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7.4 Angular Acceleration There is in fact a second type of acceleration (in addition to centripetal acceleration) that can be associated with circular motion. Angular acceleration occurs when the circular motion is not uniform; that is, if the tangential speed is increasing or decreasing. In the last lecture, we often referred to a merry-goround or a turntable. These undergo UCM most of the time, but at some point they have to come to a stop and then start up again. During these periods of time, they have a non-zero angular acceleration. Average angular acceleration is, not surprisingly, โ๐ ๐ผ= โ๐ก (thatโs a Greek lower-case โalphaโ). PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 1 7.4 Angular Acceleration In most cases, we are only concerned with constant angular acceleration. Then if ๐ = ๐0 at t = 0, we can write ๐ผ= ๐โ๐0 ๐ก or ๐ = ๐0 + ๐ผ๐ก (note the resemblance to the translational case from chapter 2, โ๐ฃ where ๐ = and ๐ฃ = ๐ฃ0 + ๐๐ก). โ๐ก As with arc length and angle (๐ = ๐๐), and tangential and angular speeds (๐ฃ = ๐๐), we can relate the tangential acceleration to the angular acceleration: ๐๐ก = ๐๐ผ. In the text (and the notes from last class), we dropped the subscript on the tangential speed ๐ฃ๐ก , since this is the only possible type of speed involved with circular motion. We can not do the same thing here, since we must distinguish ๐๐ก from ๐๐ . PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 2 7.4 Angular Acceleration Note carefully that centripetal acceleration is necessary for circular motion, but tangential acceleration is not. The two acceleration vectors are perpendicular to each other at any moment, and the total acceleration vector is a = a๐ก t + a๐ r, where t and r are unit vectors in the tangential and radial directions. The table shown here illustrates similarities between translational and angular motion. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 3 Problem #1: Fan Blades WBL EX 7.49 The blades of a fan running at low speed turn at 250 rpm. When the fan is switched to high speed, the rotation rate increases uniformly to 350 rpm in 5.75 s. a) What is the magnitude of the angular acceleration of the blades? b) How many revolutions do the blades go through while the fan is accelerating? Solution: In class PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 4 Problem #2: Dizzy Hamster The hamster shown in this video has decided to undergo some nonuniform circular motion (FOR SCIENCE!) Letโs analyze the hamsterโs kinematics during the time itโs spinning. The hamsterโs name is Steve. Steve starts spinning at the 31-second mark of the video and falls out at the 36-second mark. During this time, Steveโs angular velocity decreases in magnitude from 25.0 rad/s to 12.5 rad/s. We will assume that the angular acceleration is constant during this time and that Steveโs 0.15-kg body exists entirely at a radius of 5 cm. a) What are the signs of ฯ and ฮฑ? b) What is the constant angular acceleration, ฮฑ? c) How many revolutions does Steve make? d) What is the total distance that Steve travels while he is spinning? e) What is Steveโs maximum tangential speed? f) What is the largest magnitude of centripetal acceleration that Steve experiences? g) What is the largest magnitude of centripetal force that Steve experiences? PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 5 7.5 Newtonโs Law of Gravitation At this point in the course, our knowledge of gravity is simply that โit produces an acceleration of magnitude 9.80 m/s, directed downwardโ, and that โthis magnitude depends weakly on oneโs location on the surface of the Earth, and that it doesnโt change much up to altitudes of 100 km or soโ. That second statement is a bit annoying. Why should classical mechanics produce results that are only valid in particular regions of the universe? Wellโฆit doesnโt. Newtonโs Universal Law of Gravitation tells us what sort of gravitational forces are present among any masses at any distances from each other. Itโs valid for planets, moons, stars, and even galaxies. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 6 7.5 Newtonโs Law of Gravitation The law provides a relationship for the gravitational interaction between two point masses ๐1 and ๐2 , separated by a distance r. It states that every particle in the universe has an attractive gravitational interaction with every other particle in the universe. These forces of mutual interaction are equal and opposite Newtonโs 3rd-law pairs (F12 = โF21 ). Furthermore, the law states that the magnitude of the gravitational force depends inversely on the square of the distance between the masses (inversesquare forces tend to be rather prominent in universes with three spatial dimensions). Finally, the force depends on the product of the two masses themselves. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 7 7.5 Newtonโs Law of Gravitation All together, we can write Newtonโs law of gravitation mathematically as ๐น๐ = ๐บ๐1 ๐2 , 2 ๐ ๐บ = 6.67 × Nโm2 โ11 10 . kg 2 (๐บ is called the universal gravitational constant). From the 1/r2 dependence, we see that not only is gravity a non-contact force, it acts over an infinitely long range. Masses that are separated by hundreds of billions of km still exhibit a gravitational attraction, albeit a weak one. