Transcript Centripetal Acceleration and Centripetal Force

Centripetal Acceleration and
Centripetal Force
Centripetal Acceleration
and Centripetal Force
Circular Motion
• When an object moves
in a circle at constant
speed, we describe it
as undergoing
uniform circular
motion.
• Its speed is constant,
but its velocity is not
because velocity
includes direction and
the object’s direction
is clearly changing.
Circular Motion
• A changing velocity
means acceleration.
• The pull on the string is
always directed
perpendicular to the
velocity.
• The pull accelerates the
ball into a circular path,
even though the ball
does not speed up or
slow down.
• The pull changes only
the direction of the
velocity, not the
magnitude.
Centripetal Acceleration
• The acceleration
arising from the
change in direction of
the velocity vector is
called the centripetal
acceleration and is
determined
mathematically by:
2
v
ac 
r
Centripetal Acceleration
Centripetal Acceleration
• Centripetal means center-seeking.
• Centripetal acceleration is always directed
toward the center of the circle of motion.
• It is this centripetal acceleration that is
responsible for the change in the direction of the
velocity; the magnitude of the velocity remains
constant.
• Any change in the tangential acceleration
causes a change in the speed of the particle as
it travels around the circle. In uniform circular
motion, aT = 0, so the acceleration is completely
radial (ar) or centripetal.
Centripetal Force
• Newton’s Second Law explains that an
object undergoing acceleration is
experiencing a net force. The net force
on an object undergoing uniform circular
motion is called the centripetal force Fc.
• The centripetal force necessary for an
object of mass m to travel with constant
speed v in a circle of radius r is given by:
m   r  
mv
2
Fc 
but v  r   so Fc 
 m  r 
r
r
2
2
Centripetal Force
• The centripetal force always points
toward the center of the circle about
which the object moves with uniform
speed.
• If the centripetal force applied to the
object is removed, the object will move in
a straight-line tangent to the curved path
at the point where the centripetal force
ceases. When the centripetal force
ceases, the object has no unbalanced
forces acting upon it and thus moves in a
straight line at constant speed.
Circular Motion
• If the string breaks, the ball flies off in a
straight line. It is the force of the string
that causes the acceleration in this
example of uniform circular motion.
Centripetal Force
• Centripetal force is the name given to any force
that is directed at right angles to the path of a
moving object and that tends to produce circular
motion.
• Examples:
– the gravitational force directed toward the center of
the Earth holds the Moon in an almost circular
orbit about the Earth;
– an electromagnetic force that is directed toward the
nucleus holds the electrons that revolve about the
nucleus of the atom.
• Directions in centripetal force problems:
– Positive direction is inwards toward center of circle.
– Negative direction is outward away from center of
circle.
• Radius r is the distance from the center of
the mass to the axis of rotation.
m  v2
• The centripetal force
is not a force and
r
does not belong in a free-body diagram.
• The force F in the picture would provide the
centripetal force needed to maintain the
circular path.
Motion On A Flat Curve
• The net force on a
car traveling around
a curve is the
centripetal force.
• As a car travels
around a curve, the
net force on the car
must be the
centripetal force,
directed toward the
center of the circle
the curve is a portion
of.
Motion On A Flat Curve
• On a flat, level curve, the
friction between the tires
and the road supplies the
centripetal force.
• If the tires are worn
smooth or the road is icy
or oily, this friction force
will not be available.
• The car will not be able to
move in a circle, it will
keep going in a straight
line and therefore go off
the road.
Motion On A Flat Curve
• Accelerations:
– ay = 0 m/s2
– ax = ac = v 2
r
• Equation:
Fc = FF ;
m  v2
 μmg
r
v2
 μg
r
Motion On A Banked Curve
• Some curves are banked to compensate for
slippery conditions.
• In addition to any friction forces that may or
may not be present, the road exerts a normal
force perpendicular to its surface.
• The downward force of the car’s weight is also
present.
• These two forces add as vectors to provide a net
force Fnet that points toward the center of the
circle; this is the centripetal force.
• The centripetal force is directed toward the
center of the circle, it is not parallel to the
banked road.
Motion On A Banked Curve
• The effect of banking is to tilt
the normal force Fn toward the
center of curvature of the road
so that the inward radial
component FNsin  can supply
the required centripetal force.
• Vehicles can make a sharp turn
more safely if the road is
banked. If the vehicle
maintains the speed for which
the curve is designed, no
frictional force is needed to
keep the vehicle on the road.
Motion On A Banked Curve
• The effect of banking
is to tilt the normal
force FN toward the
center of curvature of
the road so that the
inward radial
component FNsin 
can supply the
required centripetal
force.
Motion On A Banked Curve
• There is no acceleration along the y axis,
so the sum of the forces in the y plane is
zero:
F  cos θ  F  0
N
w
Fw
FN 
cos θ
• The horizontal component of the normal
force FN, the force the road exerts against
the car, provides the necessary
centripetal force. Because the only force
in the x plane is the centripetal force:
Motion On A Banked Curve
• This equation
gives the banking
angle that allows a
car to travel in a
curve of radius r
with constant
speed v and
require no friction
force.
• Goldilocks: Just
Right!
FC  FN  sin θ
m  v2
 FN  sin θ
r
Fw
m  v2

