Transcript Document
Chapter 6: Circular Motion and Gravitation v A curved path requires an “inward” force “Center seeking” = Centripetal Dv = aDt = (F/m)dt Centripetal force is the force Dv perpendicular to the velocity of an Dv object moving along a curved path. The centripetal force is directed toward the center of curvature of the path. examples: ball on a string, car rounding a corner. Centrifugal Effect: the “fictitious force” felt by an object when the frame of reference moves along (and therefore accelerates) along a curved path. This effect is simply inertia. Stop the force and the object will undergo straight line motion. p150c6:1 Uniform Circular Motion motion in a circle at constant speed centripetal force Fc and centripetal acceleration ac is always directed towards the center centripetal force and acceleration have constant magnitudes v2 mv 2 ac Fc ma c r r the period T of the motion is the time to make one orbit 2r v T p150c6:2 Example 6.2: Find the centripetal force needed by a 1200-kg car to make a turn of radius 40m at a speed of 25 km/h (7.0 m/s). Assuming the road is level, find the minimum coefficient of static friction between the tires and the road that will permit the turn to be made without sliding. Fn ac phys Ff W p150c6:3 1 2 m v1 2 PE1 m gh m g2r KE1 minimum speed Motion in a Vertical Circle r h = 2r speed decreasing speed increasing At the bottom: 1 2 maximum speed KE 2 2 m v2 At the top: T - mg = ma PE2 0 T + mg = ma Critical speed: speed at top at which string goes slack (T=0) 2 v0 T1 m - mg 0 v 0 rg r p150c6:4 Example 6.5 (almost): An airplane pulls out of a dive in circular arc whose radius is 1000m. The speed of the airplane is a constant 200m/s. Find the force with which the 80 kg pilot presses down on his seat at the bottom of the arc. Fn ac W=mg p150c6:5 Example 6.6: A sled starts from rest and slides down a frictionless track to a vertical loop. If the radius r of the loop is 10 m, what is the minimum height must the sled start from in order to loop the loop without falling off? h-2r h p150c6:6 Gravitation A fundamental force of nature (electromagnetism, weak nuclear force, strong nuclear force) Newton’s law of universal gravitation All objects interact by virtue of having mass Force is proportional to each mass Force is inversely proportional to the square of the distance mA mB Fgrav G 2 r 2 Nm G 6.67 10 -11 kg 2 The force between two 1 kg masses separated by 1m is 6.67x10-11 N ~ 1.5x10-11 lb p150c6:7 The acceleration of gravity of a mass near the surface of the earth is due to the interaction of the mass with the earth’s mass. r re 6.37 106 m Fgrav G g G mB mearth 5.96 1024 kg m A mearth mearth m A mB G m G mg A 2 2 2 r re re mearth 2 re at other distances (r) 2 mearth re mearth gr G 2 2 G 2 r r re 2 re gr 2 g r p150c6:8 Orbital Motion: object “falls around” orbited body circular orbit v2 m mg r r mmearth v2 m G 2 r r 2 re g 0 v rg r r Gmearth v r escape spe ed v escape 2v circular orbit re g 11,000m / s p150c6:9 Example 6.9: find the altitude of a geostationary orbit. p150c6:10 Other applications of Universal Gravitation: mechanics of planetary orbits “weighing” planets by watching satellites (moons) “weighing” stars (especially our sun) by watching planets. Galaxies, Clusters, Superclusters .... p150c6:11 p150c6:12