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Gravity and Circular Motion
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First, The story of Gravity… What do we think we know?
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What is Gravity? An attractive force between all masses. No one knows why it acts. What’s is measured in? Newtons. F g = mg
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Is gravity an action/reaction pair? The earth pulls on you – do you pull on the earth? Yes. Same force? Yes. Equal in magnitude and opposite in direction? Yes.
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Why do objects fall down? Do they? What does “down” mean? Define “down” relative to Earth. Define “fall.” More accurate to say that masses on earth are pulled in a path toward the center of the earth.
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What is the Universal Gravitational Force? Newton’s Universal Law of Gravitation says that every mass in the Universe exerts a gravitational force on every other mass. This is how a solar system is born. No one had proof until astronauts went to the moon.
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How is the Universal Force of Gravity calculated? Given two objects, m1 and m2, separated by a distance r (or d), then F g = Gm 1 m 2 /d 2 where G = 6.7 x 10 -11 units for G? Nm 2 /kg 2
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What is weight? The pull you experience due to your gravitational interaction with a planet.
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g But we have two equations: F g = Gm 1 m 2 /d 2 and F g = mg Well….. Try this calculation: Gm 1 /d 2 where G = 6.7 x 10 -11 Nm 2 /kg 2 M1 = mass of Earth = 6 x 10 24 kg, d = radius of Earth = 6.38 x 10 6 m What does it equal???
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What is weight? Each planet has a little g! It’s specific to its mass and radius. Weight is mg
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Force and Weight Does not depend upon The same Mass Weight
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Gravitational Interactions
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Hw #19 F g = Gm 1 m 2 /d 2 where G = 6.7 x 10 -11 Nm 2 /kg 2 F= 8.7 x 10 -12 N
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Hw #20 M 1 = 45 kg F g =m 1 g = (45kg)(9.8 m/s 2 ) = 441 N F net = -F g +F air = -441 + 250N = -191 N F net = ma a= F net /m = -191n/45 kg = -4.24 m/s 2
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Hw #21 m = 115 kg F g on earth = (115kg)(9.8m/s 2 ) = 1127N Mass is constant everywhere Weight = 0 if it is not near any other masses.
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HW #22 F g1 = 11000N- m 1 =F g1 /9.8 = 1122kg F g2 = 3400 N - m 2 =F g2 /9.8 = 347kg r = 12 m F g = Gm 1 m 2 /d 2 F g = 1.8 x 10 -7 N
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HW #23 m A = 363 kg m B = 517 kg m C = 154 kg r ab =.5 r ac =.75 r bc =.25 Net force on A = F BA + F CA = Gm a m b /(r ab ) 2 + Gm a m c /(r ac ) 2 = 5.7x 10 -5 N Net Force on B = F AB + F CB = -Gm a m b /(r ab ) 2 + Gm b m c /(r bc ) 2 = 3.5x 10 - 5 N toward c Net force on C = F AC + F CB = 9.2x 10 -5 N toward A
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HW #24 Use Pythagorean theorem F sun on moon = Gm s m m /r sm 2 = 4.3 x 10 20 N F earth on moon = Gm e m m /r em 2 = 2 x 10 20 N Net Force on moon = 4.7 x 10 20 N
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HW #25 Acceleration due to gravity on Mars = Gm mars /r mars 2 3.75 m/s 2 Fg = mg mars = (65kg)(3.75 m/s 2 ) = 243.7 N
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Moving in Circles
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What happens to YOU when you move in a circle?
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What in our physical experience seems to move “naturally” in a circle? (By itself.) Is it true?
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Put our thinking caps on… How can we explain the motion of objects moving in a circle, compared to objects moving in a straight line? What equations can we use to solve problems and interpret the motion of objects moving in a circle? Our old friends the Kinematics equations and Newton’s Laws!
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Circular Motion Definitions 1) Axis – a straight line around which rotation takes place.
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Circular Motion Definitions 2) Rotation – when an object turns around an internal axis. Internal axis is located within the body
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Definitions continued 3) Revolution – when an object turns around an external axis. The Earth’s revolutionary period is 1 year.
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Definitions continued 4) Uniform Circular motion – describes an object travelling in a circle at a constant (uniform) speed. NOT constant velocity!
