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Circular Motion Kinematics Centripetal Acceleration PowerPoint Presentation
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Circular Motion Kinematics Centripetal Acceleration

Circular Motion Kinematics Centripetal Acceleration

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Circular Motion Kinematics Centripetal Acceleration

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  1. Circular Motion Kinematics Centripetal Acceleration v1 v = v2+ (-v1) Triangles in the same proportion l v v2 v2 v r r   -v1 An object moves around a circle at constant speed. By geometry the triangle formed by the radii and l is in the same proportion as the triangle formed by the velocity vectors and v v = l v r So… v = l v r Now… ac = v t = v l r t = v2 r Note that the centripetal acceleration (v2/r) is directed towards the center of the circle

  2. Circular Motion Dynamics Centripetal Force According to Newton’s second law (Fnet = ma) an object moving in a circle at constant speed must have a net force in the same direction as the acceleration…. This must be towards the center of the circle Fc = mv2 r The force that keeps a car moving around a bend in the road at constant speed is friction directed towards the center of the turn The force that keeps the moon moving around the earth at a constant rate is the force of gravity directed towards the center of the Earth.

  3. Circular Motion Dynamics Centripetal Force A 1000 kg car rounds a curve on a flat road of radius 50m at a speed of 14m/s. Will the car make the turn of will it skid if a) the pavement is dry (Ffr = 5880 N) and b) the pavement is icy (Ffr = 2450 N )? c) Does the ability of the car to make the turn depend on the mass of the car? m = 1000 kg r = 50m v = 14 m/s a) FFr = 5880 N Force needed is… FC = mv2 r = (1000kg)(14m/s)2 (50m) = 3920 N The maximum friction force that can be provided by the car’s tires is bigger than the force needed to keep the car turning soit will be able to make the turn b) FFr = 2450 N The maximum friction force that can be provided by the car’s tires is now smaller than the force needed to keep the car turning soit won’t be able to make the turn c) The more mass the car has, the more friction between the tires and the road but the car has proportionally more resistance to a change in motion. This will offset the increased friction, so bigger cars can’t turn quicker. (plus the higher center of gravity makes them more likely to role!)

  4. Circular Motion Dynamics Centripetal Force The Moon (m= 7.35 x 1022 kg) orbits the Earth at a distance of 384 x 106 m. a) What is the moon’s acceleration towards the Earth? b) How much does the Earth pull on the moon? c) Does the moon pull on the Earth more or less or the same as the Earth pulls on it? (The moon moves around the Earth one orbit every 27.4 days) mm = 7.35 x 1022 kg r = 384 x 106 m T = 27.4 days v = ? Fg = ? v = d / t = 2r / T Fc v v = 2 (384 x 106m) / (27.4D x 24 H/D x 3600 s/H) = 1019 m/s ag = v2 r = (1019m/s)2 (384 x 106 m) = 0.0027 m/s2 Fg = mmv2 r = (7.35 x 1022kg) (1019m/s)2 (384 x 106 m) = 1.99 x 1020 N The moon pulls equally back on the Earth according to Newton’s third law of motion. The moon moves around the Earth because of its much smaller mass (inertia).

  5. Circular Motion Dynamics Universal Gravitation We can define a field as an area of force that surrounds an object. Gravitational fields exist around all masses. The amount of gravitational force between two objects is due to the amount of mass each object has. It decreases in strength with distance in proportion to the inverse of the distance squared. M2 Fg M1 Fg Fg = G M1 M2 r122 r G = Universal Gravitational Constant = 6.67 x 10-11 Nm2/kg2 M = Mass of objects (kg) r = distance separating centers of masses (m)

  6. Circular Motion Dynamics Universal Gravitation What is the force of gravity acting on a 65kg man standing on the surface of planet Earth? (rearth = 6.38 x 106 m, MEarth = 5.97 x 1024 kg) Fg = G ME MM rE2 MM Fg rE Fg = 6.67 x 10-11Nm2/kg2 (5.97 x 1024kg)(65kg) (6.38 x 106m)2 ME Fg = 635.9 N = 636 N Check: Fg = mM g = (65kg) (9.81N/kg) = 637.6 N = 638 N

