Circular motion. F = ma = mv 2 /r. a = v 2 /r

Transcription

1 Circular Motion Circular motion is very common and very important in our everyday life. Satellites, the moon, the solar system and stars in galaxies all rotate in circular orbits. The term circular here is being used loosely since even repetitive closed motion is generally not a perfect circle. At any given instant a moving object not moving in a straight line is moving along the arc of a (maybe changing) circle. So if we understand motion in a circle we can understand more complicated trajectories. Remember at any instant the velocity is along the path of motion but the acceleration can be in any direction. 1

2 Circular motion a = v 2 /r F = ma = mv 2 /r If the velocity of an object changes direction then the object experiences an acceleration and a force is required. These are called centripetal acceleration and force and are directed toward the center of the circle. This is the effect you feel rounding a corner in a car 2

3 Circular motion F = ma = mv 2 /r a = v 2 /r Where does this come from? a = Δv/Δt = vxv/r Look at the Δv vector: it s proportional to v. (and is directed toward the center of the circle.) Also, the rate that Δv is changing is proportional to the rate that the angle between v 1 and v 2 is changing (that is, the rate that vector v is swinging around.) This rate of change is just v/r (the angle is in RADIANS) 3

4 Questions Chapter 5 Q6 A ball on the end of a string is whirled with constant speed in a counterclockwise horizontal circle. At point A in the circle, the string breaks. Which of the curves sketched below most accurately represents the path that the ball will take right after the string breaks (as seen from above)? Explain. 4 3 A 2 1. A=1, B=2, C=3 D=4 Path number 3 4

5 Balance of forces We need to understand the forces that are acting along both horizontal and vertical directions. In the case shown the tension or force exerted by the string has components which balance the weight in the vertical direction and provide the centripetal force horizontally. T v = W = mg T h = mv 2 /r 5

6 Cars F f F f Rear view When a car turns a corner it is friction between the tires and the road which provides the centripetal force. If the road is banked then the normal force also provides some centripetal force. For each banked track there is a velocity for which no friction is required. Note F f adjusts itself (up to break point) And it is the HORIZONTAL component of F f which may help supply the needed F centrip. Note: N v must balance W. Question: if v = 0, where does F f point? A. Upslope B. Downslope View from Above F f W = mg 6

7 Vertical circles N v W = mg mg N = mv 2 /r If v = 0 then N = mg As v increases N becomes smaller When v 2 /r = g The car becomes weightless. Same as weightless-training airplane, for astronauts g We choose + always toward the center of the circle If car were running on capture rails that could also hold it down, big v would cause N to point downward Ferris wheel At the bottom N - mg = mv 2 /r At the top mg N = mv 2 /r With seatbelt, at top if v 2 is big enough, seatbelt would hold you down, your weight would be upside down! 7

8 Gravitation and the planets Astronomy began as soon as man was able to observe the sky. Records exist going back several thousand years. In particular, of the yearly variation of the stars in the sky and the motion of observable objects such as planets. People observed the fixed North Star and, for example, the rising of Sirius signaling the flooding of the Nile. Copernicus was the first person to advocate a suncentered solar system. Followed by Galileo who used the first telescopes (moons of Jupiter like small solar system.) Tycho Brahe was the most famous naked eye astronomer. Kepler, his assistant used the data to draw quantitative conclusions. 8

9 Keplers Laws 1) Orbits are ellipses. For 0 eccentricity circle 2) The radius vector sweeps out equal areas in equal times 3) T 2 proportional to r 3 T is the period. For the earth, T is one year and r is the average radius. Earth s eccentricity is small, but does affect climate cyclically. For circular motion with constant velocity v the circumference of a circle is 2πr and the Period T = 2πr/v = distance (arc length) / speed 9

10 Keplers Laws 1) Orbits are ellipses. For 0 eccentricity circle 2) For the Earth, ellipse is rather close to a circle 3) T 2 proportional to r 3 T = 1 year, is the period for the earth.. For circular motion with constant velocity v the circumference of a circle is 2πr and the Period T = 2πr/v = distance (arc length) / speed So v = 2πr/T = circumf,/period = 2π 94M mi/1 yr 10

11 Newton and Gravitation Newton developed the Law of Universal Gravitation force between ANY two objects is proportional to M 1 m 2 /r 2. The (very small) constant of proportionality was measured by Cavendish more than 100 years later G = 6.67 x N.m 2 /kg 2. Since at the earths surface mg = m(gm e /r 2 ) the experiment measured the mass of the earth edu/class/applets/newtons Cannon/newtmtn.html 11

