Centripetal Motion Motion towards the center of a circle.

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Presentation transcript:

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 Period is the time needed to make one cycle around a circle Centripetal means center seeking Centrifugal means away from the center

Calculating average speed in a circle Average speed = circumference/time Circumference = 2 * pi * radius so average speed = 2 * π * R t where R = radius π = 3.14 t = time in s

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

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

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

Calculating Acceleration in Circular Motion Acceleration = 4 * π 2 * R or v 2 T R Where v= speed R = radius T = period in s

Net Force Reminder F = ma F net = m * v 2 or F net = m* 4*π 2 * R R T Where m = mass v = speed R = radius T = period in s

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 = 900kg v = 10m/s R = 25m a = v 2 /R so a = (10m/s) 2 /25m = 4m/s 2 F net = m * a = 900kg * 4m/s 2 = 3600N

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 = 95kg R = 12m traveled.25 of the circumference of 2.1s

Solution v = d/t = 2 * π * R/T v =.25 * 2 * π * 12m = 8.97m/s 2.1s a = v 2 R a = (8.97m/s) 2 = 6.71m/s 2 12m F net = m * a F net = (95kg) * (6.71m/s 2 ) F net = 637N