1. Chapter 5: Uniform Circular Motion

2. 5-1 Uniform Circular Motion
Uniform Circular Motion: The motion of an object traveling at a constant (uniform) speed on a circular path
5-1 Uniform Circular Motion

3. 5-1 Uniform Circular Motion
Since we are dealing with object moving in a circle, it is convenient to talk about the period of the motion
Period (T) = the time required to travel once around the circle (one complete revolution.
5-1 Uniform Circular Motion

4. 5-1 Uniform Circular Motion
Period (T) = the time required to travel once around the circle (one complete revolution.
The distance around the circle is just the circumference of the circle (2πr)
Speed (v) = 2πr T
Notice v is scalar – it gives the speed only (no direction)
5-1 Uniform Circular Motion

5. 5-1 Uniform Circular Motion
The blades on the
turbines at the Wild
Horse Wind Farm in
eastern WA turn at a rate of 16 revolutions/minute. Each blade is approx. 39m long. Compare the speed of the tip of the blade to the speed of the part of the blade that is 1m from the center.
5-1 Uniform Circular Motion

6. 5-1 Uniform Circular Motion
First, find the time it
Takes for one revolution.
f= 16 rev/minute
= 16 rev/60 sec
= rev/sec
T = 1/f
= 3.75 seconds
5-1 Uniform Circular Motion

7. 5-1 Uniform Circular Motion
Next, find the distance
For one revolution
r=39m
Distance = 2πr = 245m
v = distance/time = 245m/3.75s
= m/s speed of the tip
5-1 Uniform Circular Motion

8. 5-1 Uniform Circular Motion
Next, find the distance
For one revolution
r=1m
Distance = 2πr = 6.3m
v = distance/time = 6.3m/3.75s
= 1.68 m/s speed of the inner part of the blade.
5-1 Uniform Circular Motion

9. 5-1 Uniform Circular Motion
For uniform circular motion, the speed is constant, but the velocity is not because the direction is changing continually!
At any instant, the velocity
vector is always tangent to
the circle.
5-1 Uniform Circular Motion

10. Check Your Understanding
1. A tube is been placed upon the table and shaped into a three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the path of the golf ball as it exits the tube.
Check Your Understanding

11. Check Your Understanding
While the ball is in the tube, the tube exerts a force on the ball which causes the ball to accelerate in a circle. Once the ball leaves the tube, there are no forces acting on it. The ball will continue to move in a straight line.
Check Your Understanding

12. Check Your Understanding
Draw the velocity vector at each point (A,B,C) in the object’s motion.
Check Your Understanding

13. Centripetal Acceleration
Draw the velocity vector at each point (A,B,C) in the object’s motion. v
If the object is accelerating (velocity is changing), then by Newton’s 2nd Law, it must have a net force acting on it.
Centripetal Acceleration

14. Recall from the bowling ball lab that a net inward force was required to move the ball in a circle.
The force and acceleration point in the same direction. Therefore, the object’s acceleration is towards that center of the circle at any given instant.

15. Centripetal Acceleration
We call this acceleration the centripetal acceleration.
ac= 𝑣2 𝑟
The centripetal acceleration vector always points toward the center of the circle, and changes direction as the object moves.
Centripetal Acceleration

16. The bobsled track at Lake Placid, NY (site of the 1980 winter Olympics), contains turns with radii of 33m and 24m. Find the centripetal acceleration at each turn for a speed of 34m/s. How many “g’s” is this?
ac= 𝑣2 𝑟
r=33m ac=35m/s2=3.6g
r=24m ac=48m/s2=4.9g
Example 1

17. Geometry Review How do you find the length of an arc, s,? s=rθ
But θ must be in radians! r s θ
Geometry Review

18. P. 156 #1, 3, 5, 7, 8, 9, 10
Assignment