Uniform Circular Motion Section 6 5 An object moving on a circular path of radius r at a constant speed v As motion is not on a straight line the direction of the velocity vector is not constant The motion is circular Compare to 1D straight line 2D parabola v Velocity vector is always tangent to the circle Velocity direction constantly changing but magnitude remains constant v r v Vectors r and v are always perpendicular Since the velocity direction always changes this means that the velocity is not constant though speed is constant therefore the object is accelerating The acceleration ar points radially inward Like velocity its direction changes therefore ar v the acceleration is not constant though its magnitude is Vectors ar and v are also perpendicular The speed does not change since ar acceleration has no component along the velocity direction Why is the acceleration direction radially inward ar Since v 1 v1 v 2 v1 v 2 ar a v2 t 2 t1 This radial acceleration is called the centripetal acceleration 2 v ar r eq 6 15 We now look at applying Newton s 2nd law to circular motion in a plane Recall the radial or centripetal acceleration for uniform circular motion 2 v ar r This acceleration implies a force mv Fr mar r 2 centripetal force is not a force The centripetal force is the net force required to keep an object moving on a circular path Consider a motorized model airplane on a wire which flies in a horizontal circle if we neglect gravity there are only three forces the force provided by the airplane motor which tends to cause the plane to travel in a straight line air resistance and the tension force in the wire which causes the plane to travel in a circle the tension is the centripetal force air T motor Consider forces in radial direction positive to center mv Fr mar T r 2 Example similar to example 6 8 A car travels around a curve which has a radius of 316 m The curve is flat not banked and the coefficient of static friction between the tires and the road is 0 780 At what speed can the car travel around the curve without skidding y N mg mg N v r fs fs r F y may N mg 0 N mg F r mar mv fs r 2 Now the car will not skid as long as fs is less than the maximum static frictional force 2 mv 2 mv f smax sN smg r r v sgr 0 780 9 80 sm2 316 m 49 1 ms 3600 s 1 mi 49 1 110 1 hr 1609 m m s mi hr Example similar to example 6 9 To reduce skidding use a banked curve Consider same conditions as previous example but for a curve banked at the angle y n r mg F y fs may N cos f s sin mg 0 N r fs mg Choose this coordinate system since ar is radial Since acceleration is radial only Since we want to know at what velocity the car will skid this corresponds to the centripetal force being equal to the maximum static frictional force fs f max s s N Substitute into previous equation N cos sN sin mg mg N cos s sin mg N cos s sin F r mar mv 2 N sin f s cos r 2 mv N sin s cos r Substitute for N and solve for v mg mv 2 sin s cos r cos s sin sin s cos v rg cos s sin Adopt r 316 m and 31 and s 0 780 from earlier m mi v 89 7 s 200 hr Compare to example 6 8 where s 0 sin v rg rg tan 43 1 ms 96 5 mi hr cos Demo 7 1 2
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