Presentation on theme: "Circular Motion and Gravitation"— Presentation transcript:
1 Circular Motion and Gravitation Chapter 7Circular MotionandGravitation
2 Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an accelerationThe centripetal acceleration is due to the change in the direction of the velocity
3 Centripetal Acceleration, cont. Centripetal refers to “center-seeking”The direction of the velocity changesThe acceleration is directed toward the center of the circle of motion
4 Centripetal Acceleration, cont. a = Δv (eq. I)ΔtBy similar trianglesΔv = Δsv rthereforeΔv = Δs vrSub into eq. Ia = Δs v = v2r Δt rSince Δs = v
5 Centripetal Acceleration and Centripetal Force The centripetal acceleration can also be related to the angular velocityac = vt2 Fc = macrTherefore, Fc = mvt2
6 Forces Causing Centripetal Acceleration Newton’s Second Law says that the centripetal acceleration is accompanied by a forceF = maCF stands for any force that keeps an object following a circular pathTension in a stringGravityForce of friction
7 Level CurvesFriction is the force that produces the centripetal accelerationCan find the frictional force, µ, vFc= mac = mv2rFc =f !!!!!!f = μn = μmg = mv2Therefore,Example: The coefficient of friction betweenThe tires of a car and the road is 0.6. What isthe smallest circle that the car will be able toNegotiate at 60mph (26.8 m/s)?
8 Vertical Circle Consider the forces at the top of the circle If car just makes it over the top then the forces exerted by track are zero.Ftrack = 0The minimum force at the top of the circle to continue circular motion is mgThereforemg = mv2rv gr
9 Newton’s Law of Universal Gravitation (1687) Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
10 Law of Gravitation, cont. G is the constant of universal gravitationalG = x N m²/kg²This is an example of an inverse square lawExample: find the magnitude and direction of the gravitational force between the sun and the earth.
11 Escape Speed (see p. 206 in text) If an object is launched from the earth with a high enough velocity it will escape the earth’s gravity.The escape speed is the speed needed for an object to soar off into space and not returnFor the earth, vesc is about 11.2 km/sNote, v is independent of the mass of the object
12 Satellite Law Motion GMm = mv2 r2 r r = GM v2 Assuming a circular orbit is a good approximationFg = FcFGMm = mv2r rr = GMv2
13 Kepler’s Laws Based on observations made by Tycho Brahe Newton later demonstrated that these laws were consequences of the gravitational force between any two objects together with Newton’s laws of motion
14 Kepler’s Laws:All planets move in elliptical orbits with the Sun at one of the focal points.A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.
15 Kepler’s First LawAll planets move in elliptical orbits with the Sun at one focus.Any object bound to another by an inverse square law will move in an elliptical pathSecond focus is empty
16 Kepler’s Second LawA line drawn from the Sun to any planet will sweep out equal areas in equal timesArea from A to B and C to D are the same
18 Kepler’s Third LawThe square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.T = circumference of orbitorbital speedFor orbit around the Sun, KS = 2.97x10-19 s2/m3K is independent of the mass of the planetK = 42GMExample: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity.Ms=1.991 x 1030 kg
19 5-9 Kepler’s Laws and Newton's Synthesis The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.
20 Gravitational Force Chapter 7 Orbiting objects are in free fall. Section 2 Newton’s Law of Universal GravitationChapter 7Gravitational ForceOrbiting objects are in free fall.To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop.Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.
21 Torque The force can be resolved into its x- and y-components The x-component, F cos Φ, produces 0 torqueThe y-component, F sin Φ, produces a non-zero torque
22 Mechanical Equilibrium In this case, the First Condition of Equilibrium is satisfiedThe Second Condition is not satisfiedBoth forces would produce clockwise rotations
23 Kepler’s Third LawThe square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.T = circumference of orbitorbital speedFor orbit around the Sun, KS = 2.97x10-19 s2/m3K is independent of the mass of the planetK = 42GMExample: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity.Ms=1.991 x 1030 kg
24 Simple Machines Chapter 7 Section 4 Torque and Simple MachinesChapter 7Simple MachinesA machine is any device that transmits or modifies force, usually by changing the force applied to an object.All machines are combinations or modifications of six fundamental types of machines, called simple machines.These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.
25 Simple Machines, continued Section 4 Torque and Simple MachinesChapter 7Simple Machines, continuedBecause the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force.This ratio is called mechanical advantage.If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance.
26 Simple Machines, continued Section 4 Torque and Simple MachinesChapter 7Simple Machines, continuedThe diagrams show two examples of a trunk being loaded onto a truck.In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work.In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work.
27 Simple Machines, continued Section 4 Torque and Simple MachinesChapter 7Simple Machines, continuedThe simple machines we have considered so far are ideal, frictionless machines.Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat.The efficiency of a machine is the ratio of useful work output to work input.The efficiency of an ideal (frictionless) machine is 1, or 100 percent.The efficiency of real machines is always less than 1.
28 Horizontal Circle Find ac of yoyo The horizontal component of the tension causes the centripetal acceleration
29 5-9 Kepler’s Laws and Newton's Synthesis Kepler’s laws describe planetary motion.The orbit of each planet is an ellipse, with the Sun at one focus.