Circular Motion Chapter 9 in the Textbook Chapter 6 is PSE pg. 81.

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Presentation transcript:

Circular Motion Chapter 9 in the Textbook Chapter 6 is PSE pg. 81

Revolve: is to move around an EXTERNAL axis Rotate: is to move around an INTERNAL axis So the moon _________ around the earth The earth _________ once a day The earth _________ once a year ROTATES REVOLVES

The PERIOD (T) of an object is the time it takes the mass to make a complete revolution or rotation. UNITS: T in seconds f in Hz (s -1 ) The FREQUENCY (f) of an object is the number of turns per second T = 1 f Period and frequency are reciprocals

A spring makes 12 vibrations in 40 s. Find the period and frequency of the vibration. f = vibrations/time = 12vib/40s = 0.30 Hz T = 1/f = 1/(0.3Hz) = 3.33 s

The period T is the time for one complete revolution. So the linear speed or tangential speed can be found by dividing the period into the circumference: Units: m/s v = 2π r f

A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its period and linear speed m = 2 kg r = 2 m f = 3 rev/s = 0.33 s = m/s

UNIFORM CIRCULAR MOTION Uniform circular motion is motion in which there is no change in speed, only a change in direction.

CENTRIPETAL ACCELERATION An object experiencing uniform circular motion is continually accelerating. The direction and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. The vectors are the same size because the velocity is constant but the changing direction means acceleration is occurring.

To calculate the centripetal acceleration, we will use the linear velocity and the radius of the circle Or substituting for v We get acac Or a c = 4 π 2 r f 2

A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration. = 5.92 m/s 2

CENTRIPETAL FORCE The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's Second Law, the centripetal force is given by: F c = ma c Or F c = m F c = m 4 π 2 r f 2

A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/s a. How much centripetal force is required? b. Where does this force come from? Force of friction between tires and road. m = 1000 kg r = 30 m v = 9 m/s = 2700 N

Please complete PSE Practice Problems # 1-12 Due EOC next class