the left ( t1) is equalto the time the planet takes to cover the arc on theright ( t2), then the area A1 is equal to the area A2.Thus, the planettravels faster when itis closer to the sunand slower when it isfarther away. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceKepler’s Laws, continued Kepler’s third law states that T2 µ r3. The constant of proportionality is 4p2/Gm, where m isthe mass of the object being orbited. So, Kepler’s third law can also be stated as follows:2æ4p ö 32T ç rè Gm ø Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceKepler’s Laws, continued Kepler’s third law leads to an equation for the periodof an object in a circular orbit. The speed of an objectin a circular orbit depends on the same factors:r3T 2pGmmvt Gr Note that m is the mass of the central object that isbeing orbited. The mass of the planet or satellite that isin orbit does not affect its speed or period. The mean radius (r) is the distance between thecenters of the two bodies. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpacePlanetary Data Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceSample ProblemPeriod and Speed of an Orbiting ObjectMagellan was the first planetary spacecraft to belaunched from a space shuttle. During the spacecraft’sfifth orbit around Venus, Magellan traveled at a meanaltitude of 361km. If the orbit had been circular, whatwould Magellan’s period and speed have been? Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceSample Problem, continued1. DefineGiven:r1 361 km 3.61 105 mUnknown:T ?vt ?2. PlanChoose an equation or situation: Use the equations forthe period and speed of an object in a circular orbit.r3T 2pGm Houghton Mifflin Harcourt Publishing Companyvt Gmr

Chapter 7Section 3 Motion in SpaceSample Problem, continuedUse Table 1 in the textbook to find the values for theradius (r2) and mass (m) of Venus.r2 6.05 106 mm 4.87 1024 kgFind r by adding the distance between the spacecraftand Venus’s surface (r1) to Venus’s radius (r2).r r1 r2r 3.61 105 m 6.05 106 m 6.41 106 m Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceSample Problem, continued3. Calculater3(6.41 10 6 m)3T 2p 2pGm(6.673 10 –11 N m 2 /kg 2 )(4.87 10 24 kg)T 5.66 10 3 sGm(6.673 10 –11 N m 2 /kg 2 )(4.87 10 24 kg)vt r6.41 10 6 mvt 7.12 10 3 m/s4. EvaluateMagellan takes (5.66 103 s)(1 min/60 s) » 94 min to completeone orbit. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceWeight and WeightlessnessTo learn about apparent weightlessness, imagine thatyou are in an elevator:– When the elevator is at rest, the magnitude of thenormal force acting on you equals your weight.– If the elevator were to accelerate downward at 9.81m/s2, you and the elevator would both be in free fall.You have the same weight, but there is no normalforce acting on you.– This situation is called apparent weightlessness.– Astronauts in orbit experience apparentweightlessness. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 3 Motion in SpaceWeight and Weightlessness Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesPreview Objectives Rotational Motion The Magnitude of a Torque The Sign of a Torque Sample Problem Simple Machines Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesObjectives Distinguish between torque and force. Calculate the magnitude of a torque on an object. Identify the six types of simple machines. Calculate the mechanical advantage of a simplemachine. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesRotational Motion Rotational and translational motion can beanalyzed separately.– For example, when a bowling ball strikes the pins, the pinsmay spin in the air as they fly backward.– These pins have both rotational and translational motion. In this section, we will isolate rotational motion. In particular, we will explore how to measure theability of a force to rotate an object. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesThe Magnitude of a Torque Torque is a quantity that measures the ability of aforce to rotate an object around some axis. How easily an object rotates on both how muchforce is applied and on where the force is applied. The perpendicular distance from the axis of rotationto a line drawn along the direction of the force isequal to d sin q and is called the lever arm.t Fd sin qtorque force lever arm Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesThe Magnitude of a Torque, continued The applied force mayact at an angle. However, the direction ofthe lever arm (d sin q) isalways perpendicular tothe direction of theapplied force, as shownhere. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesTorqueClick below to watch the Visual Concept.Visual Concept Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesTorque and the Lever ArmIn each example, the cat is pushing on thedoor at the same distance from the axis. Toproduce the same torque, the cat mustapply greater force for smaller angles. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesThe Sign of a Torque Torque is a vector quantity. In this textbook, we willassign each torque a positive or negative sign,depending on the direction the force tends to rotatean object. We will use the convention that the sign of the torqueis positive if the rotation is counterclockwise andnegative if the rotation is clockwise.Tip: To determine the sign of a torque, imagine that the torqueis the only one acting on the object and that the object is free torotate. Visualize the direction that the object would rotate. Ifmore than one force is acting, treat each force separately. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesThe Sign of a TorqueClick below to watch the Visual Concept.Visual Concept Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSample ProblemTorqueA basketball is being pushed by two players during tipoff. One player exerts an upward force of 15 N at aperpendicular distance of 14 cm from the axis ofrotation.The second player applies a downward force of11 N at a distance of 7.0 cm from the axis of rotation.Find the net torque acting on the ball about its center ofmass. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSample Problem, continued1. DefineGiven:F1 15 Nd1 0.14 mF2 11 Nd2 0.070 mUnknown:tnet ?Diagram: Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSample Problem, continued2. PlanChoose an equation or situation: Apply thedefinition of torque to each force,and add up theindividual torques.t Fdtnet t1 t2 F1d1 F2d2Tip: The factor sin q is not included in the torqueequation because each given distance is theperpendicular distance from the axis of rotation to aline drawn along the direction of the force. In otherwords, each given distance is the lever arm. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSample Problem, continued3. CalculateSubstitute the values into the equation andsolve: First,determine the torque produced by eachforce.Use the standard convention for signs.t1 F1d1 (15 N)(–0.14 m) –2.1 N mt2 F2d2 (–11 N)(0.070 m) –0.77 N mtnet t1 t2 –2.1 N m – 0.77 N mtnet –2.9 N m4. EvaluateThe net torque is negative,so the ball rotates in aclockwise direction. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSimple Machines A machine is any device that transmits or modifiesforce, usually by changing the force applied to anobject. All machines are combinations or modifications of sixfundamental types of machines, called simplemachines. These six simple machines are the lever, pulley,inclined plane, wheel and axle, wedge, and screw. Houghton Mifflin Harcourt Publishing Company

Chapter 7Section 4 Torque and SimpleMachinesSimpl

Centripetal Acceleration" The acceleration of an object moving in a circular path and at constant speed is due to a change in direction." An acceleration of this nature is called a centripetal acceleration. CENTRIPETAL ACCELERATION ac vt 2 r centripetal acceleration (tangential speed)2 radius of circular path Section 1 Circular Motion

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