 sin θ
r
cos θ
m  v2
 m  g  tan θ
r
v2
tan θ 
rg
2 

v

θ  tan-1 
rg


Motion On A Banked Curve
• A banked curve is designed for one specific
speed, called the “design speed”.
– If the banked curve is icy so that there is no
friction force at all, then traveling at a speed
higher than the design speed means the car will
slide out, up, and over the edge.
– Traveling at a lower speed than the design speed
means the car will slide in, down, and off the
bank.
• When a banked curve has to be negotiated
at a speed above the design speed (or if you
are asked to find the maximum speed),
friction is required. The frictional force Ff
acts parallel to the road surface.
Motion On A Banked Curve (w/ Friction)
Goldilocks: Too Fast!
Motion On
A Banked
Curve (w/
Friction)
Goldilocks:
Too Fast!
Motion On A Banked Curve (w/ Friction)
• Resolve the normal force FN and friction FF into
horizontal and vertical components.
(up) Fy  FN  cos θ
(down ) Fy  FF  sin θ
(in) Fx  FN  sin θ
(in) Fx  FF  cos θ
• The horizontal components of FN and FF are both
directed inward toward the center of the curve,
therefore, these two force components combine to
determine the centripetal force; ΣFx = m·ac.
m  v2
FN  sinθ  FF  cos θ 
r
Motion On A Banked Curve (w/ Friction)
• Because there are no unbalanced forces
in the vertical direction, the upward
forces must equal the downward forces,
therefore: F  cos θ  F  sinθ  m  g
N
F
• Substitute FF  μ  FN into both
equations.
m  v2
FN  sin θ  μ  FN  cos θ 
r
FN  cos θ  μ  FN  sin θ  m  g
Motion On A Banked Curve (w/ Friction)
• Solve both equations for Fn and set these
equations equal to each other (because FN = FN
as there is only one force acting normal to the
surface). From this equation, the unknown
variable can be determined.
m  v2
FN  (sin θ  μ  cos θ) 
r
FN  cos θ - μ  sin θ  m  g
mv
mg

r  (sin θ  μ  cos θ) cos θ  μ  sin θ
2
Motion On A Banked Curve (w/ Friction)
• For problems involving a minimum speed for
the vehicle to travel around the curve without
skidding, the frictional force is directed up the
incline to keep the vehicle from sliding down to
the bottom of the banked curve.
• Resolve the normal force FN and friction FF into
horizontal and vertical components.
(up) Fy  FN  cos θ
(in) Fx  FN  sin θ
(up) Fy  FF  sin θ
(out) Fx  FF  cos θ
Motion On A Banked Curve (w/ Friction)
Goldilocks: Too Slow!
Motion On
A Banked
Curve (w/
Friction)
Goldilocks:
Too Slow!
Motion On A Banked Curve (w/ Friction)
• The horizontal component of FN is
inward toward the center of the curve
and is positive; FF is directed outward
away from the center of the curve and
is negative. These two force
components combine to determine the
centripetal force; ΣFx = m·ac.
mv
FN  sin θ  FF  cos θ 
r
2
• Because there are no unbalanced forces
in the vertical direction, the upward
forces must equal the downward forces,
therefore:
FN  cos θ  FF  sin θ  m  g
• Substitute FF  μ  FN into both
equations.
mv
FN  sin θ  μ  FN  cos θ 
r
2
FN  cos θ  μ  FN  sin θ  m  g
• Solve both equations for Fn and set these
equations equal to each other (because
FN = FN as there is only one force acting
normal to the surface). From this
equation, the unknown variable can be
determined.
m  v2
FN  (sin θ  μ  cos θ) 
r
FN  cos θ  μ  sin θ  m  g
mv
mg

