VIEWS: 3 PAGES: 17 POSTED ON: 12/5/2011
Chapter (6) Circular Motion 1 Circular Motion Consider an object moving at constant speed in a circle. The direction of motion is changing, so the velocity is changing (even though speed is constant). Therefore, the object is accelerating. The direction of the acceleration is toward the center of the circle and so we call it centripetal acceleration. The magnitude of the acceleration is v2 ac r 2 Centripetal Acceleration • The best estimate of the acceleration at P is found by calculating the average acceleration for the symmetric interval 12. v1 (v cos ) x (v sin ) y ˆ ˆ v2 (v cos ) x (v sin ) y ˆ ˆ Distance travelled 2 r if measuredin radians 3 Elapsed time t d/v 2θr/v Components of Acceleration ax v cos v cos 0 2θr/v ay v sin v sin v 2 sin 2θr/v r v2 ay r if 0 Then, v2 ac r 4 Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? Solution: r (earth) 1.496 1011 m 2 ac t 1year r 2r But t Then 4 r2 ac 2 5.93 10 3 m / s t 5 Tangential and Radial acceleration Tangential acceleration The tangential acceleration component causes the change in the speed of the particle. This component is parallel to the instantaneous velocity, and is given by Note: If the speed is d constant then the tangential at acceleration is zero dt (uniform Circular Motion) Radial acceleration The radial acceleration component arises from the change in direction of the velocity vector and is given by 2 a r ac 6 r Total acceleration The total acceleration vector a can be written as the vector sum of the component vectors: a at a r a a 2t a 2 r Since the component perpendicular to other 7 Example: A car exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this instant? 8 2 36 ar 0.072 m / s 2 r 500 a a t a 2 2 r 0.309m / s 2 If the angle between ar tan 1 13 .5 at 9 Problem: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to round the bend. The radius of the curve is 150m. Compute the acceleration at the train. Problem: A particle moves in a circular path 0.4m in radius with constant speed. If the particle makes five revolution in each second of its motion, find (a) the speed of the particle and (b) its acceleration. 10 Centripetal Force • A string cannot push sideways or lengthwise. • A string in tension only pulls. • The string pulls the ball inward toward the center of the circle 11 What if we cut the sting? The ball should move off with constant velocity This means the ball will continue along the tangent to the circle. 12 Centripetal Force If there is a centripetal acceleration, then the net force must also be a centripetal force: v2 Fc ma c m r 13 Example: • The Conical Pendulum • As the ball revolves faster, the angle increases • What’s the speed for a given angle? 14 2 mv T sin .......... 1) .......... .....( r T cos mg.......... ........( ) .......... 2 then v2 tan rg v rg tan but (r L sin ) Lg sin tan 15 Problem: I rotate a ball at an angle of 30o. What is the centripetal acceleration? If the string is 1 meter long, how fast is it rotating? 16 Problem Driving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the Figure. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump. a=v2/r N mg 17