1. Circular Motion 6.2 In this section you will:
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.

2. Section
Circular Motion
6.2
Describing Circular Motion
ch6_2_movanim

3. Circular Motion 6.2 Centripetal Acceleration
Section
Circular Motion
6.2
Centripetal Acceleration
centripetal – center seeking; toward the center
centripetal acceleration (ac) – acceleration toward the center of circular motion.
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.

4. time T Circular Motion 6.2 Centripetal Acceleration
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

5. T Circular Motion 6.2 Centripetal Acceleration Since and v = 2r then
Section
Circular Motion
6.2
Centripetal Acceleration
Since
and v = 2r T
then

6. Circular Motion 6.2 Centripetal Acceleration
Section
Circular Motion
6.2
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.

7. Section
Circular Motion
6.2
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.

8. Circular Motion 6.2 Centripetal Acceleration
Section
Circular Motion
6.2
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

9. Circular Motion 6.2 A Nonexistent 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.

10. Section Check 6.2 Question 1
Explain why an object moving in a circle at a constant speed is accelerated.

11. 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.

12. Section Check 6.2 Question 2
What is the relationship between the magnitude of centripetal acceleration (ac) and an object’s speed (v)?

13. 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.

14. Section Check 6.2 Question 3
What is the direction of the velocity vector of an accelerating object?
Toward the center of the circle.
Away from the center of the circle.
Along the circular path.
Tangent to the circular path.

15. 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.

16. Section
Circular Motion
6.2
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.
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) = N

17. Circular Motion 6.2 Practice Problems p. 156: 12,13,14,15.
Section
6.2
Circular Motion
Practice Problems p. 156: 12,13,14,15.
HW 6.B handout

18. Circular Motion Review
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.