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Definitions continued 5) Linear Speed (also called Tangential speed) – distance travelled per time. 6) Rotational speed (also called angular speed) - # of rotations per time.
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Definitions continued 7) Centripetal Force – Center seeking Force necessary to keep an object in uniform circular motion 8) Centrifugal Force – center fleeing Fake! Fictitious!
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Definitions continued 9) Centripetal Acceleration – acceleration that points in to the center of the circle. It is a change in direction
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Definitions 10) Period of uniform circular motion – Time required to complete one revolution.
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To move in a circle, a force must point in toward the center. What supplies that force? No new forces! It is gravity, friction, tension or a normal force. How does the object accelerate? In a direction in toward the center. Is an object in circular motion in equilibrium? Good way to get ketchup out of a bottle?
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Handout Examples Rubber stopper attached to a string. m=0.013 kg r = length of string = 0.93m Distance = 2πr T = period = 1.18s A) speed = distance/time = 2πr/T = 4.95 m/s B) a c = v 2 /r = (4.95 m/s) 2 /0.93m= 26.4 m/s 2 C) F c = ma c = mv 2 /r = 0.34N directed radially inward
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Jupiter orbiting the Sun What supplies the centripetal force? F g What is the magnitude of this force? First let’s use F c = mv 2 /r Convert T = 3.74 x 10 8 s Then v = 2πr/T = 13,070 m/s F c = mv 2 /r = 3.95 x 10 22 N
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Jupiter continued: Now let’s try by calculating F g : F g = Gm 1 m 2 /d 2 where G = 6.67 x 10 -11 F g = (6.67x 10 -11 )(2 x 10 30 )(1.8 x 10 26 ) (7.78 x 10 11 m) 2 F g = 3.97 x 10 22 N
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Try some problem solving. #1. v = 2πr/T Given v and r T = 2πr/v T = 160 (rounded from 162.8s) #2. v = 21m/s, r = 0.053m T = time for one revolution T = 2πr/v = 0.016s
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#3. T = 118 s, v = 17 m/s r = vT/2 π r= 320m #5. v = 98.8 m/s a c = 3.00g = 3.00 x 9.8 m/s 2 a c = v 2 /r r = v 2 /a c = r = 332 m
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Windshield Wiper #7. T = 4(.28s) r = 0.76m a c = v 2 /r a c = (2πr/T) 2 /r v = 2π (.76)/4(.28) = 4.26 m/s a c = (4.26m/s) 2 /0.76 a c = 23.9 m/s 2
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Helicopter blades #9. R = 6.7 m r = 3m a cR = (v 2 /R) = (2πR/T) 2 (1/R) a cr (v 2 /r) (2πr/T) 2 (1/r) R 2 /R = R/r = 6.67m/3m = 2.2 r 2 /r
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Crate on Flat bed Truck #11. Force is supplied by friction f = F c = mv 2 /r f = 426 N
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#12. f old = mv 2 /r f new = 1/3(f old ) mv new 2 /r = 1/3 (mv old 2 /r) m and r don’t change, so cancel them out Then… v new 2 = 1/3(v old 2 ) v new = square root of v old 2 /3 v new = 12.12 m/s
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Rotor-Ride Just you and the wall!
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Rotor Ride Do we need a sign that says you need to be a certain mass to ride this? Vertical: f = mg Horizontal: Fn = mv 2 /r Since f also = μFn, then μ(mv 2 /r) = mg μ(v 2 /r) = g Mass cancels! So the answer is NO!
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Swing Ride
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Homework Ch5 #19 Swing Ride Angle is 65 to the y axis, Length = 12m, m=220 kg Find Tension, T, and linear speed, v. Horizontal: Tsin65 = mv 2 /r Vertical: Tcos65 = mg T = mg/cos65 = 5101 N v 2 = (Tsin65)( r) / m r = 12sin65 = 10.9 m v 2 = (5101 N)sin(65)(10.9 m) / 220kg=229 v = 15.1 m/s
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Banked Curves
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A banked section of Highway #427 in Toronto.
