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Transcript
This Week • • • • Circular motion Going round the bend Riding in a ferris wheel, the vomit comet Gravitation Our solar system, satellites (Direct TV) • The tides, Dark matter, Space Elevator 5/2/2017 Physics 214 Fall 2010 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 an object that is not moving in a straight line is moving along the arc of a 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. 5/2/2017 Physics 214 Fall 2010 2 Circular motion a = v2/r F = ma = mv2/r If the velocity of an object changes direction then the object experiences an acceleration and a force is required. This is centripetal acceleration and force and is directed toward the center of the circle. This is the effect you feel rounding a corner in a car 5/2/2017 Physics 214 Fall 2010 3 Balance of forces We need to understand the forces that are acting horizontally and vertically. 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. Tv = W = mg 5/2/2017 Th = mv2/r Physics 214 Fall 2010 4 Cars Ff Ff Rear Above 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 a force. For a banked track there is a velocity for which no friction is required. Ff W = mg 5/2/2017 Physics 214 Fall 2010 5 Vertical circles N v g W = mg mg – N = mv2/r If v = 0 then N = mg + is always toward the center of the circle At the bottom N - mg = mv2/r As v increases N becomes smaller When v2/r = g the car becomes weightless. Same as the “vomit comet” 5/2/2017 Ferris wheel Physics 214 Fall 2010 At the top Mg – N = mv2/r 6 Gravitation and the planets Astronomy began as soon as man was able to observe the sky and records exist going back several thousand years. In particular 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 signalling the flooding of the Nile. Copernicus was the first person to advocate a sun centered solar system. Followed by Galileo who used the first telescopes Tycho Brahe was the most famous naked eye astronomer. Kepler, his assistant used the data to draw quantitative conclusions. 5/2/2017 Physics 214 Fall 2010 7 Keplers Laws 1) Orbits are ellipses 2) The radius vector sweeps out areas in equal times equal 3) T2 proportional to r3 T is the period which for the earth is one year and r is the average radius For circular motion with constant velocity v The circumference of a circle is 2πR and the Period T = 2πr/v 5/2/2017 Physics 214 Fall 2010 8 Newton and Gravitation Newton developed the Law of Gravitation force between two objects is F = GM1m2/r2 . The constant of proportionality was measured by Cavendish after more than 100 years G = 6.67 x 10-11 N.m2/kg2. Since at the earths surface mg = GmMe/r2 the experiment measured the mass of the earth 5/2/2017 Physics 214 Fall 2010 http://www.physics.purdue. edu/class/applets/Newtons Cannon/newtmtn.html 9 Planetary orbits For a simple circular orbit GmM/r2 = mv2/r T2/r3 = 4π2/GMs 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 a satellite. For a geosynchronous orbit period is 24 hours and height above the earths surface is 22,000miles 5/2/2017 Physics 214 Fall 2010 10 Summary of Chapter 5 Circular motion and centripetal acceleration and force. Fc = mv2/r Ferris wheel, car around a corner or over a hill. Gravitation and Planetary orbits For a simple circular orbit GmM/r2 = mv2/r where M is the mass of the sun and m the mass of the earth. v2 = GM/r T = 2πr/v T2 = 4π2r2/v2 = 4π2r3/GMs T2/r3 = 4π2/GMs 5/2/2017 Physics 214 Fall 2010 11 Examples of circular motion Vertical motion Looking down N N v v W = mg mg – N = N = mv2/r mv2/r Side N mg N - mg = mv2/r 5/2/2017 v T mg Ff mg mg = Ff Physics 214 Fall 2010 mg + T = mv2/r top T - mg = mv2/r bottom 12 1D-02 Conical Pendulum Could you find the NET force? T sin(θ) = mv2/R T cos(θ) = mg v = sqrt( gR tan(θ) ) 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 5/2/2017 Physics 214 Fall 2010 13 1D-03 Demonstrations of Central Force What will happen T T THE SHAPES/SURFACES OF SEMI-RIGID OBJECTS BECOME MORE when it is CURVED TO PROVIDE subjected to forcesGREATER CENTRAL FORCES DURING ROTATION. during rotation ? θ θ 2T cos (θ)= mv2/R 5/2/2017 Physics 214 Fall 2010 14 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. 5/2/2017 Physics 214 Fall 2010 15 1D-05 Twirling Wine Glass m v WHAT IS THE PHYSICS THAT KEEPS THE WINE FROM SPILLING ? g Same as N + mg = mv2/R string 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 5/2/2017 Physics 214 Fall 2010 16 1D-07 Paper Saw THE RADIAL FORCES HOLDING THE PAPER TOGETHER MAKE THE PAPER RIGID. Is paper more rigid than wood ? 5/2/2017 Physics 214 Fall 2010 17 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. 5/2/2017 Physics 214 Fall 2010 18 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 after the string breaks (as seen from above)? Explain. 