  7. Circular Motion Dynamics Universal Gravitation Tides are created by the gravitational attraction of the sun and moon on Earth. Calculate the net force pulling on Earth during a) a new moon and b) a full moon and c) a first quarter moon. (Mmoon = 7.35 x 1022 kg, MEarth = 5.98 x 1024 kg, MSun = 1.99 x 1030 kg, rME = 3.84 x 108 m, rSE = 1.50 x 1011 m) Earth New Moon Full Moon Sun 1st Quarter FSE = G MS ME rSE2 = 6.67 x 10-11Nm2/kg2 (1.99 x 1030kg)(5.98 x 1024kg) (1.50 x 1011m)2 = 3.53 x 1022N FME = G MM ME rME2 = 6.67 x 10-11Nm2/kg2 (7.35 x 1022kg)(5.98 x 1024kg) (3.84 x 108m)2 = 1.99 x 1020N

  8. Circular Motion Dynamics Universal Gravitation Earth New Moon Full Moon Sun 1st Quarter FSE = 3.53 x 1022N FME = 1.99 x 1020N a) During a new moon FE = FSE + FME = 3.53 x 1022 N + 1.99 x 1020 N = 3.55 x 1022 N b) During a full moon FE = FSE + FME = 3.53 x 1022 N - 1.99 x 1020 N = 3.51 x 1022 N c) During 1st Quarter FE = FSE + FME FSE FE = (FSE2 + FME2) = ((3.53 x 1022 N)2 + (1.99 x 1020 N)2)  FME FE = 3.53 x 1022 N FE  = tan-1 (FME / FSE) = tan-1 (1.99 x 1022 / 3.53 x 1022 ) = 0.320

  9. Circular Motion Dynamics Gravitational Field Strength Field strength is defined as the amount of force per quantity. For a gravitational field around an object it is the number of Newton’s of force acting on every kg of a second object placed in this field. g = Fg M2 = G M1 r122 g What is the gravitational field strength on the surface of the moon? (Mm = 7.35 x 1022 kg, rm = 1.74 x 106 m) r12 M1 gM = G MM rM2 gM = 6.67 x 10-11Nm2/kg2 7.35 x 1022 kg (1.74 x 106m)2 = 1.62 N/kg (about 1/6 that on the surface of Earth)

  10. Circular Motion Dynamics Gravitational Field Strength A typical white dwarf star, which once was an average star like our sun but is now in the last stages of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity (g) of this star? (Ms = 1.99 x 1030 kg) g gWD = G MS rM2 rM gWD = 6.67 x 10-11Nm2/kg2 1.99 x 1030 kg (1.74 x 106m)2 = 4.38 x 107 N/kg MS (This is about 4.5 million times our surface gravity on Earth)

  11. Circular Motion Dynamics Gravitational Potential Gravitational Potential Energy is the energy an object has due to its position in a gravitational field. The change in GPE (GPE) between two points at different distances from an object we have already expressed as - M2gdv according to the work-energy theorem now….g = G M1 r122 GP The gravitational potential energy of mass M2 at distance r12 can be defined as…. GPE = - G M1 M2 r12 r12 M1 Like field strength, the amount of potential energy per kg of a second object placed at this position in the gravitational field is called gravitational potential GP = - G M1 r12 The GPE and GP get smaller the further you go from the mass M1. They both are reduced to zero at infinity. It should also be noted that the work done to move a mass between two points in a gravitational field is independent of the path taken. (proved by calculus)

  12. Circular Motion Dynamics Escape Speed How much speed is needed to send a rocket (M2) soaring into space so that it escapes the pull of a planet (M1)? If the rocket just makes it to infinity and slows down all the way its final velocity (vf) will be zero and its final GPE will be zero so the total energy will be zero ve KEi + PEi = KEf + PEf = 0 rp 1/2 M2 ve2 - G M1 M2 = 0 r M1 ve =  2G M1 rp Determine the escape speed from the surface of earth rearth = 6.38 x 106 m MEarth = 5.97 x 1024 kg G = 6.67 x 10-11 Nm2/kg2 ve =  2G ME rearth =  (2(6.67 x 10-11 Nm2/kg2) (5.97 x 1024 kg )) 6.38 x 106 m = 11.2 x 103m/s (25000 mi/h)