(r  sin θ)  r  μ  cos θ cos θ  μ  sin θ
2
Vertical Circles
• The force of gravity causes the speed of an
object in a vertical circular path to vary. The
object accelerates on the downward portion of
its circular path and decelerates on the upward
portion of the circular path.
• At the top and bottom of a vertical circular path,
the weight and the normal force (or an
equivalent supporting force, such as tension)
are the only forces acting on an object. The
centripetal force is supplied by the resultant of
the weight and a supporting force (often the
normal force).
Vertical Circles
• The forces acting on a
person sitting in a
roller coaster car are
shown. The person’s
weight FW is present
and so is the normal
force FN that the seat
exerts on him (this is
your apparent
weight).
FC  FW  FN
m  v2
FN 
 mg
r
Vertical Circles
• The normal force FN, the force you feel on
the seat of your pants, can be positive,
negative, or zero.
• A negative value for FN means the
passenger has to be strapped in, with the
straps exerting an upward force. Such a
situation would be dangerous, and roller
coaster designers avoid this.
• If FN = 0 N, the person seems to be
weightless as well as upside down.
Vertical Circles
• The forces on an
airplane pilot at the
bottom of a dive can be
quite large.
• Gravity pulls downward
and the seat exerts its
usual normal force FN,
this time upward.
FC  FN  FW
m v
FN 
m g
r
2
Vertical Circles
• At the bottom of the dive, the normal force can
only be positive, must be greater than the
weight, and can become very large. A roller
coaster at the bottom of the loop provides the
same forces.
• The acceleration can be expressed as:
2
v
a
g
r
• This acceleration can be expressed in terms of
g’s, where g’s are determined by dividing the
centripetal acceleration by gravity. One g is
9.8 m/s2. The number of g’s represents the
relative pull of gravity on the body that the
person experiences.
Vertical Circles
• Experiencing a significant number of g’s
makes the work of the heart more
difficult. Accelerations of eight to ten g’s
make it difficult for the circulatory
system to get enough blood to the brain
and may result in blackouts. Pressure
suits that squeeze on the legs push blood
back into the rest of the body, including
the brain, and help prevent blackouts.
Vertical Circles
• For the Ferris wheel,
the only difference
occurs at the top
where the seat is
facing upward.
• Top: FC  FW  FN
m  v2
FN  m  g r
– This equation is also
for a car passing over
the top of a curve.
• Bottom: F  F  F
C
N
W
m  v2
FN 
 mg
r
FC  FW  FN
FN  m  g 
m  v2
r
Tension
• For an object
attached to a
string and moving
in a vertical circle,
the centripetal
force is at a
minimum at the
top of its vertical
path and at a
maximum at the
bottom of its
vertical path.
Tension
• Top: the centripetal force on the
object equals the tension of the
string plus the weight of the ball,
both acting toward the center of the
vertical circle. Mathematically:
FC  FT  FW
m  v2
FT 
mg
r
• Bottom: the centripetal force on the
object is equal to the difference
between the tension of the string
FC
and the weight of the object. The
tension is exerted inward toward
the center of the vertical circle,
FT
while the weight is directed away
from the center of the vertical circle.
Mathematically:
 FT  FW
m  v2

 mg
r
Critical Velocity (vmin)
• Critical velocity: velocity below which an
object moving in a vertical circle will not
describe a circular path.
• Critical velocity depends on the
acceleration due to gravity and the
radius of the vertical circle, not on the
mass of the object.
v min  r  g
Tension at an Angle & Horizontal Circles
• Vertical component of the tension is equal to
the weight.
• Horizontal component of the weight is equal to
the centripetal force.
• the radius is the distance from the center of the
mass to the dot at the center of the horizontal
circle.
Conical Pendulum
L  cos θ
T = 2
g
is
L
r
T
•For conical
pendulums,
centripetal
force is
provided by a
component of
the tension.
• is the angle
between the
vertical and
the cord.
Fw = m·g
Conical Pendulum
Fy  0; T  cos   m  g  0
mg
T  cos   m  g; T 
cos 
mv
Fx 
; T  sin   Fc
r
m  v2
mg
m  v2
T  sin  
;
 sin  
r
cos 
r
2
2
v
g  tan  
r
Fn Difference?
• In every previous inclined plane problem such
as a skier on a hill or a block on a ramp, the
normal force is given by Fn = m·g·cos .
• In every banked curve problem the normal
force is given by Fn=m·g/cos .
• Both equations are correct!
• The difference is due to the direction of the
acceleration.
– the direction of the acceleration in the inclined plane
problem is down toward the bottom of the incline.
There is no component of the acceleration in the
normal direction.
– the direction of the acceleration in the banked curve
is horizontally toward the center of the circle. This
produces a component of acceleration in the normal
direction.
• In the inclined plane problems, I taught you to
rotate the x- and y- axes so that the x-axis is
parallel to the surface and the y-axis is
perpendicular to the surface. We did not do
this with the banked turn problems.
– This results in the acceleration being zero on the y
axis for both cases.
– In the inclined plane problems, the acceleration is
zero along the normal perpendicular to the incline;
thus the normal force equals a component of
gravity.)
– In the frictionless banked turn,
the acceleration is zero
vertically; thus the force of
gravity equals a component of
the normal force.
– In the inclined plane problems,
one component of gravity
causes the acceleration (the
other component cancels out
the normal force).
– In the banked turn, one
component of the normal force
causes the acceleration (the
other component cancels out
the force of gravity).
For Uniform Circular Motion:
d n 2  r
v 
t
t
n = number of revolutions (rotations)
r = radius
t = time
Period = time for one revolution (rotation)
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