Page 1Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 1Lecture 10Goals:Goals:Employ Newton’s Laws in 2D problems with circular motionAssignment: HW5, (Chapters 8 & 9, due 10/15, Wednesday)For this Wednesday: Read Chapter 9Physics 207: Lecture 10, Pg 2Uniform Circular MotionFor an object moving along a curved trajectory, with non-uniform speeda = ar(radial only)arv|ar|= vT2rPerspective is importantPage 2Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 3Non-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 10, Pg 4Circular motion Circular motion implies one thing|aradial| =vT2 / rPage 3Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 5Key steps Identify forces (i.e., a FBD) Identify axis of rotation Apply conditions (position, velocity, acceleration)Physics 207: Lecture 10, Pg 6ExampleThe pendulumConsider a person on a swing:When is the tension on the rope largest? And at that point is it :(A) greater than(B) the same as(C) less thanthe force due to gravity acting on the person?axis of rotationPage 4Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 7ExampleGravity, Normal Forces etc.vTmgTat bottom of swing vTis maxFr= m ac= m vT2 / r = T - mgT = mg + m vT2 / rT > mgmgTat top of swing vT= 0Fr= m 02 / r = 0 = T – mg cos θT = mg cos θT < mgθaxis of rotationyxPhysics 207: Lecture 10, Pg 8Conical Pendulum (very different) Swinging a ball on a string of length L around your head(r = L sin θ)axis of rotationΣ Fr= mar= T sin θΣ Fz= 0 = T cos θ – mgsoT = mg / cos θ (> mg)mar= mg sin θ / cos θar= g tan θ = vT2/r vT= (gr tan θ)½Period:t = 2π r / vT=2π (r cot θ /g)½Page 5Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 9A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed vtat the top so that it can manage the loop successfully ?Loop-the-loop 1 Physics 207: Lecture 10, Pg 10To navigate the top of the circle its tangential velocity vTmust 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/rvT= (gr)1/2mgvTPage 6Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 11The 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 2mgvNFr= mar= mvB2/r = N - mgN = mvB2/r + mgPhysics 207: Lecture 10, Pg 12Once again the car is going to execute a loop-the-loop. What must be its minimum speed at the bottom so that it can make the loop successfully?This is a difficult problem to solve using just forces. We will skip it now and revisit it using energy considerations later on…Loop-the-loop 3Page 7Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 13Example ProblemSwinging around a ball on a rope in a “nearly”horizontal circle over your head. Eventually the rope breaks. If the rope breaks at 64 N, the ball’s mass is 0.10 kg and the rope is 0.10 mHow fast is the ball going when the rope breaks?(neglect mg contribution, 1 N << 40 N)(mg = 1 N)T = 40 NFr= m vT2/ r ≈ TvT= (r Fr/ m)1/2 vT= (0.10 x 64 / 0.10)1/2 m/svT= 8 m/sPhysics 207: Lecture 10, Pg 14Example, Circular Motion Forces with Friction (recall mar= m |vT|2 / r Ff≤µµµµsN ) How fast can the race car go? (How fast can it round a corner with this radius of curvature?)mcar= 1600 kgµS= 0.5 for tire/roadr = 80 m g = 10 m/s2rPage 8Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 15 Only one force is in the horizontal direction: static frictionx-dir: Fr= mar= -m |vT|2 / r = Fs= -µsN (at maximum)y-dir: ma = 0 = N – mg N = mgvT= (µs m g r / m )1/2vT= (µs g r )1/2 = (0.5 x 10 x 80)1/2vT= 20 m/s ExampleNmgFsmcar= 1600 kgµS= 0.5 for tire/roadr = 80 m g = 10 m/s2yxPhysics 207: Lecture 10, Pg 16 A horizontal disk is initially at rest and very slowly undergoes constant angular acceleration. A 2 kg puck is located a point 0.5 m away from the axis. At what angular velocity does it slip (assuming aT<< arat that time) if µs=0.8 ? Only one force is in the horizontal direction: static frictionx-dir: Fr= mar= -m |vT|2 / r = Fs= -µsN (at ω)y-dir: ma = 0 = N – mg N = mgvT= (µs m g r / m )1/2vT= (µs g r )1/2 = (0.8 x 10 x 0.5)1/2vT= 2 m/s ω = vT/ r = 4 rad/sAnother Examplempuck= 2 kgµS= 0.8r = 0.5 m g = 10 m/s2yxNmgFsPage 9Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 17UCM: Acceleration, Force, VelocityFavPhysics 207: Lecture 10, Pg 18Zero Gravity RideA rider in a “0 gravity ride” finds herself stuck with her back to the wall.Which diagram correctly shows the forces acting on her?Page 10Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 19Banked CurvesIn the previous car scenario, we drew the following free body diagram for a race car going around a curve on a flat track.nmgFfWhat differs on a banked curve?Physics 207: Lecture 10, Pg 20Banked CurvesFree Body Diagram for a banked curve.Use rotated x-y coordinatesResolve into components parallel and perpendicular to bankNmgFfFor very small banking angles, one can approximate that Ffis parallel to mar. This is equivalent to the small angle approximation sin θ = tan θ, but very effective at pushing the car toward the center of the curve!!marxxy y θPage 11Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 21Banked Curves, Testing your understandingFree Body Diagram for a banked curve.Use rotated x-y coordinatesResolve into components parallel and perpendicular to bankNmgFfAt this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you accelerate in that direction.How does the radial acceleration change? maryyx x θPhysics 207: Lecture 10, Pg 22Navigating a hillKnight concept exercise: A car is rolling over the top of a hill at speed v. At this instant,A. n > w.B. n = w.C. n < w.D.We can’t tell about n without knowing v.This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion.Fc= mg = m v2/r v = (gr)1/2 ½(just like an object in orbit)Note this approach can also be used to estimate the maximum walking speed.At what speed does the car lose contact?Page 12Physics 207 – Lecture 10Physics 207: Lecture 10, Pg
View Full Document
Unlocking...