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Transcript
CHAPTER 5 Uniform Circular Motion Uniform circular motion is the motion of an object traveling at a constant speed on a circular path. If T (period) is the time it takes for an object to travel once around a circle, of radius, r, then the velocity of the object is given by: 2r vt T Where vt is the tangential velocity of the object Acceleration in Uniform Circular Motion An object that moves at constant speed in a circular path is accelerating. This acceleration is known as Centripetal Acceleration (ac) and it is directed towards the center of the circular path. This object traveling in a circular path also experiences a force that is also directed towards the center of the circular path This force is known as Centripetal Force (Fc) Therefore change in velocity of this object is also directed toward the center of the circular path © 2012 OpenStax College Centripetal Acceleration The magnitude of the centripetal 2 t v aC r Where : vt= tangential speed r= radius of the circle. The direction of the force is toward the center of the circle. © 2012 OpenStax College acceleration is given by the formula Centripetal Force The force causing constant circular motion is given by: mvt Fc maC r 2 This force is not a new force. It just tells us the amount of force that must be provided, by a tension, gravity, etc., in order for an object to move in a circle. © 2012 OpenStax College Banked Curves: © 2012 OpenStax College Banked Curves 2 v Fc FN sin m r FN cos mg Putting the equations together: 2 v FN sin m 2 v r FN sin m r FN cos mg FN cos mg v2 tan rg or v rg tan ***This equation can be used to calculate the speed of an object traveling on a banked curve Example problem #1 Trebuchet project: 2-part project Both test grades Part 1 (research, design, build) due 12/07/16 Part 2 (launch and review) due 12/09 or 12/12 3 to 4 people in a team Opportunities for extra credit points Example problem #2 The Moon (mass = 7.36 x 1022 kg) orbits the Earth at a range of 3.84 x 105 km, with a period of 28 days. What is the magnitude of the force that maintains the circular motion of the Moon? 𝑮: 𝑚 = 7.6𝑥1022𝑘𝑔, 𝑟 = 3.84𝑥108𝑚, 𝑇 = 2.42𝑥106𝑠 𝑼: 𝐹𝑐 =? 𝑬: 𝐹𝑐 = 𝑚𝑣𝑡2 𝑟 S: 𝐹𝑐 = 𝑚(2𝜋𝑟)2 𝑟𝑇2 7.36×1022 (2𝜋×3.84×108 )2 3.84×108 2.42×106 2 𝐹𝑐 = S: 𝟏. 𝟗 𝒙 𝟏𝟎𝟐𝟎 𝑵. Newton’s Law of Universal Gravitation Gravitational force is the mutual force of attraction between particles of matter There is a gravitational force between any 2 objects The larger the object…the more pull it has For example, there is a gravitational force between 2 pencils If the objects are larger, they will have a larger gravitational field Our gravitational field is extremely small (due to our small mass) relative to the earth’s gravitational field (which has a much larger mass) As a result, we don’t feel the gravitational force between us and “small” objects Gravitational force is the force that keeps planets in orbit and keeps them from coasting off in a straight line Gravitational force is an attractive force It depends on the distance between two objects and The magnitude of the masses Increase the distance-decrease the gravity Gravitational force is directly proportional to the product of the two masses involved m1m2 Fg G 2 r Gravitational force is inversely proportional to the square of the distance of separation 𝐹𝑔 = 𝑚1 𝑚2 𝐺 2 𝑟 Fg = the gravitational force m1 and m2 = the masses being observed r = the distance between the two masses ( in meters) G = the universal gravity constant The constant is: G = 6.67 x 2 𝑁∙𝑚 10-11 2 𝑘𝑔 Inverse square of distance – complete the chart Original New distance distance 10 20 20 10 50 5 8 64 25 5 𝑵𝒆𝒘 reduce 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 20 2 10 1 10 1 20 2 inverse 1 2 2 1 Square Change in Fg ¼x 4x Example problem #1 Calculate the gravitational force between Mrs. Akibola and her computer monitor when they are 0.45m apart. Mrs. Akibola’s weight is 736 N and the weight of the computer is 98.1 N G: r = 0.45 m, m1 = U: Fg = ? E: 𝐹𝑔 = 𝑆: 𝐹𝑔 = 𝑚1 ×𝑚2 𝐺 𝑟−2 6.67 x 10 11 × 75 x 10 0.452 S: 2.4 x 10-7 N 736 = 9.81 75 kg , m2 = 98.1 9.81 = 10 kg Satellites in Circular Orbits There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. Finding the speed of an Earth Satellite write these equations at the bottom of your notes!! Centripetal acceleration: Centripetal force: The force is the force of Gravity The mass of the satellite divides out, so it doesn’t matter. v2 ac r mSatellite v 2 F mSatellite a r GM EarthmSatellite mSatellitev 2 2 r r GM Earth v 2 2 r r Finding the speed of an Earth Satellite (Continued) Solving for v: GM Earth GM Earth v or v r r 2 Newton’s Cannon Newton compared the motion of a falling object to the motion of the Moon. An object in orbit is actually falling toward the Earth, at just the same rate that the Earth curves away from it. Source: Brian Brondel, Newton Cannon.svg, Wikimedia Commons, http://en.wikipedia.org/wiki/File:Newton_Cannon .svg Gravitational Field Strength of a Point Mass (or Spherically Symmetric Mass) write these equations at the bottom of your notes!! Mm FG G 2 r Fg Mm g G 2 m r m M g G 2 r The gravity field around a point mass, or a spherically symmetric mass, depends only on the mass and the distance away from the center. Satellites in Circular Orbits Example #2: Orbital Speed of the Hubble Space Telescope Determine the speed of the Hubble Space Telescope orbiting at a height of 598 km above the earth’s surface. GM E v r v 6.67 10 5.98 10 11 6.38 10 598 10 6 24 3 7.56 10 m s 3 16900 mi h Vertical Circular Motion 2 1 v FN 1 mg m r 2 v2 FN 2 m r FN 4 2 4 v m r 2 3 v FN 3 mg m r Kepler’s contributions to astronomy Used Tycho’s data to formulate three laws of planetary motion. The planets move in elliptical orbits with the Sun at one center of the ellipse. The planets trace out equal areas in equal times. The square of a planet’s period is proportional to the cube of its distance from the sun. T2 ~ r3 Kepler’s First and Second Law Here is an animated visualization of Kepler’s first two laws. http://www.surendranath.org/Applets/Dynamics/Kepler/Kepler1Applet. html