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UW-Madison PHYSICS 207 - PHYSICS 207 Lecture Notes

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Lecture 11Chapter8 Reprisal of : Uniform Circular MotionNon-uniform Circular MotionUniform or non-uniform circular motionKey stepsExample The pendulumExample Gravity, Normal Forces etc.Conical Pendulum (Not a simple pendulum)Conical Pendulum (very different)Another example of circular motion Loop-the-loop 1Loop-the-loop 1Loop-the-loop 2Orbiting satellites vT = (gr)½Geostationary orbitSlide 16RecapPhysics 207: Lecture 11, Pg 1Lecture 11Goals:Goals: Employ Newton’s Laws in 2D problems with circular motion Relate Forces with accelerationAssignment: HW5, (Chapter 7, 8 and 9 due 10/18)For Wednesday: Reading through 1st four sections in Ch. 9Physics 207: Lecture 11, Pg 3Chapter8 Reprisal of : Uniform Circular MotionFor an object moving along a curved trajectory with constant speeda = ar (radial only)|ar |= vt2rarvPhysics 207: Lecture 11, Pg 4Non-uniform Circular MotionFor an object moving along a curved trajectory, with non-uniform speeda = ar + aT (radial and tangential)araTdtd| |v|aT|= |ar |= vT2rPhysics 207: Lecture 11, Pg 5Uniform or non-uniform circular motionimplies and if there is acceleration there MUST be a net force|aradial | =vT2 / rPhysics 207: Lecture 11, Pg 6Key stepsIdentify forces (i.e., a FBD)Identify axis of rotationApply conditions (position, velocity &acceleration)Physics 207: Lecture 11, Pg 7Example The pendulumConsider a person on a swing: When is the tension equal to the weight of the person + swing? (A) At the top of the swing (turnaround point)(B) Somewhere in the middle(C) At the bottom of the swing(D) Never, it is always greater than the weight (E) Never, it is always less than the weight axis of rotation(A)(B)(C)Physics 207: Lecture 11, Pg 8Example Gravity, Normal Forces etc.at bottom of swing vT is maxFr = m ac = m vT2 / r = T - mgT = mg + m vT2 / rT > mgat top of swing vT = 0Fr = m 02 / r = 0 = T – mg cos T = mg cos T < mgaxis of rotationvTmgTmgTyxPhysics 207: Lecture 11, Pg 9Conical Pendulum (Not a simple pendulum)Swinging a ball on a string of length L around your head(r = L sin )axis of rotation Fr = mar = T sin  Fz = 0 = T cos – mg so T = mg / cos  (> mg)mar = (mg / cos ) (sin  )ar = g tan  vT2/r  vT = (gr tan )½ Period:T= 2 r / vT =2 (r cot /g)½ = 2 (L cos /g)½ rLPhysics 207: Lecture 11, Pg 10Conical Pendulum (very different)Swinging a ball on a string of length L around your headaxis of rotationPeriod:t = 2 r / vT =2 (r cot /g)½ = 2 (L cos / g )½ = 2 (5 cos  / 9.8 )½ = 4.38 s = 2 (5 cos  / 9.8 )½ = 4.36 s = 2 (5 cos 1/ 9.8 )½ = 4.32 srLPhysics 207: Lecture 11, Pg 11A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed vt at the top so that it can manage the loop successfully ?Another example of circular motionLoop-the-loop 1Physics 207: Lecture 11, Pg 12To navigate the top of the circle its tangential velocity vT must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching). Loop-the-loop 1Fr = mar = mg = mvT2/r vT = (gr)1/2mgvTPhysics 207: Lecture 11, Pg 13The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point?Hint: The car is constrained to the track. Loop-the-loop 2mgv NFr = mar = mvB2/r = N - mgN = mvB2/r + mgPhysics 207: Lecture 11, Pg 14Orbiting satellites vT = (gr)½ Net Force: ma = mg = mvT2 / r gr = vT2 vT = (gr)½The only difference is that g is less because you are further from the Earth’s center!Physics 207: Lecture 11, Pg 15Geostationary orbitPhysics 207: Lecture 11, Pg 16Geostationary orbitThe radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth. Fgravity is proportional to r -2 and so little g is now ~10 m/s2 / 50 vT = (0.20 * 42000000)½ m/s = 3000 m/s At 3000 m/s, period T = 2 r / vT = 2 42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) Great for communication satellites (1st pointed out by Arthur C. Clarke)Physics 207: Lecture 11, Pg 18RecapAssignment: HW5, For Wednesday: Finish reading through 1st four sections in Chapter


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UW-Madison PHYSICS 207 - PHYSICS 207 Lecture Notes

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