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CIRCULAR DYNAMICS
How Force Components Change Velocity
Learning Objectives
After you complete the homework associated with this
lecture, you should be able to:
• Explain why forces parallel to velocity change the speed
of an object and forces perpendicular change its
direction of motion.
• State the magnitude and direction of the acceleration of
an object undergoing circular motion.
• Use free-body diagrams and force laws to analyze
circular dynamics.
C
Force vector P
F causes acceleration P
a of mass m.
C
This produces a change in its velocity vector P
v.
C
Can see effects of P
F on P
v by decomposing force P
F
into components parallel ( || ) and perpendicular ( z )
to velocity
vector.
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Parallel Force Component Changes Speed
Perpendicular Force Component Changes
ΔPv ' (ΔPv|| % ΔPvz ) ' (Pa|| % Paz ) Δt '
ΔPv|| '
P
F
||
m
P
F
||
m
%
P
F
z
m
Direction of Motion
Δt
P changes the "length )) of the
Δt Y F
||
ΔPvz'
P
F
z
m
P changes the direction of the velocity
Δt Y F
z
vector, i.e. where it )s headed
velocity vector, i.e. it )s speed
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Summary of Effects of Force Components
C
C
The force component parallel to velocity P
F ||
Some Consequences
C
If you apply a force parallel to the velocity vector
changes the speed of the object.
you can only change an object’s speed, not its
The force component perpendicular to velocity P
Fz
direction.
changes the direction in which it is headed.
C
If you apply a force perpendicular to an object’s
velocity vector, you will change its direction of
motion BUT NOT ITS SPEED !
C
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Kinematics of Circular Motion
ACCELERATION COMPONENTS IN CIRCULAR MOTION
If an object goes in a circular arc, the following is true
irrespective of anything else:
C
Pv is always tangent to the path of a particle
Pv will be tangent to the circular arc
C
The component of P
a perpendicular to the velocity
(P
a z) equals
az '
P
A fine example is circular motion
v2
, toward the center
R
of the circle
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C
From the properties of a circle, we can make the
following inferences:
1) The velocity vector is tangent to the circular path
at the position of the mass.
2) A perpendicular to tangent on a circle is directed
radially toward circle’s center
C Therefore:
ar = az
atang = a||
Note: If speed constant, a|| = 0
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LASER DISC : Acceleration is toward center at uniform
speed using floating cork accelerometer.
Force needed is toward center (with NO
such thing as outward centrifugal force).
Demo
Demo:
inverted water-filled flask with tethered cork
IMPORTANT : The magnitude of radial acceleration for
an object moving at speed v in circle of radius R is
given by:
v2
ar '
R
regardless of whether it’s changing its speed or not.
C This must be generated by net radial force 3Fr
directed TOWARD the center:
j Fr ' m ar ' m
v2
R
web resource http://www.walter-fendt.de/ph11e/carousel.htm
change period to 3.0 seconds, click sketch than with forces
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Car taking a flat curve with friction
IMPORTANT :
The acceleration
component tangent to
the circle changes the
speed of the object.
adirection Pv '
Fdir Pv
m
'
dv
' rate of change of speed
dt
To change the speed requires an Fdirection of velocity !
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R2 is where the mass would
“want to go” if it were not for
friction. In a radial
coordinate system, R appears
to want to increase even
though the mass really wants
to go in a straight line. This
is what we identify as the
centrifugal pseudo-force.
Since velocity vector changes direction, friction forceP
f s acts
perpendicular to it, i.e., toward the center of circular motion.
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Hard example:
What is the maximum safe speed
around a banked curve of radius R
and incline θ, with road coefficient
of friction µ ?
j Fx ' Nx % fx % Fg x ' sum of forces in radial direc
v2
' m ax ' m ac ' m %
R
How many forces act on the car?
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3Fx = Nx + fx + Fgx = max
+N sinθ + (+f cosθ) + 0 = m (+v2/R)
1 eq., 3 unknowns (N, f, v) Y
(Eq. 1)
Need 2 more eqs.
3Fy = Ny + fy + Fgy = may
+N cosθ + ( S f sinθ) + ( S mg) = m(0) = 0
v = vmax
Y
f = fmax = µN
(Eq. 2)
(Eq. 3)
Y
(3) into (2):
N cosθ S µN sinθ= mg
(3) into (1):
N sinθ + µN cosθ + 0 = m (+v2/R)
(4) into (5):
v2 '
R
N (sinθ%µcosθ) '
m
N '
mg
cosθ&µsinθ
(Eq. 4 )
(Eq. 5)
Rmg sinθ%µcosθ
sinθ%µcosθ
' Rg
m
cosθ&µsinθ
cosθ&µsinθ
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