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 8 7.5 Newtonโs Law of Gravitation Of course, most interesting masses are not point masses; they have a spatial extent. For homogeneous spheres, the preceding equation is valid if we consider all of the mass to exist at the center of the sphere (and if we only wish to calculate ๐น๐ outside of the spheresโ surfaces). Proving this requires calculus, but the figure should at least convince you that the direction of F๐ makes sense. It should be noted that Newton didnโt have a good way of determining the value of the proportionality constant ๐บ. It was first measured by Henry Cavendish about 100 years after Newton first published the inverse-square law. Even today, ๐บ remains one of the โleast-preciseโ physical constants. Some physicists would argue that even the three significant digits shown on the previous slide are one too many. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 9 7.5 Newtonโs Law of Gravitation Well, now we have two equations for the gravitational force: ๐น๐ = ๐บ๐1 ๐2 ๐2 and ๐น๐ = ๐๐ How are they related? Letโs assume that the two masses are a (relatively small mass) ๐ and (a planetโs mass) ๐. Newtonโs 2nd law can then be used to analyze the acceleration of ๐ due to the gravitational force: ๐บ๐๐ ๐น๐๐๐ก = ๐๐ โ 2 = ๐๐๐ ๐ Thus, the acceleration due to gravity at any distance ๐ from the planetโs center (assuming that itโs outside of the planetโs surface) is ๐๐ = PC141 Intersession 2013 ๐บ๐ ๐2 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 10 7.5 Newtonโs Law of Gravitation What if this acceleration is being measured by someone at the Earthโs surface? In this case, ๐ = ๐๐ธ = 5.98 × 1024 kg (the mass of the Earth), and ๐ = ๐ ๐ธ = 6.371 × 106 m (the radius of the Earth). Plugging in the numbers, we get ๐๐ = ๐บ๐๐ธ 2 ๐ ๐ธ = (6.67×10โ11 Nโm2 /kg 2 )(5 .98×1024 kg) 6.371×106 m 2 = 9.82 m/s 2 This is the value of g that we have been using since chapter 2. Well, almost. We use 9.80. Thereโs various reasons โ primarily, the fact that the Earth isnโt perfectly spherical, so the value of ๐ ๐ธ shown above is just an approximation. Note that our equation for ๐๐ does not involve m, the mass of the object. This agrees with our experimental observation that acceleration due to freefall is independent of the mass of the falling object. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 11 7.5 Newtonโs Law of Gravitation We are now in a position to calculate the magnitude of gravitational acceleration at higher altitudes. If h is the altitude, then the distance from the Earthโs center is ๐ ๐ธ + โ, and ๐บ๐๐ธ ๐๐ = ๐ ๐ธ + โ 2 Since ๐ ๐ธ โ 6370 km, an altitude of a few hundred km is required in order to achieve a noticeable change in ๐๐ . Later, we will learn why astronauts appear โweightlessโ, despite the data below. Altitude (km) Altitude Example ๐๐ (m/s2) 0 8.8 36.6 400 35,700 PC141 Intersession 2013 9.82 9.80 9.71 8.70 0.225 Mean Earth surface Mt. Everest Highest crewed balloon Space shuttle orbit Communications satellite Day 15 โ June 11 โ WBL 7.4-7.6 Slide 12 7.5 Newtonโs Law of Gravitation In chapter 5, we learned that an object at a height h above a (zero-potential-energy) reference level has a potential energy ๐ = ๐๐โ. However, this was valid only if the gravitational force was constant over the distance h. As we now know, if h is very large, Fg is not constant; it varies with position. Using fairly simple calculus (not repeated here), it can be shown that the gravitational potential energy of two point masses separated by a distance r is ๐บ๐1 ๐2 ๐=โ ๐ (this definition requires that we set U = 0 when the objects are infinitely far apart; i.e. ๐ โ โ). If we consider a mass m at an altitude of h above the Earthโs surface, then ๐บ๐๐๐ธ ๐=โ ๐ ๐ธ + โ 2 PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 13 7.5 Newtonโs Law of Gravitation We say that the Earth (in fact, any mass) exists in a negative gravitational potential energy well. As h increases, U becomes less negative. Thus, when gravity does negative work (an object moves higher in the well) or positive work (an object falls lower in the well), there is a change in potential energy. This energy change will be beneficial in solving some upcoming problems. Our equation for U can also be extended to the case of three or more masses. For example, ๐ = ๐12 + ๐23 + ๐31 = PC141 Intersession 2013 ๐บ๐1 ๐2 โ ๐12 Day 15 โ June 11 โ WBL 7.4-7.6 ๐บ๐2 ๐3 โ ๐23 ๐บ๐3 ๐1 โ ๐31 Slide 14 7.5 Newtonโs Law of Gravitation For example, when considering the total mechanical energy of a mass m1 in orbit about a mass m2, we have ๐ธ =๐พ+๐ = 1 ๐1 ๐ฃ 2 2 โ ๐บ๐1 ๐2 . ๐ Rearranging, we have ๐ฃ= 2 ๐บ๐1 ๐2 ๐ธ+ ๐1 ๐ Although we like to think that the Earthโs orbit about the sun is circular, it is actually slightly elliptical (more on that later). At perihelion (the point of closest approach), the speed of the Earth in its orbit is