Download Section 6.2 Circular Motion Acceleration

Section
6.2 Circular Motion
In this section you will:
● Explain why an object moving in a circle at
a constant speed is accelerated.
● Describe how centripetal acceleration
depends upon the object’s speed and the
radius of the circle.
● Identify the force that causes centripetal
acceleration.
Section
6.2 Circular Motion
Suppose that you were driving a car with the
steering wheel turned in such a manner that your
car followed the path of a perfect circle with a
constant radius. And suppose that as you drove,
your speedometer maintained a constant reading of
10 mi/hr. In such a situation as this, the motion of
your car could be described as experiencing
uniform circular motion.
Uniform circular motion is the motion of an object
in a circle with a constant or uniform speed.
Section
6.2 Circular Motion
The distance of one complete cycle around the
perimeter of a circle is known as the circumference.
With a uniform speed of 5 m/s, a car could make a
complete cycle around a circle that had a circumference
of 5 meters. At this uniform speed of 5 m/s, each cycle
around the 5-m circumference circle would require 1
second. At 5 m/s, a circle with a circumference of 20
meters could be made in 4 seconds; and at this uniform
speed, every cycle around the 20-m circumference of
the circle would take the same time period of 4 seconds
Section
6.2 Circular Motion
Circumference = 2*pi*Radius
Section
6.2 Circular Motion
The Direction of the Velocity Vector
Objects moving in uniform circular motion will have
a constant speed. But does this mean that they will
have a constant velocity?
The direction of the
velocity vector is directed
in the same direction that
the object moves. Since
an object is moving in a
circle, its direction is
continuously changing.
Section
6.2 Circular Motion
Acceleration
As mentioned, an object moving in uniform circular motion is moving in a
circle with a uniform or constant speed. The velocity vector is constant in
magnitude but changing in direction. Therefore the object is accelerating in
a circular motion path.
Section
6.2 Circular Motion
The Centripetal Force Requirement
As mentioned earlier, an object moving in a circle is experiencing an
acceleration. Even if moving around the perimeter of the circle with a
constant speed, there is still a change in velocity and subsequently an
acceleration.
This acceleration is directed towards the center of the circle. And in
accord with Newton's second law of motion, an object which experiences
an acceleration must also be experiencing a net force. The direction of
the net force is in the same direction as the acceleration. So for an object
moving in a circle, there must be an inward force acting upon it in order to
cause its inward acceleration. This is sometimes referred to as the
centripetal force requirement.
The word centripetal (not to be confused with the F-word centrifugal)
means center seeking. For object's moving in circular motion, there is a
net force acting towards the center which causes the object to seek the
center.
Section
6.2 Circular Motion
Centripetal Acceleration
The acceleration of an object moving in a circle is
always in the direction of the net force acting on
it, there must be a net force toward the center of
the circle. This force can be provided by any
number of agents.
When an Olympic hammer thrower swings the
hammer, the force is the tension in the chain
attached to the massive ball.
Section
6.2 Circular Motion
Centripetal Acceleration
When an object moves in a circle, the net force
toward the center of the circle is called the
centripetal force.
To analyze centripetal acceleration situations
accurately, you must identify the agent of the
force that causes the acceleration. Then you
can apply Newton’s second law for the
component in the direction of the acceleration
in the following way.
Section
6.2 Circular Motion
Centripetal Acceleration
Newton’s Second Law for Circular Motion
The net centripetal force on an object moving in
a circle is equal to the object’s mass times the
centripetal acceleration.
Section
6.2 Circular Motion
Centripetal Acceleration
When solving problems, it is useful to choose a
coordinate system with one axis in the direction
of the acceleration.
For circular motion, the direction of the
acceleration is always toward the center of the
circle.
Section
6.2 Circular Motion
Centripetal Acceleration
Rather than labeling this axis x or y, call it c, for
centripetal acceleration. The other axis is in the
direction of the velocity, tangent to the circle. It is
labeled tang for tangential.
Centripetal force is just another name for the net
force in the centripetal direction. It is the sum of
all the real forces, those for which you can
identify agents that act along the centripetal axis.
Section
6.2 Circular Motion
A Nonexistent Force
According to Newton’s first law, you will continue
moving with the same velocity unless there is a net
force acting on you.
The passenger in the car
would continue to move
straight ahead if it were
not for the force of the car
acting in the direction of
the acceleration.
Section
6.2 Circular Motion
A Nonexistent Force
The so-called centrifugal, or outward force, is a
fictitious, nonexistent force.
Section
6.2 Section Check
Question 1
Explain why an object moving in a circle at a
constant speed is accelerating.
Section
6.2 Section Check
Answer 1
Acceleration is the rate of change of velocity, the
object is accelerating due to its constant change
in the direction of its motion.
Section
6.2 Section Check
Question 2
What is the relationship between the magnitude
of centripetal acceleration (ac) and an object’s
speed (v)?
A.
B.
C.
D.
Section
6.2 Section Check
Answer 2
Reason: From the equation for centripetal
acceleration:
Centripetal acceleration always points to
the center of the circle. Its magnitude is
equal to the square of the speed divided
by the radius of the motion.
Section
6.2 Section Check
Question 3
What is the direction of the velocity vector of an
accelerating object?
A. toward the center of the circle
B. away from the center of the circle
C. along the circular path
D. tangent to the circular path
Section
6.2 Section Check
Answer 3
Reason: While constantly changing, the velocity
vector for an object in uniform circular
motion is always tangent to the circle.
Vectors are never curved and therefore
cannot be along a circular path.
Section
6.2 Section Check
Sample Problem #1
A 900-kg car moving at 10 m/s takes a turn around a circle
with a radius of 25.0 m. Determine the acceleration and
the net force acting upon the car.
Known Information:
Requested Information:
m = 900 kg
a = ????
v = 10.0 m/s
Fnet = ????
R = 25.0 m
Section
6.2 Section Check
To determine the acceleration of the car, use the equation a = v2 / R.
The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation
Fnet = m•a.
The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
Section
6.2 Section Check
Sample Problem #2
A 95-kg halfback makes a turn on the football field. The halfback sweeps
out a path that is a portion of a circle with a radius of 12-meters. The
halfback makes a quarter of a turn around the circle in 2.1 seconds.
Determine the speed, acceleration and net force acting upon the halfback.
Known Information:
Requested Information:
m = 95.0 kg
v = ????
R = 12.0 m
a = ????
Traveled 1/4-th of the
circumference in 2.1 s
Fnet = ????
Section
6.2 Section Check
To determine the speed of the halfback, use the equation v = d / t where the d is
one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v=d/t
v = (0.25 • 2 • pi • R) / t
v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = v2 / R. The
solution is as follows:
a = v2 / R
a = (8.97 m/s)2 / (12.0 m)
a = (80.5 m2/s2) / (12.0 m)
a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m•a.
The solution is as follows.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N