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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 gravitationally bound orbits can be elliptical (more or less “squashed circles”.) At any given instant a moving object, if it’s not 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. Week 4 Physics 214 Spring 2010 1 Circular motion F = ma = mv2/r a = v2/r If the velocity of an object changes direction then the object experiences an acceleration. This requires a force. These are called “centripetal” acceleration and centripetal force and are directed toward the center of the circle. This is the force on you rounding a corner in a car Week 4 Physics 214 Spring 2010 2 Circular motion F = ma = mv2/r a = v2/r Where does this come from? a = Δv/Δt = v.v/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 v1 and v2 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) Week 4 Physics 214 Spring 2010 3 Angular Measure in Radians R Arc Length The “natural” measure of an angle is in RADIANS, which is just the arc length subtended by the angle, divided by the radius of the circle. A full circle, which is 360 degrees, has an arc length (circumference) of 2πR. Divide this arc by R and you get 2π radians is a full circle is 360 degrees. So 360 deg/2π rad = 57.296 [deg/rad] = 1.00000 You can multiply or divide by this number and units to convert back and forth. Just use algebra on the [units] to know which to do. Week 4 Physics 214 Spring 2010 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 2 A 1 A=1, B=2, C=3 D=4 Path number 3 Week 4 Physics 214 Spring 2010 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. Th = mv2/r Tv = W = mg r is the HORIZONTAL distance to the center of the circle. Week 4 Physics 214 Spring 2010 6 Cars Ff Ff Rear view When a car turns a corner it is friction betweenView 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. The horizontal component of N exactly matches the needed Fc Note Ff adjusts itself (up to break point) And it is the HORIZONTAL component of Ff Ff which may help supply the needed Fcentrip. Note: Nv must balance W. Question: if v = 0, where does Ff point? Week 4 Physics 214 Spring 2010 A. Upslope from Above W = mg B. Downslope 7 Vertical circles N v W = mg mg – N = mv2/r If v = 0 then N = mg As v increases N becomes smaller When v2/r = g The car becomes weightless. Same as weightless-training airplane, for astronauts Week 4 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 Physics 214 downward Spring 2010 Ferris wheel At the bottom N - mg = mv2/r At the top mg – N = mv2/r With seatbelt, at top if v2 is big enough, seatbelt would hold you down, your “weight” would 8 be upside down! Response to a thoughtful question Monday a student asked me how we can know if the rotating disks had moved 7o or maybe one or more full revolutions + 7o as the bullet went half a meter. Logically, that could be the case, but quantitatively you can pretty easily rule it out by extra outside knowledge – namely, that the bullet is travelling pretty fast. At 17 revolutions per second, if the disk made one full revolution, the bullet’s speed is about 0.5m x 17 “per s” [d = v t and v = d/t and t is 1/17 second for one rev.] This is about 8.5 m/s, and if you fire a bullet at a can maybe 25.5 m away, it sure takes a lot less than 3 seconds for the can to jump, as you can observe. Week 4 Physics 214 Spring 2010 9 Response to another thoughtful question Last Friday, several students wanted a better explanation of the inertia ball (where you can break several strings while the single upper suspending string survives!) I think the easiest way to think about it is to use the concept of IMPULSE. The hammer causes the rod in the lower string loops to speed up VERY briskly (very large acceleration). The heavy ball would need a fair amount of impulse to reach full speed. Impulse is Ft, and if the interval t is very short, that would mean a LOT of force in the lower loops while the upper loop doesn’t exert a lot of extra force (it’s the NET motion and impulse.) Week 4 force that generatesPhysics 214 10 Spring 2010 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 (the moons of Jupiter are like a small solar system.) Tycho Brahe was the most famous naked-eye astronomer. Kepler, his assistant used the data to draw quantitative Week 4 Physics 214 11 conclusions. Spring 2010 Keplers Laws 1) Orbits are ellipses. For 0 “eccentricity” Æ circle 2) The radius vector sweeps out equal areas in equal times 3) T2 proportional to r3 T is the period. For the earth, T is one year and r is the average radius (distance to the sun). 