1. Uniform Circular Motion
Unit 4
Uniform Circular Motion

2. Definitions and Equation
Uniform Circular Motion: Motion of an object in a circle with a constant or uniform speed
Period (T): The time required to travel once around the circle (1 revolution)
The distance the object covers is the circumference of the circle (𝐶=2𝜋𝑟)
Therefore, the equation for speed in circular motion (comes from 𝑣= 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 )
𝒗= 𝟐 𝝅 𝒓 𝑻

3. Example
A ball is being whirled around at the end of a rope. The ball travels around in a circle 5 times in 4 seconds.
Q1: What is the period of the motion of the ball?
𝑇= 4 5 =0.8 𝑠
Q2: If the rope is 1.2 m long, what is the ball’s speed?
𝑣= 2𝜋𝑟 𝑇 = 2𝜋∗ ≈9.42 𝑚 𝑠

4. Velocity and Acceleration
What is the difference between speed and velocity?
Therefore in circular motion if the speed isn’t changing what is ?
What are the three ways to cause acceleration?
In circular motion the change in direction causes an acceleration

5. Centripetal Acceleration
Centripetal Acceleration (ac) is the acceleration caused by the object moving in a circular path. The change in motion is due to the change in direction.
𝒂 𝒄 = 𝒗 𝟐 𝒓
v = constant speed, r = radius
Drawings on board.
Units ac is in m/s2 or ft/s2
v is till m/s or ft/s
r is in m or ft

6. Centripetal Acceleration
The centripetal acceleration vector always points toward the center of the circle and continually changes direction as the object moves

7. Centripetal Acceleration
If the radius of the curve or path is increased the acceleration decreases
Acceleration is directly proportional to 𝑣 2
If the speed is doubled, the acceleration increases by a factor of 4
There can be no equilibrium for an object in circular motion because of the always changing direction which causes acceleration
Cars and bobsled in turns
Car moving along straight path into a turn, in a turn and coming out of the turn along a straight line. (in turn no equilibrium)

8. Example – Centripetal Acceleration
A merry-go-round has a diameter of 30 ft and completes 9 revolutions during a 90 second ride.
Q1: At what speed is someone on the outer edge of the merry-go-round traveling?
𝑟=15 𝑓𝑡, 𝑇=90÷9=10 𝑠
𝑣= 2𝜋𝑟 𝑇 = 2𝜋∗15 10 ≈9.42 𝑓𝑡 𝑠

9. Example – Centripetal Acceleration
A merry-go-round has a diameter of 30 ft and completes 9 revolutions during a 90 second ride.
Q2: What is the centripetal acceleration on the outer edge?
𝑎 𝑐 = 𝑣 2 𝑟 = ≈5.92 𝑓𝑡 𝑠 2

10. Example – Centripetal Acceleration
A merry-go-round has a diameter of 30 ft and completes 9 revolutions during a 90 second ride.
Q3: Would a person who is standing only 2 feet from the center of the merry-go-round experience more or less centripetal acceleration than someone on the edge? Justify your answer with a calculation.
Less centripetal acceleration, since speed is smaller.
𝑣= 2𝜋∗2 10 ≈1.26 𝑓𝑡 𝑠 so 𝑎 𝑐 = 𝑣 2 𝑟 = ≈0.79 𝑓𝑡 𝑠 2

11. Demonstration of Centripetal Acceleration

12. Review Which direction does the centripetal acceleration vector point?
What is period?
How do you find the distance the object covers in a circular path?
If the radius of the curve increases what happens to the centripetal acceleration?

13. Unit 4 Continued: Centripetal Force
Newton’s Laws of motion apply to objects moving in a curved path.
What does inertia say objects want to do?
If an object is released from its circular path, it will continue in straight line tangent to the circle.
How do we keep a body moving in a circle?
Inertia: says an object wants to continue in a straight line.
Keep in circle by applying a force perpendicular to the line of motion of the body. And have it moving at a constant speed (uniform)

14. Centripetal Force
Think about a rock or a ball on a string, how do you keep it whirling around?
Keep the object in line by applying a force on the string.
ircmotion/

15. Centripetal Force The force is called the centripetal force.
Centripetal force is the center seeking force
Causes the object to stay in the circular path
Force is exerted toward the center of the circle
If Force is removed then the object would move off tangent to the path

16. Examples of Centripetal Force
Centripetal force isn’t a new kind of force! The name just applies to any force that keeps something moving in a circle.
Whirling an object on the end of a string
The force is tension in the string.
Planets orbiting the sun
The force is gravity.
A car turning a corner
The force is the friction between the tires and the road.

17. To find Centripetal Force
𝑭 𝒄 =𝒎 𝒂 𝒄 = 𝒎 𝒗 𝟐 𝒓
F is force in Newtons (or pounds)
m is mass in kilograms (or slugs)
ac is centripetal acceleration in m/s2 (or ft/s2
v is circular speed (tangential speed) in m/s or ft/s
r is the radius of the curve in m or ft

18. Example 1 – Centripetal Force
A 850-kg car traveling a constant speed of 40 m/s enters a turn. The turn is on an arc of radius 50 m.
Q1: What is the centripetal force needed to keep the car on the road?
𝐹 𝑐 = 𝑚 𝑣 2 𝑟 = 850∗ =27,200 𝑁
Q2: What must be providing this force?
The friction of the tires on the road.

19. Example 2 – Centripetal Force
A 5-kg sphere is to be spun around at the end of a 0.75 m rope. The rope can withstand up to 500 N of tension force. Q1: What is the maximum speed at which the sphere can be spun horizontally without breaking the rope? 𝑟=0.75 𝑚, 𝑚=5 𝑘𝑔, max 𝐹 𝑐 =500 𝑁 𝐹 𝑐 =𝑚 𝑎 𝑐 = 𝑚 𝑣 2 𝑟 500= 5 𝑣 so 𝑣≈8.66 𝑚 𝑠

20. Example 2 – Centripetal Force
A 650-kg car is traveling around a curve. The coefficient of kinetic friction for the scenario is 0.70, and the maximum speed at which the car can safely take the turn is 40 m/s. Find the radius of the curve. 𝑚=5 𝑘𝑔, 𝑣=40 𝑚 𝑠 𝐹 𝑐 =𝑓=𝜇𝑁= =4459 𝑁 4459= 650 (40) 2 𝑟 so 𝑟= 650 (40) ≈233 𝑚

21. Example 3 – Centripetal Force
A 5-kg sphere is to be spun around at the end of a 0.75 m rope. The rope can withstand up to 500 N of tension force. Q2: How does the problem change if the sphere is spun vertically? What is the maximum speed at which the sphere can be spun vertically? At the bottom, the rope must withstand both the centripetal force and the tension from the weight. 𝐹 𝑐 =500−5 9.8 =451 N 451= 5 𝑣 so 𝑣≈8.22 𝑚 𝑠