  13. Circular Motion Dynamics Escape Speed - Black Hole When a huge star many times our own sun’s mass runs out of nuclear fuel gravity causes it to collapse in on itself eventually resulting in an incredibly dense piece of matter. Light has a speed (c) of 3 x 108 m/s. If the escape speed is greater than this not even light will escape. Karl Schwarzchild calculated the radius around a black hole at which light wouldn’t be able to escape. This boundary is called the event horizon (R) c =  2G MBH R So ….R = 2G MBH c2 R MBH If a black hole has a mass of 3 times our own sun’s mass and a radius of 2km determine its event horizon (Ms = 1.99 x 1030 kg) c = 3 x 108 m/s MBH = 5.97 x 1030 kg G = 6.67 x 10-11 Nm2/kg2 R = 2G MBH c2 = 2(6.67 x 10-11 Nm2/kg2) (5.97 x 1030 kg ) (3 x 108)2 m = 8850m (8.9 km)

  14. Circular Motion Dynamics Kepler’s Third Law Kepler’s third law relates the force of gravity using Newton’s Universal Law of Gravitation with centripetal force. This force of gravity that the earth exerts on the moon is the centripetal force that the moon requires to orbit the earth. That means that… Fg = Fc G ME m/ r2 = m vt2 / r or…..vt2 = GME / r With this equation, we can find the velocity of the moon around the earth given the mass of the earth and the radius of the moon’s revolution. This relationship is the same for any planet revolving around the sun. Since the period of a planet, T, can be related to velocity by v = 2r / T, we’re all set. Now we just solve for T. (2r / T)2 = GM / r So…..T2 = (42 r3) / (GM). or…..T2  r3 This is Kepler’s third law. Using this equation, we can find the period of a planet if we know the radius of its revolution and the mass of the planet it revolves around. Find the time for the earth to revolve around the sun, given that the sun’s mass is 1.991 • 1030 kg and that the earth is 1.496 • 1011 m from the sun. = 9.953 x 1014 s2. T2 = [4 • 2 • (1.496x1011 m )3] / [(6.673x10-11 N•m2/kg2) • (1.991x1030 kg)] T = 3.155 x 107 s = 365.14 days. (as expected!!)

  15. Circular Motion Dynamics Satellites and Weightlessness Satellites are freefalling towards the earth just as a dropped ball falls towards earth but they have such high tangential velocities that as they fall they follow the curvature of the planet. The earth’s surface drops about 5m for every 8km horizontally. How fast should a projectile be fired to go into a low orbit around the earth? A projectile falls about 5m in 1s so it should be fired at 8km/s (18000 mi/h) Motion of projectile without gravity If the tangential speed is high enough the projectile will fall “around the earth”

  16. Circular Motion Dynamics Satellites and Weightlessness How can someone in the space shuttle be experiencing weightlessness when the spacecraft is experiencing nearly as much gravitational acceleration (g) as someone on the surface? An airborne athlete and an astronaut both experience weightlessness

  17. Circular Motion Dynamics Satellites and Weightlessness Circular Motion Dynamics Satellites and Weightlessness Our experience of weight is really the normal force of the surface we are in contact with pushing back on our body You are made aware of this idea when traveling on a high speed elevator or riding on an amusement park ride. You are made aware of this idea when traveling on a high speed elevator or riding on an amusement park ride. The normal force has a small magnitude at the top of the loop (where the rider often feels weightless) and a large magnitude at the bottom of the loop (where the rider often feels heavy).

