Lecture 16: Rotational MotionPowerPoint PresentationSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Lecture 16: Rotational MotionQuestions of Yesterday1) You are going through a vertical loop on roller coaster at a constant speed. At what point is the force exerted by the tracks on you (and the cart you are in) the greatest? a) at the highest pointb) at the lowest pointc) halfway between the highest and lowest pointd) the force is equal over the whole loop2) You are on a merry-go-round moving at constant speed. If you move to the outer edge of the merry-go-round, what happens to the net centripetal force keeping you on the merry-go-round? a) it increases b) it decreasesc) it stays the samed) there is no net centripetal force acting on youRotational Motion: Angular Quantitiessr=Angular Position:=f - iAngular Displacement:av=f - itf - tit=Average Angular Velocity:limt -> 0=tInstantaneous Angular Velocity:av=f - itf - tit=Average Angular Acceleration:limt -> 0=tInstantaneous Angular Acceleration:Motion with Constant : = 0 + t = 0t + 1/2t22 = 02 + 2Angular and Linear Quantitiesrtisr=tfDisplacement:Direction of linear velocity v of an object moving in a circular path is always TANGENT to the pathsvT=rTangential Speed:aT=rTangential Acceleration:Centripetal AccelerationrvfCentripetal Acceleration always points towards the CENTER of the circle vf - vi tf - tiaav =vi-vivfvCentripetal ForceF = maIf an object is accelerating what do know about (think Newton’s 2nd law)?Can an object be moving in a circular path if no forces are acting on?If an object is undergoing constant speed circular motion what direction is the net force acting on the object?mv2rFc = mac =Centripetal AccelerationWhat if your tangential speed is NOT constant? rvfvi-vivfvrAcceleration has both tangential and centripetal components!v2rac =a = (ac2 + aT2)1/2vvcvTaT=rCentripetal ForceIn what direction is the net force if an object is undergoing circular motion and changing its tangential speed?-vivfvaacaTFFTFCFT = maTmv2rFc = mac = F = maJust like linear motion (∑Fx = max, ∑Fy = may)…must split vector equation in the perpendicular components!!Practice ProblemAn air puck of mass 0.5 kg is tied to a string and allowed to revolve in a circular radius of 1.0 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table and a mass of 1.0 kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves.What is the tension in the string?What is the horizontal force acting on the puck?What is the speed of the puck?Practice ProblemTarzan (m = 100 kg) tries to cross a river by swinging from a 10-m-long vine. His speed at the bottom of the swing (as he just clears the water) is 8.0 m/s. Tarzan doesn’t know that the vine has a breaking strength of 1500 N.Does he make it safely across the river? If not, what is the maximum speed he can have to make it?If Tarzan continues swinging on the vine what is the highest point he reaches?What is the tension in the vine at this highest point?What is the net force on Tarzan at this point?Planetary MotionWhy do the planets revolve around the sun, and the moon revolve around the Earth? Is there a net force acting on the planets and moons?How do you know?What is the direction of the force?Gravitational ForceForce of attraction between any two objects in the Universe. Gravitational force causes….Objects in free fall near the Earth’s surface to accelerate towards the Earth the moon to orbit the earth & the planets to orbit the sunAn astronaut to be able to jump higher on the Moon than on EarthGravitational ForceIf gravity is an attractive force why doesn’t the moon crash into the Earth? The moon is constantly falling towards Earth The planets are constantly falling towards the sunGravitational ForceNewton’s Law of Gravitational Force m1,m2= mass of objects attracting each otherr = distance between the objectsUniversal gravitational constant = G = 6.67*10-11 N*m2/kg2Fg = Gm1m2 r2 Gravitational Force between two objects is felt equally by both objects Fg of m1 exerted by m2 = Fg of m2 exerted by m1Gravitational ForceWhat if you have many objects near each other? The net gravitational force felt on an object is equal to the sum of the gravitational forces exerted by all the surrounding objects ∑FgE = FgSE + FgME SEMPractice ProblemObjects with masses of 200 kg and 500 kg are separated by 0.500 m.Find the net gravitational force exerted by these objects on a 50.0 kg object placed midway between them.At what position can the 50.0 kg object be placed so as to experience a net force of zero?Questions of the DayYou are riding on a Ferris wheel moving at constant speed. 1a) At what point is the net force acting on you the greatest?a) the top b) the bottomc) halfway between top and bottomd) the force is the same over the whole motion 1b) Is the net force doing work on you?a) YESb) NO2) If the mass of the moon were doubled, what would happen to its centripetal acceleration? a) it would increaseb) it would decreasec) it would stay the samePractice ProblemA 0.500-kg pendulum bob passes through the lowest part of its path at a speed of 5.00 m/s.What is the tension in the pendulum cable at this point if the pendulum is 100.0 cm long?When the pendulum reaches its highest point, what angle does the cable make with the vertical? What is the tension in the pendulum cable when the pendulum reaches its highest point?What is the net force acting on the pendulum at this point?What is the direction of the