Centripetal Force. Acceleration in a Circle  Acceleration is a vector change in velocity compared to time.  For small angle changes the acceleration.

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Presentation transcript:

Centripetal Force

Acceleration in a Circle  Acceleration is a vector change in velocity compared to time.  For small angle changes the acceleration vector points directly inward.  This is called centripetal acceleration. dd

Centripetal Acceleration  Uniform circular motion takes place with a constant speed but changing velocity direction.  The acceleration always is directed toward the center of the circle and has a constant magnitude.

Buzz Saw  A circular saw is designed with teeth that will move at 40. m/s.  The bonds that hold the cutting tips can withstand a maximum acceleration of 2.0 x 10 4 m/s 2.  Find the maximum diameter of the blade.  Start with a = v 2 / r. r = v 2 /a.  Substitute values: r = (40. m/s) 2 /(2.0 x 10 4 m/s 2 ) r = m.  Find the diameter: d = 0.16 m = 16 cm.

Law of Acceleration in Circles  Motion in a circle has a centripetal acceleration.  There must be a centripetal force. Vector points to the center  The centrifugal force that we describe is just inertia. It points in the opposite direction – to the outside It isn’t a real force

Conical Pendulum  A 200. g mass hung is from a 50. cm string as a conical pendulum. The period of the pendulum in a perfect circle is 1.4 s. What is the angle of the pendulum? What is the tension on the string? FTFT 

Radial Net Force  The mass has a downward gravitational force, -mg.  There is tension in the string. The vertical component must cancel gravityThe vertical component must cancel gravity  F Ty = mg  F T = mg / cos   F Tr = mg sin  / cos  = mg tan   This is the net radial force – the centripetal force. mg FTFT F T cos  F T sin  

Acceleration to Velocity  The acceleration and velocity on a circular path are related. mg FTFT mg tan   r

Period of Revolution  The pendulum period is related to the speed and radius. FTFT mg tan   r L cos  =  = 13 °

Radial Tension  The tension on the string can be found using the angle and mass.  F T = mg / cos  = 2.0 N  If the tension is too high the string will break! next