  18. Circular Motion Dynamics Apparent Weightlessness Consider a roller coaster track that has a series of hills and dips as shown below. The black arrows show that the centripetal acceleration is directed towards the center of the circular shaped arcs as the car moves along the track. The forces acting on the car at positions A, B and D are shown below

  19. Circular Motion Dynamics Apparent Weightlessness Consider a roller coaster moving over a series of hills as shown on the previous slide. At the bottom of a hill the track has do do two things: i) It has to support the car’s weight which means that it has to provide a normal force (FN) upwards ii) It also has to change the motion of the car because the path that the car follows is curved (not straight!). Because the direction of motion changes constantly the track has tp provide a centripetal force towards the center of the turn. Therefore…… FN = Fg + FC = mg + mv2 / r = m (g + v2 / r)

  20. Circular Motion Dynamics Apparent Weightlessness At the top of the hill the track can push upwards or pull downwards depending on the car’s speed. FN i) If the car travels slowly over the top FN is directed upwards but is a little bit smaller than Fg (down) because the car starts to lift just a little bit off the track because it wants to continue in a straight line path as it goes over the top ii) If the car travels fast over the top FN is directed downwards because the car actually lifts a lot off the track but a second set of wheels below the track stop the car from moving upwards so the track applies a force back down (3rd Law) on the car (Fapp) Notice that both the top and bottom of the hill there is an unbalanced force (FNET) towards the center of the turn. This is the centripetal force (FC) that allows the car to follow the curved track Also note that at some speed between fast and slow the passengers in the car (and the car itself) will not receive a normal force and so will feel weightless.

  21. Circular Motion Dynamics Torque The ability of a force to rotate an object about some axis is measured by quantity called torque, . Torque like force is a vector quantity. Its magnitude is:  = Frperp where rperp is the lever arm and is the perpendicular distance from the axis of rotation to a line drawn along the direction of force. If there are several forces acting on the object then the net torque is obtained by summing the torques produced by each of these forces., thus net =   = 1 + 2 +…..= F1r1perp + F2r2perp +….. A student pushes on a door furthest from the hinges (1.5m) with a force of 15N at an angle of 350 to the plane of the door. Determine the torque on the door. F   r rperp  = Frperp = F r sin = (15N) (1.5m) sin350  = 13 Nm Note: Positive torque is assigned counterclockwise and negative torque clockwise

  22. Circular Motion Dynamics Rotational Equilibrium The first condition for equilibrium that we have discussed is namely, the sum of all the external forces is zero. With no net forces acting on an object it obeys Newton’s first law, I.e. no accelerations and thus no changes in motion. (F= 0) The object could still rotate and change its rate of rotation if the torques don’t add to zero. The second condition for equilibrium, therefore, is that there are no changes in rotation. This happens when the sum of the external torques adds to zero, so we have  = 0 Position of the Axis of Rotation If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque; the location of the axis is completely arbitrary. The Center of Gravity The center of gravity is that point in or near an object where all the torques due to the weight of the object add to zero no matter how the object is orientated. It is what we usually call the balance point. The x component is found from the equation xcg = m1x1 + m2x2 +… = i mi+xi m1 + m2 +… M The y and z components of the center of gravity are found in a completely analogous manner. The center of gravity of a symmetrical body that is homogenous must lie on the axes of symmetry.

  23. Circular Motion Dynamics Rotational Equilibrium A hungry 700 N bear walks out on a beam in an attempt to retrieve some “goodies” hanging at the end. The beam is uniform, weighs 200 N, and is 6.00 m long; the goodies weigh 80.0 N. a) Draw a FBD for the beam. b) When the bear is at x=1.00 m, find the tension in the wire and the components of the reaction force at the hinge. c) If the wire can withstand a maximum tension of 900 N, what is the maximum distance the bear can walk before the wire breaks? Ty = T sin600= 0.866T a) 700 N X Tx = T cos600 = 0.5 T H 3m 3 m V 80 N 200 N b) If x = 1 m then left end = (-700 N)(1m) - (200 N)(3 m) - (80 N)(6 m) + (0.866T)(6 m) Equating this to zero gives: T = 342 N From Fx = 0, H = 0.5 T = 171 N From Fy = 0, V = 980 N - 0.866T = 683 N c) If T = 900 N left end = (-700 N)(X) - (200 N)(3 m) - (80 N)(6 m) + (779.4 N)(6 m) Equating this to zero and solving for x gives: x = 5.13 m