12 Newton and Gravitation The force between ANY two objects is ATTRACTIVE and = GM 1 m 2 /r 2. N3 (third law) happens here too: each M pulls the other, equal and opposite. For a spherical object, all its mass acts as if at the CENTER. 12

13 Planetary orbits For a simple circular orbit GmM/r 2 = mv 2 /r where M is the mass of the sun and m the mass of the earth [or M is the mass of the earth and m the mass of an earth satellite.] Cancel m & solve GM/r 3 =v 2 /r 2 but v/r = 2π/T GM/r 3 =(2π/T) 2 T 2 /r 3 = 4π 2 /GM s Use average radius for elliptical orbits. Works for other attractors: Earth, Jupiter. 13 Use right Mass

14 Satellite Orbits For a geosynchronous orbit period is 24 hours and height above the earths surface is ~22,000 miles (add ~4000 miles to get the distance to the CENTER) T 2 /r 3 = 4π 2 /GM e R 3 =T 2 GM e /4π 2 Take square root both sides Geosynchronous orbit must be CIRCULAR, and will be off-scale on the above picture. Also is equatorial. 14

15 Summary of Chapter 5 Circular motion and centripetal acceleration and force. F c = mv 2 /r Ferris wheel, car around a corner or over a hill. Circular pendulum. Gravitation and Planetary orbits For a simple circular orbit GmM/r 2 = mv 2 /r where M is the mass of the sun and m the mass of the earth. v 2 = GM/r T = 2πr/v T 2 = 4π 2 r 2 /v 2 = 4π 2 r 3 /GM s At any given instant, F G is given by the distance BETWEEN CENTERS OF MASS Wrong r!! use AVG. T 2 /r 3 = 4π 2 /GM s 15

16 Examples of circular motion Vertical motion N v Looking down on stunt driver s cylindrical cage N v W = mg mg N = mv 2 /r N mg N - mg = mv 2 /r N = mv 2 /r Side view Red force arrows should F f be equal, if enough static friction is N available. If μ s is too small, not mg enough F f and car will slide downwards. mg = F f v T mg Swing object on a string in a vertical circle mg + T = mv 2 /r top T - mg = mv 2 /r bottom 16

17 Quiz question. Two objects each of mass 1 Metric ton are 2 meters apart (between centers). What is the force between them? G = 6.67x10-11 and one Metric ton = 1,000kg A. 1.7x10-11 N B. 1.7x10-17 N C. 1.7x10-5 N. D. 1.7x10-8 N E. 1.7x10 6 N 17

18 Moon and tides anim0012.mov Earth tides are dominantly due to the gravitational force exerted by the moon. Since the Earth turns faster than the Moon orbits, the tidal bulges sweep around the Earth, with high tides happening approx. twice a day. Because of the friction generated by tides the Moon is gaining energy from the Earth s spin, and moving away from the earth. whytides.gif X E+M Centerof Mass is 80 times closer to Earth since Earth is 80 times heavier than Moon Earth & Moon both orbit around their mutual CofM 18

19 Moon and tides The tidal bulge away from the moon is where centrifugal force is bigger than gravitational force. The bulge towards the moon is where the gravity outweighs the centrifugal force. This works because the entire earth moves with the same angular velocitiy around the CofM. The outward bulge is where F G is less and F c (because of v 2 ) is more, than required for the orbit of the Earth s center. And vice versa for the Moon-ward bulge. X anim0012.mov ~250,000 mi ~384,000 km E+M Centerof Mass is 80 times closer to Earth s center, since Earth is 80 times heavier than Moon Earth & Moon both orbit around their mutual CofM whytides.gif 19

20 Moon and tides The Moon itself has a tidal bulge, due to ITS orbit. The Earth pulls on this bulge and keeps the Moon locked with the bulge facing the Earth. [This required the Moon to lose some of its spin energy early-on in the life of the solar system.] The Earth spins faster than the Moon orbits, so the Earth s tidal bulges torque the Moon s orbit to move the orbit further away and slow down the Earth s rotation. This ongoing effect is measurable (3.8 cm/yr). Also geologically measurable (Devonian seashells with growth layers showing ~400 days per year!!) Someday far in the future, the Earth s spin may get locked to the Moon s orbit! X anim0012.mov E+M Centerof Mass is 80 times closer to Earth since Earth is 80 times heavier than Moon Earth & Moon both orbit around their mutual CofM whytides.gif 20

