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Motion, Forces and Energy Lecture 5: Circles and Resistance Fr m r Fr A particle moving with uniform speed v in a circular path of radius r experiences an acceleration which has magnitude a = v2/r (see Serway Section 4.4 for derivation). Acceleration is ALWAYS directed towards the centre of the circle, perpendicular to the velocity vector. Fr 2 mv Fr mar r m When string breaks, ball moves along tangent. NB The centripetal force is a familiar force acting in the role of a force that causes a circular motion. Tension as a centripetal force Fr r Fr m If the ball has a mass of 0.5kg and the cord is 1.5m long, we can find the maximum speed of the ball before the cord breaks (for a given tension). We assume the ball remains perfectly horizontal during its motion:. The TENSION in the cord PROVIDES the CENTRIPETAL FORCE: mv2 T r so Tr v m For a tension of say 100 N (breaking strain), the Maximum speed attainable is (100x1.5/0.5)1/2 = 17.3 ms-1. Conical Pendulum A conical pendulum consists of a “bob” revolving in a horizontal plane. The bob doesn’t accelerate in the vertical direction: T cos mg may 0 T (1) The centripetal force is provided by the tension component: mv2 T sin mar r L (1) / (2) gives v2 tan rg (2) so v rg tan Finally we note that r = L sin so that: mg v gL sin tan A car moving round a circular curve Fr The centripetal force here is provided by the frictional force between the car tyres and the road surface: mv Fr f s r 2 The maximum possible speed corresponds to the maximum frictional force, fsmax = ms n. Since here n = mg, fsmax = ms mg vmax m s gr So for ms = 0.5 and r = 35 m, vmax = 13.1 ms-1. Non-uniform circular motion Sometimes an object may move in a circular path with varying speed. This corresponds to the presence of two forces: a centripetal force and a tangential force (the latter of which is responsible for the change in speed of the object as a function of time. An example of such motion occurs if we whirl a ball tied to the end of a string around in a VERTICAL circle. Here, the tangential force arises from gravity acting on the ball. vtop m mg Ttop R centre T Tbot vbot mgsin mgcos mg mg We must consider both radial and tangential forces! mg sin mat at g sin mv2 T mg cos R v2 T m g cos R Analysis Tangential acceleration General equation At the bottom of the path where = 0o: vbottom2 Tbottom m g R At the bottom of the path where = 180o: vtop2 Ttop m g R Therefore, the maximum tension occurs at the bottom of the circle, where the cord is most likely to break. Motion with resistive forces We’ll take a quick look at fluid resistance (such as air resistance or Resistance due to a liquid). Resistive forces can depend on speed In a complex way, but here we will look at the simplest case: R=bv ie the resistance is proportional to the speed of the moving object. R v mg With fluid (air) resistance, the falling object does not continue to accelerate, but reaches terminal Velocity (see Terminal Velocity Analysis notes).