slightly faster than at aphelion (the farthest point). PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 15 Problem #3: High-Altitude Gravitational Acceleration WBL LP 7.17 Compared with its value on the Earthโs surface, the value of the acceleration due to gravity at an altitude of one Earth radius is A the same B two times as great C one-half as great D one-fourth as great PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 16 Problem #4: Orbit of the Moon WBL EX 7.59/7.63 The moon orbits the Earth with an approximately circular orbit of radius 3.85 x 105 km, and an orbital period of 27.5 days. a) b) c) d) e) Use this data to calculate the mass of the Earth What is the moonโs tangential speed? What is the moonโs kinetic energy? What is the Earth-moon systemโs potential energy? What is the systemโs total energy? Solution: In class PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 17 Problem #5: Gravitational Potential Energy WBL EX 7.64 a) What is the gravitational potential energy of the configuration shown in the figure, if all of the masses are 1.0 kg? b) What is the gravitational force per unit mass at the center of the configuration? Solution: In class PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 18 7.6 Keplerโs Laws and Earth Satellites In the very early 1600s (pre-dating Newton), Johannes Kepler produced three empirical laws that governed planetary motion. With the benefit of Newtonโs law of gravitation, Keplerโs laws can now be derived using just a few lines of algebra. We present the laws here. Note that they do not only apply to planetary orbits. Any system consisting of a body revolving about a much more massive body due to an inverse-square force conforms to Keplerโs laws. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 19 7.6 Keplerโs Laws and Earth Satellites Keplerโs First Law (the law of orbits) Planets move in elliptical orbits, with the sun at one of the focal points. This fact was alluded to a few slides ago. Absolutely nothing about Newtonโs Law of Gravitation suggests that the orbits need to be circular; the orbits are in fact ellipses (at least in the case of closed orbits. For open orbits, in which a body approaches the sun and then leaves, never to return, the path is a hyperbola or parabola). As you may recall, a circle is a special case of ellipse for which the two focal points coincide. Most planets have orbits that are nearly circular; the Earthโs perihelion and aphelion differ by only about 3% of the average Earth-sun separation. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 20 7.6 Keplerโs Laws and Earth Satellites Keplerโs Second Law (the law of areas) A line from the sun to a planet sweeps out equal areas in equal times. This is illustrated in the figure below. Since the speed of the planet is greatest when it is closer to the sun (slide 15), the arc lengths ๐ 1 and ๐ 2 swept out during the same time duration are not equal in length. However, the two areas ๐ด1 and ๐ด2 are equal. This can be proven fairly easily using techniques from chapter 8, but we wonโt attempt it here. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 21 7.6 Keplerโs Laws and Earth Satellites Keplerโs Third Law (the law of periods) The square of the orbital period of a planet is directly proportional to the cube of the average distance of the planet from the sun. That is, ๐ป๐ โ ๐๐ . This law can be proven fairly easily in the special case of a circular orbit. Since the centripetal force is supplied by the force of gravity, these forces are equal. Using Ms and mp as the masses of the sun and the planet, respectively, ๐๐ ๐ฃ 2 ๐บ๐๐ ๐๐ = ๐ ๐2 and thus the speed of the orbiting planet (constant, in a circle) is ๐ฃ= PC141 Intersession 2013 ๐บ๐๐ ๐ Day 15 โ June 11 โ WBL 7.4-7.6 Slide 22 7.6 Keplerโs Laws and Earth Satellites Keplerโs Third Law (the law of periods) contโ However, we also know from last lecture that the speed is related to the period as ๐ฃ = 2๐๐/๐. Therefore, 2๐๐ = ๐ ๐บ๐๐ ๐ Squaring both sides and solving for T gives ๐2 = 4๐ 2 3 ๐ ๐บ๐๐ Which is a mathematical statement of the law. Itโs important to note that the proportionality constant (the term in the brackets) depends only on the sunโs mass. In this way, we can easily calculate the masses of various planets by observing the orbital radii and periods of their moons. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 23 7.6 Keplerโs Laws and Earth Satellites Keplerโs Third Law (the law of periods) contโ This law can be used to determine the required orbital altitude of a geosynchronous satellite. This is a satellite whose orbital period is one day, such that itโs always overhead of the same point on the Earthโs surfaceโ . These orbits are useful for communications satellites, since the relay stations on the Earthโs surface donโt have to continuously re-aim. A good example can be found on p. 242 of the text. โ Iโve skipped a whole lot of physics here. A geosynchronous satellite can only remain overhead of the same point on the Earthโs surface if that point lies on the equator โ this is called a geostationary orbit. A geosynchronous orbit that isnโt geostationary does move relative to this point during each day, but not by a large degree. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 24 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites We conclude this chapter with a few words about man-made satellites and the physics that governs their flight and orbits. First, consider this โthought experimentโ. From chapter 2, we know that when an object is projected upward, it slowly loses speed (since acceleration is downward). Eventually, it will come to an instantaneous stop, after which it will fall back downward (accelerating as it goes). In this case, we assumed that the acceleration was constant. What if acceleration became weaker as the object climbed to higher altitudes? Is it possible for the object to have a sufficiently high initial speed for it to โoutrunโ the weakening acceleration, such that it never comes to a stop? The answer is yes! PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 25 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites contโ We can show this using conservation of energy. Suppose that a spacecraft is initially on the surface of the Earth, and that it is given a speed which is sufficient for it to escape the Earthโs potential well โ that is, it only slows to a stop when it is infinitely far away. Conservation of energy tells us that ๐พ๐ + ๐๐ = ๐พ๐ + ๐๐ 1 2 (the terminology will become clear in a In this case, ๐พ๐ = ๐๐ฃ๐๐ ๐ 2 ๐บ๐๐๐ธ moment), and ๐๐ = โ . Furthermore, ๐พ๐ = ๐๐ = 0 (why?) ๐ ๐ธ Rearranging, we find that ๐ฃ๐๐ ๐ = PC141 Intersession 2013 2๐บ๐๐ธ = ๐ ๐ธ Day 15 โ June 11 โ WBL 7.4-7.6 2๐๐ ๐ธ Slide 26 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites contโ ๐ฃ๐๐ ๐ is called the escape speed (roughly 11.2 km/s) - the initial speed needed to escape from the surface of the Earth (without falling back toward it). It depends only on the Earthโs parameters, not those of the spacecraft. Of course, we can calculate escape speeds from any other planet / moon / asteroid / etc., as long as we know its mass and radius. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 27 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites - Tangential Speed and โZero-Gravityโ Earlier in this lecture, we saw that ๐๐ โ 8.70 m/s2 at the altitude of the space shuttle (โ 400 km). This may seem strange, because video of astronauts aboard the shuttle seems to show that they are weightless. Whatโs going on here? Remember that any orbiting body must have a tangential speed โ if we simply placed the space shuttle at an altitude of 400 km and released it, it would come plummeting back to Earth. The tangential speed depends on the desired orbital altitude. Recycling our Keplerโs 3rd law math, ๐๐ฃ 2 ๐ = ๐บ๐๐๐ธ ๐2 โ๐ฃ= ๐บ๐๐ธ ๐ (as usual, r is the distance from the center of the Earth). PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 28 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites - Tangential Speed and โZero-Gravityโ contโ This tangential speed sees to it that the satellite continuously โfallsโ around the Earth. If an insufficient tangential speed is reached, the satellite will spiral inward toward the Earth, eventually burning up in the atmosphere (due to the rather extreme drag forceโฆkinetic energy is converted to thermal energy!). If the tangential speed is too great, the satellite simply achieves an elliptical orbit. If ๐ฃ > ๐ฃ๐๐ ๐ , the satellite leaves its orbit and moves off into space. PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 29 7.6 Keplerโs Laws and Earth Satellites Earthโs Satellites - Tangential Speed and โZero-Gravityโ contโ The reason that astronauts aboard the space shuttle appear to be weightless is simply that the shuttle and its occupants are accelerating toward the Earth at the same rate. Itโs the same effect that would occur if a person was inside an elevator when the cable snapped. The effect can be safely replicated near the Earthโs surface in an airplane that flies in a parabolic arc (or unsafely replicated in an airplane that has stalled!) PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 30 Problem #6: Kepler Orbits WBL LP 7.19 As a planet moves in its elliptical orbitโฆ A โฆits speed is constant B โฆits distance from the sun is constant C โฆit moves faster when it is closer to the sun D โฆit moves slower when it is closer to the sun PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 31 Problem #7: Satellite About Venus WBL EX 7.71 Venus has a rotational period of 243 days and a mass of 4.87 x 1024 kg. What would be the altitude of a โvenosynchronousโ satellite? Solution: In class PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 32 Problem #8: Asteroid Belt WBL EX 7.69 The asteroid belt that lies between Mars and Jupiter may be the debris of a planet that broke apart, or that was not able to form as a result of Jupiterโs strong gravitation. An average asteroid in the belt has an orbital period (about the sun) of about 5.0 years. Approximately how far from the sun would this โfifthโ planet have been? Solution: In class PC141 Intersession 2013 Day 15 โ June 11 โ WBL 7.4-7.6 Slide 33