21 Dark Matter For the orbit of a body of mass m about a much more massive body of mass M: GmM/r 2 = mv 2 /r and GM/r = v 2. M is the mass inside the orbit. If we look at stars in motion in galaxies we find that the orbital speeds do NOT fall off as quickly as you d expect from the VISIBLE (shining) stars. There s extra invisible mass boosting the effective M inside each orbit. We now know that 25% of our Universe is Dark Matter, only abaout 5% is baryonic matter (made of nuclei and atoms) and only about 1% 2% of the Universe is made of shining stars!! v~1/sqrt(r) 21

22 Space Elevator (courtesy of Arthur C. Clarke) station v counterweight 100,000km The Space Elevator is a thin ribbon, with a cross-section area roughly half that of a pencil, extending from a ship-borne anchor to a counterweight well beyond geosynchronous orbit. and over ¼ of the way from Earth to Moon The ribbon is kept taut due to the rotation of the earth (and that of the counterweight around the earth). At its bottom, it pulls up on the anchor with a force of about 20 tons. Electric vehicles, called climbers, ascend the ribbon using electricity generated by solar panels (i.e. photovoltaic) facing a ground based booster light beam (laser.) In addition to lifting payloads from earth to orbit, the elevator can also release them directly into lunar-injection or earth-escape trajectories. The baseline system weighs about 1500 tons (including counterweight) and can carry up to 15 ton payloads, easily one per day. The ribbon is 62,000 miles long, about 3 feet wide, and is thinner than a sheet of paper. It is made out of a carbon nanotube composite material. The climbers travel at a steady 200 kilometers per hour (120 MPH), do not undergo accelerations and vibrations, can carry large and fragile payloads, and have no propellant stored onboard. The climbers are driven by earth based lasers. Orbital debris are avoided by moving the floating anchor ship on Earth, and the ribbon itself is made resilient to local space debris damage. The elevator can increase its own payload capacity by adding ribbon layers to itself. There is no limit on how large a Space Elevator can be! 22

23 Space Elevator (courtesy of Arthur C. Clarke) station v counterweight 100,000km The Space Station can be beyond geosynchronous height (as shown) and can launch objects with greater than escape velocity. This is because at its location, it is LESS gravitationally bound to the Earth AND it has MORE than the Velocity for a circular orbit AT THAT DISTANCE. Basically, the free orbit from the blue Space Station is not actually an orbit, it won t even close into an ellipse, the path is too fast and straight for that. If the ribbon breaks, the station and the counterweight will go flying off away from the earth and not return. 23

24 Space Elevator (courtesy of Arthur C. Clarke) station v counterweight 100,000km Engineering note: The strongest steel cable is much too heavy, for its strength, to do this job. A 10 mile long steel cable suspended (magically) near the Earth s surface, would break under its own weight. We need something WAY lighter (per unit of strength) to make the space elevator. Carbon nanotubes are incredibly strong. They are like rolled-up tubes of single-layer graphite (a hexagonal network of Carbon atoms, like chicken wire.) The Carbon-Carbon bond achieves this (somewhat the way diamond, which is a crystal of pure carbon, is so hard.) 24

25 Ch 5 QUIZ Question On a rotating platform, spinning at a certain constant rate, you move three times farther from the center. How many times more (or less) friction does it take to keep you from slipping off the turntable? A. 1/9 B. 1/3 C. same D. 3 E. 9 Note that both v and r change when you move. Think about how the distance around the circle changes when r increases, and remember that you go around once in the same amount of time wherever you are on the turntable. 25

26 Ch 5 QUIZ Question On a rotating platform, spinning at a certain constant rate, you move three times farther from the center. How many times more (or less) friction does it take to keep you from slipping off the turntable? A. 1/9 B. 1/3 C. same D. 3 E. 9 Note that both v and r change when you move. Think about how the distance around the circle changes when r increases, and remember that you go around once in the same amount of time wherever you are on the turntable. New v (call it v ) is three times faster. New r (call it r ) is three times bigger New mv 2 /r = m(3v) 2 /3r = (9/3)mv 2 /r : Three times as much F c is needed. 26

27 1C-05 Velocity of Rifle Bullet MEASURING SPEEDS OF OBJECTS MOVING VERY FAST MAY NOT BE DIFFICULT. THIS TECHNIQUE IS THE SAME ONE USED TO MEASURE SPEEDS OF MOLECULES. How can we measure the speed of a bullet? We know that the distance between two disks is L. If the second disk rotates an angle Δθ before the bullet arrives, the time taken by this rotation is t = Δθ / 2πn, where 2πn is the angular frequency of the shaft, in radians/s. Therefore we come up with v = L / t = 2πn L / Δθ. Note: θ is in radians, and n by itself is the revolution rate, in revols./s or Hz. 27