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The force equation for the y direction is F N cos θ - mg= 0 F N cos θ= mg F N = mg/cos θ Then looking at the x force equation: F N sin θ= mv 2 /r F N sin θ = mg sin θ/cos θ = mv 2 /r tan θ= mv 2 /r / mg = v 2 /gr θ= tan-1[v 2 /(gr)] This is the banking angle!
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Quiz Topics Gravitational Interactions Big G vs. little g Three sphere in a line – finding Force of gravity on one. Find centripetal acceleration, v, T, & force Flat bend, rotor ride, banked curves Vocabulary
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Satellite Homework #27. Satellite over Jupiter Placed 6 x 10 5 m above surface, given mass of Jupiter, find v. So r = radius of Jupiter + height over surface r = 7.2 x 10 7 m V 2 = GM J /r v 2 = 1.76 x 10 9 m/s v = 4.195 x 10 4 m/s
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#28. Find r so astronauts on space station weigh half of their weight on Earth, if v = 35.8 m/s. Fn = Fc = mv 2 /r Fn = (1/2) mg (1/2) mg = mv 2 /r r = 2v 2 /g r = 262 m
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#29. Satellites circling unknown planet. Sat1 has v1 = 1.7 x 10 4 m/s, and r = 5.25 x 10 6 m. Sat2 has r = 8.6 x 10 6 m. Find v for Sat2. V 2 = GM PLANET /r So M PLANET G = constant = v 2 r V 1 2 r 1 = v 2 2 r 2 V 2 = 1.33 x 10 4 m/s
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Vertical Circular Motion
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Ball on a string: At the bottom: T – mg = ma = mv 2 /r Max speed at the bottom means T = mg + mv 2 /r Max T
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Ball on a string: At the Top T + mg = ma = mv 2 /r T = mv 2 /r – mg Min speed at the top, occurs if T = 0, then Vmin = √rg
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Minimum Velocity Occurs always at the top! Fn + mg = mv 2 /r Vmin occurs if Fn goes to zero for an instant. So Fn + mg = mv 2 /r becomes mg = mv 2 /r Or v = √(rg)
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Ch. 5: Homework #37. Plane flies over the top of a vertical circle, passengers experience weightlessness. Given v = 215 m/s, what’s r? At the top, -Fn + mg = mv 2 /r Fn = 0 N, so -Fn + mg = mv 2 /r becomes mg = mv 2 /r Or r = v 2 /g r = 4717 m
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#38. Fighter Pilot Dives to ground at 230 m/s. http://physics.bu.edu/~duffy/semester1/c8_vertical.html Find r so Fn = 3mg. At bottom of circle: Fn – mg = mv 2 /r 3mg – mg = mv 2 /r 2g = v 2 /r r = v 2 / 2g = 2699 m
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#39. Ball spinning on string. Given r, T max,and m, find v at top and bottom. Speed at the bottom: Max T at bottom, T – mg = ma = mv 2 /r v 2 = (T – mg)(r/m) = 23.1 V = 4.8 m/s
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#39. continued Speed at Top: NOTE – It doesn’t ask for min velocity Assume max T at top too T + mg = ma = mv 2 /r v 2 = (T + mg)(r/m) = 32.9 v = 5.7m/s at top
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#40. Downhill Skiier Bottom of the curve, Fn – mg = mv 2 /r Fn = mg + mv 2 /r, given r, v, m, Fn=? Fn = 1306 N
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#41. Demolition Ball Given m, r, and v, and motion is at the bottom of swing, find Tension in cable. T – mg = mv 2 /r T = mg + mv 2 /r T = 2.87 x 10 4 N
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#42. Roller-Blader Given r, find v. At the top of the hill, -Fn + mg = mv 2 /r V min occurs if Fn goes to zero for an instant. -Fn + mg = mv 2 /r mg = mv 2 /r v 2 = rg v = 14 m/s
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#43. Spinning stone 2 ways. Vertical equation: Given T max = 1.1 T H At bottom T max – mg = mv 2 /r 1.1(mv 2 /r) – mg = mv 2 /r 1.1 (mv 2 /r) - mv 2 /r = mg 0.1(mv 2 /r) = mg v 2 = (gr)/0.1 V= 9.6 m/s Given r= 0.95m T max = 1.1 T H Horizontal eq: T H = mv 2 /r Find v (same for both)
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