Download Circular Motion

Chapter
Circular Motion
6.2
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.
Read Chapter 6.2.
HW 6.B: Handout
Circular Motion Study Guide, due before Test.
Section
6.2
Circular Motion
Describing Circular Motion
ch6_2_movanim
Section
Circular Motion
6.2
Centripetal Acceleration
centripetal
– center seeking; toward the center
centripetal acceleration
acceleration toward the center of circular motion.
(ac) –
The formula for centripetal acceleration is:
where v is the tangential velocity and
r is the radius of the circle
uniform circular motion – the movement of an object at a constant
speed around a circle with a fixed radius.
Section
Circular Motion
6.2
Centripetal Acceleration
One way of measuring the speed of an object moving in a circle
is to measure its period, T, the time needed for the object to
make one complete revolution. The frequency, f, is how many
revolutions per second.
T = 1/f
period is the reciprocal of frequency
During this time, the object travels a distance equal to the
circumference of the circle, 2πr. The object’s speed, then, is
represented by v = 2πr/T.
v = distance = 2r
time
T
Section
Circular Motion
6.2
Centripetal Acceleration
Since
and
then
v = 2r
T
Section
6.2
Circular Motion
Centripetal Acceleration
Because 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 a hammer thrower
swings the hammer, as in the
adjoining figure, the force is
the tension in the chain
attached to the massive ball.
Section
6.2
Circular Motion
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
(such as tension on a string). Then you can apply Newton’s
second law for the component in the direction of the
acceleration in the following way.
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
r is the radius of the circle
v is the tangential velocity
a (or ac) is the centripetal
acceleration
Fc is the centripetal force
Section
Circular Motion
6.2
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 door
acting in the direction of the
acceleration.
The so-called centrifugal force
or apparent outward force, is
a fictitious, nonexistent force
centrifugal
,
.
– center fleeing; away from the center.
Section
Section Check
6.2
Question 1
Explain why an object moving in a circle at a constant speed is
accelerated.
Section
Section Check
6.2
Answer 1
Because acceleration is the rate of change of velocity, the object
accelerates due to the change in the direction of motion and not
speed.
Section
Section Check
6.2
Question 2
What is the relationship between the magnitude of centripetal
acceleration (ac) and an object’s speed (v)?
A.
B.
C.
D.
Section
Section Check
6.2
Answer 2
Answer: C
Reason: From the equation for centripetal acceleration,
That is, 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
Section Check
6.2
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
Section Check
6.2
Answer 3
Answer: D
Reason: The displacement, ∆r, of an object in a circular motion
divided by the time interval in which the displacement
occurs is the object’s average velocity during that time
interval. If you notice the picture below, ∆r is in the
direction of tangent to the circle and, therefore, is the
velocity.
Section
6.2
Circular Motion
Example: Shawna swung a 10.0 g mass on a 0.50 m string in a circle
over her head. The period of the motion is 0.50 s.
a. What is the centripetal acceleration of the mass?
ac = 42r = 42 (0.50 m) = 79.0 m/s2
T2
(0.50 s)2
b. What is the tension in the string?
FT = FC = m ac = (0.01 kg) (79.0 m/s2) = 0.79 N
Section
Circular Motion
6.2
Practice Problems p. 156: 12,13,14,15.
HW 6.B handout
Section
6.2
Circular Motion Review
Formulas:
Fc = m ac = mv2
r
ac = v2 = 42r
r
T2
T is the period (time for one revolution around the circle.)
The distance around the circle is 2r.
Uniform circular motion has constant speed and constant radius.
The net force toward the center of the circle causes the centripetal acceleration.