Download 1.1 UCM AP

Uniform Circular Motion AP
• Uniform circular
motion is motion
in a circle at the
same speed
• Speed is
constant, velocity
direction
changes
• the speed of an object moving in a circle is given
by

circumference 2 r r= radius of
v 

T circle
time
T = period, time for 1
complete circle

• Is called the
2 r
v 
tangential velocity
T
Moon setting
over Rio de
Janerio
Pictures every
6.5 minutes
showing the
moon is moving
at a uniform
speed
Centripetal Acceleration
• The direction of
motion is changing so
the object is
accelerating



 v v f  vi
a

t
t

• If you draw the vectors tailtail,



 v  v f  vi
• the centripetal acceleration of the object is
 v2
ac 
r
ac = centripetal acceleration
(always to centre of circle)
2 ac Equations
 2 r 



2
v
T


a 

c
r
r
4 r
2
 T
r
2 2
2
4 r
 2
T
2
On data sheet too!
•the velocity vector is
always directed along the
tangent of the circle (at
right angles to the
acceleration vector)
•centripetal means centreseeking, the acceleration
vector always points to
the centre of the circle
•Acceleration is radially
inward
Centripetal Force
• inertia tends to keep objects moving in a
straight line, a force is needed to cause a
circular motion (2nd Law of Motion)
• the force causing the motion is called the
centripetal force and it always acts
towards the centre of the circle (parallel to
the acceleration vector)
Centrifuge Training


F net  m a


2
mv
Fc 
r

&
2
v
ac 
r
On data
sheet
NOT on data sheet
• an object will move in a circle whenever a
constant magnitude force acts on the
object at right angles to its direction of
motion
• The force and
acceleration are
radially inward
• The velocity is
tangential
• The string pulls ball
towards the centre of
the circle
• 3rd Law of Motion
states that a reaction
force will act outward
(ball pulls string)
• this is sometimes
called centrifugal force
Important!!!
• centripetal force is not a fundamental force
like gravity, it is the net force acting on an
object moving in a curved path
• The force will not change the speed of the
object because the force has NO
component parallel to the velocity
vector
Example
• A centrifuge for pilot training is 11 m in
radius. At what speed must it rotate in
order to inflict 7.0 g on a pilot?
Solution
2
v
ac 
r
v
ac r
v  7.0  9.81m / s 11m
2
Frequency
• sometimes speed is given in revolutions
per minute (rpm) or revolutions per second
(rps)
• this is how many times the object goes
around the circle in 1 minute or 1 second
• frequency is the number of cycles per
second measured in hertz (Hz)
1
1
Hz 
s
second
1
T
f

2 r
v 
 2 rf
T
Example
• Find the period of an object that rotates at 35.0
rps.
35.0circles 1circle

1s
T
T  0.0286s
Example
• A 3.50 kg object is swung in a 1.50 m radius
horizontal circle at 40.0 rps (40.0 Hz). What
magnitude force acts on the object?


F net  m a 

2
4 rm
Fc 
2
T
Solution

 4 2r 
2
m
4 rm
F

F net  m a
c  2 Fc 
2
T
T
1
f 
T
2
 4 (1.50 m) (3.50 kg)
Fc 
2
2
2
so F  4
rmf
(0.0250 s)
Fc  3.32 x 105 N
Effect of radius on speed and forces
• all points on a rotating
solid have the same
period, but different
speeds
• because the inner points
have smaller distances to
travel, their speeds are
less
• the speed and force
depends on the radius
Rotating
disc
• NASA and other space agencies use this to help
launch satellites
• at the equator, the speed is about 1667 km/h, at
Edmonton, it is about 900 km/h
Angular displacement & velocity
• The rotational
displacement of a
point, , in radians
• 2 radians in 1 circle
• Angular velocity, 
(lower case omega,)
radians/s
ave


t
Angular & linear speed
• Linear speed = r
• Linear speed is
sometimes called
tangential speed
Period & angular speed
2
1
T


f
Direction of Angular velocity
Equations of Rotational Motion
1 2
1
   0  0t   t x  x0  vx 0t  axt 2
2
2
  0   t
vx  vx 0  a x t
Example
• A diver completes 1.5
rotations in 2.3 s.
Determine the
average angular
speed of the diver
Solution
• 1.5 rotations x 2 rad
= 9.425 radians
ave


 4.1rad / s
t
Example
• A bullet is fired
through 2 discs
rotating at 99.0 rad/s.
The discs are 0.955
m apart. The angular
displacement
between the holes is
0.260 rad. Calculate
the speed of the
bullet.
Solution
•  = 99.0 rad/s.
• D = 0.955 m
•  = 0.260 rad.
ave


t
t = 2.626 x 10-3 s
v = 364 m/s
Small Angle Approximation
• For small angles,  < 0.5 rad,   sin  tan
O