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Circular Motion. Position on a Circle  Motion in a circle is common.  The most important measure is the radius ( r ).  The position of a point on the.

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Presentation on theme: "Circular Motion. Position on a Circle  Motion in a circle is common.  The most important measure is the radius ( r ).  The position of a point on the."— Presentation transcript:

1 Circular Motion

2 Position on a Circle  Motion in a circle is common.  The most important measure is the radius ( r ).  The position of a point on the circle is described by a radial vector. Origin is at the center. Magnitude is equal everywhere. r r

3 Velocity on a Circle  Velocity is a vector change in position compared to time.  As the time gets shorter, the velocity gets closer to the tangent.

4 Direction of Motion  In the limit of very small angular changes the velocity vector points along a tangent of the circle.  This is perpendicular to the position.  For constant rotation rate, the magnitude stays the same, but the direction always changes.

5 Period and Frequency  Movement around a circle takes time.  The period (T) is the time it takes to complete one revolution around the circle.  The frequency (f) is the number of cycles around completed in a time. Cycles per second (cps or Hz)Cycles per second (cps or Hz) Revolutions per minute (rpm)Revolutions per minute (rpm)  Frequency is the inverse of period (f = 1/T).

6 Speed on a Circle  The circumference of a circle is 2  r.  The period is T.  The speed is distance over time. v = 2  r/T v = 2  rf r s = 2  r

7 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

8 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.

9 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 = 0.080 m.  Problem wants the diameter d = 0.16 m = 16 cm. next


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