28 1D-02 Conical Pendulum Could you find the NET force? T sin(θ) = mv 2 /R T cos(θ) = mg v = sqrt( gr tan(θ) ) cuz tan() = v 2 /gr Note that T s cancel, and m s cancel. Which of the m s is inertial? Which of the m s is grav.? Period of the pendulum τ= 2πR/v, where R = L / sin(θ) τ= 2πsqrt( Lcos(θ)/g ) NET FORCE IS TOWARD THE CENTER OF THE CIRCULAR PATH 28

29 1D-03 Demonstrations of Central Force THE SHAPES/SURFACES OF SEMI-RIGID OBJECTS BECOME MORE CURVED TO PROVIDE GREATER CENTRAL FORCES DURING ROTATION. What will happen when it is subjected to forces during rotation? T θ T θ F c = mv 2 /R The force comes from the strips wanting to un-flex 29

30 1D-04 Radial Acceleration & Tangential Velocity Once the string is cut, where is the ball going? AT ANY INSTANT, THE VELOCITY VECTOR OF THE BALL IS DIRECTED ALONG THE TANGENT. AT THE INSTANT WHEN THE BLADE CUTS THE STRING, THE BALL S VELOCITY IS HORIZONTAL SO IT ACTS LIKE A HORIZONTALLY LAUNCHED PROJECTILE AND LANDS IN THE CATCH BOX. 30

31 1D-05 Twirling Wine Glass WHAT IS THE PHYSICS THAT KEEPS THE WINE FROM SPILLING? Same as m string v g N + mg = mv 2 /R N > 0 THE GLASS WANTS TO MOVE ALONG THE TANGENT TO THE CIRCLE AND THE REACTION FORCE OF THE PLATE AND GRAVITY PROVIDE THE CENTRIPETAL FORCE TO KEEP IT IN THE CIRCLE (and the wine in the glass). 31

32 Q8 For a ball twirled in a horizontal circle at the end of a string, does the vertical component of the force exerted by the string produce the centripetal acceleration of the ball? Explain. Vertical component balances the weight Horizontal component provides the acceleration Q9 A car travels around a flat (unbanked) curve with constant speed. A. Show all of the forces acting on the car. B. What is the direction of the net force act. F f N mg Rear The force acts toward the center of the turn circle 32

33 Q10 Is there a maximum speed at which the car in question 9 will be able to negotiate the curve? If so, what factors determine this maximum speed? Explain. Yes. The friction between the tires and the road Q11 If a curve is banked, is it possible for a car to negotiate the curve even when the frictional force is zero due to very slick ice? Explain. Yes there is just one speed. If the car moves too slowly it will slide down. If it moves to fast it will slide up. 33

34 Q12 If a ball is whirled in a vertical circle with constant speed, at what point in the circle, if any, is the tension in the string the greatest? Explain. (Hint: Compare this situation to the Ferris wheel described in section 5.2). The tension is the greatest at the bottom because the string has to support the weight and provide the force for the centripetal acceleration. Q19 Does a planet moving in an elliptical orbit about the sun move fastest when it is farthest from the sun or when it is nearest to the sun? Explain by referring to one of Kepler s laws. When it is nearest 34

35 Q20 Does the sun exert a larger force on the earth than that exerted on the sun by the earth? Explain. The magnitude of the forces is the same they are a reaction/action pair Q23 Two masses are separated by a distance r. If this distance is doubled, is the force of interaction between the two masses doubled, halved, or changed by some other amount? Explain. The force reduces by a factor of 4 35

36 Ch 5 E 14 The acceleration of gravity at the surface of the moon is about 1/6 that at the surface of the Earth (9.8 m/s 2 ). What is the weight of an astronaut standing on the moon whose weight on earth is 180 lb? W earth = m g earth = 180 lb W moon = m g moon g moon g moon = 1/6 g earth W moon = m 1/6 g earth = 1/6 m g earth = 1/6 (180 lb) = 30 lb 36

37 Ch 5 E 16 Time between high tides = 12 hrs 25 minutes. High tide occurs at 3:30 PM one afternoon. a) When is high tide the next afternoon b) When are low tides the next day? a) 3:30 PM + 2 (12 hrs 25 min) high tide T= 12hrs 25min = 3:30 PM + 24 hrs + 50 min t = 4:20 PM low tide T= 12hrs 25min b) Low tide the next day = 4:20 PM - 6 hr 12 min 30 s = 10:07:30 AM 2nd Low tide = 10:07:30 AM + 12 hrs 25 min = 10:32:30 PM 37