4 • 3 2 A 1 Path number 3 5/2/2017 Physics 214 Fall 2010 19 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 N Q9 A car travels around a flat (unbanked) curve with constant speed. Rear Ff A. Show all of the forces acting on the car. B. What is the direction of the net force act. mg The force acts toward the center of the turn circle 5/2/2017 Physics 214 Fall 2010 20 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. 5/2/2017 Physics 214 Fall 2010 21 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 5/2/2017 Physics 214 Fall 2010 22 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 5/2/2017 Physics 214 Fall 2010 23 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/s2). What is the weight of an astronaut standing on the moon whose weight on earth is 180 lb? Wearth = m gearth = 180 lb gmoo Wmoon = m gmoon n gmoon = 1/6 gearth Wmoon = m 1/6 gearth = 1/6 m gearth = 1/6 (180 lb) 5/2/2017 Physics 214 Fall 2010 24 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? T= 12hrs 25min a) 3:30 PM + 2 (12 hrs 25 min) high tide = 3:30 PM + 24 hrs + 50 min = 4:20 PM t 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 5/2/2017 Physics 214 Fall 2010 25 Ch 5 CP 2 A Ferris wheel with radius 12 m makes one complete rotation every 8 seconds. a) Rider travels distance 2r 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? 5/2/2017 Physics 214 Fall 2010 26 Ch 5 CP 2 (con’t) a) S = d/t = 2r/t = 2(12m)/8s = 9.42 m/s Fcent b) acent = v2/r = s2/r = (9.42m/s)2/12m = 7.40 m/s2 c) Fcent = m v2/r = m acent = (40 kg)(7.40 m/s2) = 296 N W = mg = (40 kg)(9.8 m/s2) = 392 N Yes, his weight is larger than the centripetal force required. d) W – Nf = 296 N = 96 newtons N e) rider is ejected W 5/2/2017 Physics 214 Fall 2010 27 Ch 5 CP 4 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 = 2r, what is speed of passenger? b) What is centripetal acceleration? Compare it to gravity (9.8 m/s2) c) Passenger has mass = 60 kg, what is centripetal force required to produce the acceleration? Compare it to passengers weight. a) s = d/t = 2(3.0m)/1 = 19m/s b) a = v2/r = s2/r = (19 m/s)2/3m = 118 m/s2 = 12g 3m c) F = ma = (60 kg)(118 m/s2) = 7080 N F = ma = m (12 g) = 12 mg = 12 weight 5/2/2017 Physics 214 Fall 2010 28 Ch 5 CP 6 The period of the moons orbit about the earth is 27.3 days, but the average time between full moons is 29.3 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 = 0E Moon orbital period = 27.3 days = 0M 0M/0E = 27.3/364 0.075 5/2/2017 Physics 214 Fall 2010 29 Ch 5 CP 6 (con’t) b) (i) (ii) S E M E M S E (iii) M Day 0 Full Moon 27.3 Days Later This is not a full moon. This is the next full moon. S 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 0.075 of its full orbit in 27.3 days (see diagram (ii)). To be inline w/ sun and earth, moon must move thru same fraction of orbit (see diagram (iii)). 0.075 (27.3 days) 2 days. 5/2/2017 Physics 214 Fall 2010 30 Moon and tides anim0012.mov Tides are dominantly due to the gravitational force exerted by the moon. Since the earth and moon are rotating this effect also plays a role. The moon is locked to the earth so that we always see the same face. Because of the friction generated by tides the moon is losing energy and moving away from the earth. 5/2/2017 http://www.sfgate.com/getoutside/1996/jun/tides.html whytides.gif Physics 214 Fall 2010 31 Dark Matter For the orbit of a body of mass m about a much more massive body of mass M GmM/r2 = mv2/r and GM/r = v2. In fact M is the mass inside the orbit, that is the sun could be nearly as big as the orbit of the earth and it would not change anything. If we look at stars in motion in galaxies we find there is not enough normal matter to provide the necessary gravitational force. We believe this is caused by a new form of matter, which we call dark matter, and it comprises 25% of the energy in our Universe.(normal matter = 4.4%) 5/2/2017 Physics 214 Fall 2010 32 Space Elevator http://www.youtube.com/watch?v=F2UZDHHDhog http://www.pbs.org/wgbh/nova/sciencenow/3401/02.html station v counterweight 100,000km 5/2/2017 •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. •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 and a ground based booster light beam. •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 anchor ship, 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! Physics 214 Fall 2010 33 Trajectories to other planets To launch a space craft from earth to say Mars or Jupiter is quite complicated since all the planets are moving including the earth. One needs to be able to calculate a trajectory that minimizes the amount of fuel required. That is why in nearly all launches there are specific time windows which are optimum. 5/2/2017 Physics 214 Fall 2010 34