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Centripetal Motion

Centripetal Motion. Motion towards the center of a circle. Terms. Uniform circular motion is the motion of an object in a circle with constant or uniform speed Circumference is the distance of one complete cycle around the perimeter of a circle

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Centripetal Motion

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  1. Centripetal Motion Motion towards the center of a circle

  2. Terms • Uniform circular motion is the motion of an object in a circle with constant or uniform speed • Circumference is the distance of one complete cycle around the perimeter of a circle • Period is the time needed to make one cycle around a circle • Centripetal means center seeking • Centrifugal means away from the center

  3. Calculating average speed in a circle • Average speed = circumference/time • Circumference = 2 * pi * radius so average speed = 2 * π * R twhere R = radiusπ = 3.14 t = time in s

  4. Tangential • The direction of the velocity vector of a circle is constantly changing since the direction is constantly changing • The direction of the velocity vector at one instant is the direction of a tangent line • A tangent line is a line that touches a circle at one point but does not intersect it tangent circle tangent

  5. Why an object moving in a circle is accelerating • An object moving in a circle is accelerating since the velocity is changing due to direction • Reminder that velocity is speed and direction and acceleration is a change in velocity • The speed of the object may be constant but the velocity is changing due to a change in direction

  6. Centripetal Force • Any object moving in a circle is experiencing centripetal force • There is a physical force pushing or pulling the object toward the center of the circle • Centripetal and centrifugal have very different meanings

  7. Calculating Acceleration in Circular Motion • Acceleration = 4 * π2 * R or v2 T R • Where v= speed R = radius T = period in s

  8. Net Force • Reminder F = maFnet = m * v2 or Fnet = m* 4*π2 * R R T • Where m = mass v = speed R = radius T = period in s

  9. Sample Problem 1 • A 900kg car moving at 10m/s takes a turn with a radius of 25m. Determine the acceleration and net force acting on the car.m = 900kgv = 10m/sR = 25m • a = v2/R so a = (10m/s)2/25m = 4m/s2 • Fnet = m * a = 900kg * 4m/s2 = 3600N

  10. Sample Problem 2 • A 95kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12m. The halfback makes a quarter of a turn around the circle in 2.1s. Determine the speed, acceleration and net force acting on the halfback. • m = 95kgR = 12mtraveled .25 of the circumference of 2.1s

  11. Solution • v = d/t = 2 * π * R/Tv = .25 * 2 * π * 12m = 8.97m/s 2.1s • a = v2 R • a = (8.97m/s)2 = 6.71m/s2 12m • Fnet = m * a • Fnet = (95kg) * (6.71m/s2) • Fnet = 637N

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