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 Week 4 Physics 214 Spring 2010 12 Keplers Laws Period T = 2πr/v = distance (arc length) / speed So v = 2πr/T = circumf./period = 2π 94M mi/1 yr v ≈ 2π . 94 . 106 mi/ π . 107 s = 18.4 mi/s Compare with escape velocity from earth, which I think is around 7 mi/s Week 4 Physics 214 Spring 2010 13 Newton and Gravitation Newton developed the Law of Universal Gravitation force between ANY two objects is proportional to M1m2/r2 . The (very small) constant of proportionality was measured by Cavendish more than 100 years later G = 6.67 x 10-11 N.m2/kg2. Since at the earths surface mg = m(GMe/r2) the experiment measured the mass of the earth Week 4 Physics 214 Spring 2010 http://www.physics.purdue. edu/class/applets/Newtons Cannon/newtmtn.html 14 Newton and Gravitation The force between ANY two objects is ATTRACTIVE and = GM1m2/r2 . 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 (provided the observer is outside the sphere!). Week 4 Physics 214 Spring 2010 15 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 D. 1.7x10-8 N C. 1.7x10-5 N E. 1.7x106 N F=GMM/r2 = 6.67E-11Nm2/kg2 x 10E3kgx1E3kg/(2m)2 F = 6.67/4 x E(-11+6)N Week 4 exponent is -5 Physics 214 Spring 2010 answer is C. 16 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 [or M is the mass of the earth and m the mass of an earth satellite.] Cancel m & solve GM/r3=v2/r2 but v/r = 2π/T GM/r3=(2π/T)2 Week 4 T2/r3 = 4π2/GMs Use average radius for elliptical orbits. Works for other attractors: Earth, Jupiter. Physics 214 Spring 2010 Use right Mass17 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) T2/r3 = 4π2 /GMe R3 =T2GMe /4π2 T is 24 . 3600 s, G=6.6710-11, Me= 5.971024 kg, take cube root both sides to find R in meters, etc. Geosynchronous orbit must be CIRCULAR, and will be quite off-scale on the above picture. Also is equatorial. Week 4 Physics 214 Spring 2010 18 Geosynchronous Orbits The notion of a geosynchronous satellite for communication purposes was first published in 1928 (but not widely so) by Herman Potočnik.[1] The idea of a geostationary orbit was first published on a wide scale in a paper entitled "Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?" by Arthur C. Clarke, published in Wireless World magazine in 1945.[2] In this paper, Clarke described it as a useful orbit for communications satellites. As a result this is sometimes referred to as the Clarke Orbit.[3] Similarly, the Clarke Belt is the part of space approximately 36,000 km (22,000 mi) above sea level, in the plane of the equator, where near-geostationary orbits may be implemented. Geostationary orbits are useful because they cause a satellite to appear stationary with respect to a fixed point on the rotating Earth. As a result, an antenna can point in a fixed direction and maintain a link with the satellite. The satellite orbits in the direction of the Earth's rotation, at an altitude of 35,786 km (22,236 mi) above ground. This altitude is significant because it produces an orbital period equal to the Earth's period of rotation, known as the sidereal day. Week 4 Physics 214 Spring 2010 19 Summary of Chapter 5 Circular motion and centripetal acceleration and force. Fc = mv2/r Ferris wheel, car around a corner or over a hill. “Circular” pendulum. 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 At any given instant, FG is given by the distance BETWEEN CENTERS OF MASS Week 4 Physics 214 Spring 2010 Wrong r !! use AVG. T2/r3 = 4π2/GMs 20 Examples of circular motion Vertical motion N Looking down on stunt driver’s cylindrical cage N v v W = mg mg – N = mv2/r N mg N - mg = mv2/r Week 4 N = mv2/r Side view Red force arrows should Ff be equal, if enough static friction is N available. If μs is too small, not mg enough Ff and car will slide mg = Ff downwards. Physics 214 Spring 2010 v T mg Swing object on a string in a vertical circle mg + T = mv2/r top T - mg = mv2/r bottom 21 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 mv2/r = m(3v)2/3r = (9/3)mv2/r : Three times as much Fc is needed. Week 4 Physics 214 Spring 2010 22 Moon and tides 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. anim0012.mov 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 http://www.sfgate.com/getoutside/1996/jun/tides.html Week 4 whytides.gif Physics 214 Spring 2010 23 Moon, Sun, 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 FG is less and Fc (because of v2) is more, than required for the orbit of the Earth’s center. And vice versa for the Moon-ward bulge. The Sun’s effect is equally important!! Week 4 anim0012.mov ~240,000 mi ~384,000 km X E+M Centerof Mass is 76 times closer to Earth’s center, since Earth is 76 times heavier than Moon. Earth & Moon both orbit around their mutual CofM, ~3000 mi from