38 1D-07 Paper Saw THE RADIAL FORCES HOLDING THE PAPER TOGETHER MAKE THE PAPER RIGID. Is paper more rigid than wood? 38

39 1D-08 Ball in Ring Is the ball leaving in a straight line or continuing this circular path? THE FORCE WHICH KEEPS THE BALL MOVING CIRCULAR IS PROVIDED BY THE RING. ONCE THE FORCE IS REMOVED, THE BALL CONTINUES IN A STRAIGHT LINE, ACCORDING TO NEWTON S FIRST LAW. 39

40 Ch 5 CP 2 A Ferris wheel with radius 12 m makes one complete rotation every 8 seconds. a) Rider travels distance 2πr every rotation. What speed do riders move at? b) What is the magnitude of their centripetal acceleration? c) For a 40 kg rider, what is magnitude of centripetal force to keep him moving in a circle? Is his weight large enough to provide this centripetal force at the top of the cycle? d) What is the magnitude of the normal force exerted by the seat on the rider at the top? e) What would happen if the Ferris wheel is going so fast the weight of the rider is not sufficient to provide the centripetal force at the top? 40

41 Ch 5 CP 2 (con t) a) S = d/t = 2πr/t = 2π(12m)/8s = 9.42 m/s b) a cent = v 2 /r = s 2 /r = (9.42m/s) 2 /12m = 7.40 m/s 2 F cent r = 12m c) F cent = m v 2 /r = m a cent = (40 kg)(7.40 m/s 2 ) = 296 N W = mg = (40 kg)(9.8 m/s 2 ) = 392 N Yes, his weight is larger than the centripetal force required. d) W N f = 296 N = 96 newtons e) rider is ejected N W 41

42 A passenger in a rollover accident turns through a radius of 3.0m in the seat of the vehicle making a complete turn in 1 sec. a) Circumference = 2πr, what is speed of passenger? b) What is centripetal acceleration? Compare it to gravity (9.8 m/s 2 ) c) Passenger has mass = 60 kg, what is centripetal force required to produce the acceleration? Compare it to passengers weight. a) s = d/t = 2π(3.0m)/1 = 19m/s Ch 5 CP 4 b) a = v 2 /r = s 2 /r = (19 m/s) 2 /3m = 118 m/s 2 = 12g 3 m c) F = ma = (60 kg)(118 m/s 2 ) = 7080 N F = ma = m (12 g) = 12 mg = 12 weight 42

43 Ch 5 CP 6 The period of the moon s orbit about the earth is 27.3 days, but the average time between full moons is 29.5 days. The difference is due to the Earth s rotation about the Sun. a) Through what fraction of its total orbital period does the Earth move in one period of the moons orbit? b) Sketch the sun, earth & moon at full moon condition. Sketch again 27.3 days later. Is this a full moon? c) How much farther does the moon have to move to be in full moon condition? Show that it is approx. 2 days. a) Earth orbital period = 365 days = 0 E (I d have called it T E ) Moon orbital period = 27.3 days = 0 M ( T M ) 0M/0E = 27.3/

44 Ch 5 CP 6 (con t) b) Day 0 E M Full Moon (i) S (ii) S E M 27.3 Days Later This is not a full moon. (iii) S E M This is the next full moon. c) For moon to achieve full moon condition, it must sit along the line connecting sun & earth. In part (a) we found that the earth has moved thru of its full orbit in 27.3 days (see diagram (ii)). To be in-line w/ sun and earth, moon must move thru same fraction of orbit (see diagram (iii)). This answer is only approx. (but not too bad) since in 2 days the earth-sun line has advanced even further! (27.3 days) 2.04 days. Apparent lunar month days 44

45 Ch 5 CP 6 (con t) b) Day 0 E M Full Moon (i) S (ii) S E M 27.3 Days Later This is not a full moon. (iii) S E Fraction of 2 pi radians angle This is the next full moon. c) Exact solution: θ E = 2π(t/365.25) = θ M = 2π(t/27.3-1) with t in days Here, earth makes a fraction of a revolution, and the moon makes one full revolution plus the SAME fraction of the next revolution, to line up with the Sun-Earth direction. We have to subtract off the full revolution to compare angles. Solve: 1 + t/ = t/27.3; 1 = t t = t t = days compare with approx days previous slide M Fraction of a revolution