Circular Motion

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Transcript Circular Motion

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
ROTATES once a day
The earth _________
REVOLVES once a year
The earth _________
REVOLVES around the earth
So the moon _________
The PERIOD (T) of an object is the time it
takes the mass to make a complete revolution
or rotation.
The FREQUENCY (f) of an object is the
number of turns per second
UNITS:
T in seconds
f in Hz (s-1)
Period and frequency are reciprocals
1
f 
T
T=1
f
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:
2r
v
T
v = 2π r f
Units: m/s
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=2m
f = 3 rev/s
1
1 = 0.33 s
T 
f 3rev / s
2r 2 (2m)
v

T
0.33s
= 37.70 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
2
v
ac 
r
Or substituting for v
4

r
ac 
2
T
We get
2
Or
ac = 4
2
2
π rf
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.
2
v
ac 
r
4 r
 2
T
2
4 (0.6m)

2
(2s)
2
= 5.92 m/s2
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:
Or
Fc = mac
Fc = m 4 r
mv
Fc 
r
2
T
2
Fc = m 4
2
2
2
π rf
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?
m = 1000 kg
r = 30 m
v = 9 m/s
mv 2 (1000kg)(9m / s) 2
Fc 

= 2700 N
30m
r
b. Where does this force come from?
Force of friction between tires and
road.
Please complete
PSE Practice Problems # 1-12
Due EOC next class