Earth’s center http://www.sfgate.com/getoutside/1996/jun/tides.html whytides.gif Physics 214 Spring 2010 24 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.] anim0012.mov X The Earth spins faster than the Moon orbits, and in the same direction, so the Earth’s tidal bulges “torque” the Moon’s orbit. The Earth gives up “angular momentum” to the Moon’s orbit. This actually moves the Moon’s orbit further away and slows 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!!) Because the Earth spun faster then, days were shorter, year has not changed. E+M Centerof Mass is 76 times closer to Earth since Earth is 76 times heavier than Moon Earth & Moon both orbit around their mutual CofM, ~3000mi from Earth’s center Someday far in the future, the Earth’s spin may get locked to the Moon’s orbit! Week 4 http://www.sfgate.com/getoutside/1996/jun/tides.html whytides.gif Physics 214 Spring 2010 25 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 = 0E (I’d have called it TE) Moon orbital period = 27.3 days = 0M ( “ TM) 0M/0E = 27.3/365.25 ≈ 0.0747 Week 4 Physics 214 Spring 2010 26 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 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! 0.0747 (27.3 days) ≈ 2.04 days. Apparent lunar month ≈ 29.34 Week 4 Physics 214 27 days Spring 2010 Ch 5 CP 6 (con’t) b) (i) (ii) S E M E M S M E (iii) S Day 0 Full Moon 27.3 Days Later This is not a full moon. This is the next full moon. Fraction of 2 pi radians angle Fraction of a revolution ------------- c) Exact solution: θE = 2π(t/365.25) = θM = 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/365.25 = t/27.3; 1 = -0.002738 t + 0.036630 t = 0.033892 t t = 29.505 days compare with approx. 29.34 days previous slide Week 4 Physics 214 Spring 2010 28 Dark Matter For the orbit of a body of mass m about a much more massive M is the body of mass M: GmM/r2 = mv2/r and GM/r = v2. 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 about 5% is “baryonic” matter (made of nuclei and atoms) and only about 1% – 2% of the Universe is made of shining stars!! v (and v2) constant v~1/√r Week 4 Physics 214 Spring 2010 Total M within r from center must grow linearly with r Î Green curve 29 Space Elevator (courtesy of Arthur C. Clarke) station v counterweight 100,000km Week 4 •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! Physics 214 Spring 2010 30 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. Week 4 Physics 214 Spring 2010 31 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.) Week 4 Physics 214 Spring 2010 32 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. Week 4 Physics 214 Spring 2010 33 1D-02 Conical Pendulum T sin(θ) = mv2/R T cos(θ) = mg v = sqrt( gR tan(θ) ) ‘cuz tan() = v2/gR Could you find the NET force? 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 Week 4 Physics 214 Spring 2010 34 1D-03 Demonstrations of Central Force THE SHAPES/SURFACES OF SEMI-RIGID OBJECTS BECOME MORE CURVED TO PROVIDE GREATER CENTRAL FORCES DURING ROTATION. T What will happen when it is subjected to forces during rotation ? θ T θ Fc= mv2/R Week 4 Physics 214 Spring 2010 The force comes from the strips wanting to un-flex 35 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. Week 4 Physics 214 Spring 2010 36 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 (and the wine in the glass). Week 4 Physics 214 Spring 2010 37 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 Week 4 Physics 214 Spring 2010 38 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. Week 4 Physics 214 Spring 2010 39 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 Week 4 Physics 214 Spring 2010 40 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 Week 4 Physics 214 Spring 2010 41 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 gmoon Wmoon = m gmoon gmoon = 1/6 gearth Wmoon = m 1/6 gearth = 1/6 m gearth = 1/6 (180 lb) = 30 lb Week 4 Physics 214 Spring 2010 42 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 Week 4 Physics 214 Spring 2010 43 1D-07 Paper Saw THE RADIAL FORCES HOLDING THE PAPER TOGETHER MAKE THE PAPER RIGID. Is paper more rigid than wood ? Week 4 Physics 214 Spring 2010 44 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. Week 4 Physics 214 Spring 2010 45 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? Week 4 Physics 214 Spring 2010 46 Ch 5 CP 2 (con’t) a) S = d/t = 2πr/t = 2π(12m)/8s = 9.42 m/s Fcent r= 12m 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 Week 4 Physics 214 Spring 2010 47 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 = 2πr, 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 = 2π(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 Week 4 Physics 214 Spring 2010 48