Section 6 2 Circular Motion Presentation by Jose

Section 6. 2 - Circular Motion Presentation by Jose Aviles, Chris Collazo, and Mellany Gomez

Circular Motion

Circular Motion • Uniform circular motion: The movement of an object or particle trajectory at a constant speed around a circle with a fixed radius. • The position of an object in uniform circular motion is given by the position vector r • Object's average velocity formula: v =�� r/�� t • Object's average acceleration formula: a = �� v/�� t

Centripetal Acceleration

Centripetal Acceleration • Centripetal acceleration: The acceleration vector of an object in uniform circular motion always points in toward the center of the circle. • Centripetal acceleration formula: ac = 4 pi 2 r/T 2 • Newton's Second Law for Circular Motion Formula: Fnet = mac

Example problem 1 What is the direction of the force that acts on Jose while he rides Six Flag's Sky Screamer? What exerts the force?

Answer and explanation Ø The direction of the force is toward the center of the ride. The seat connected to the ride exerts force on Jose. Explanation: We know that the net force in circular motion is an inward or centripetal force. The object is always changing its direction and undergoing an inward acceleration. Without this inward force, an object would continue in a straight line, never changing its direction.

Example problem 2 A 13 -gram rubber stopper is attached to a 0. 93 -meter string. The stopper swung in a horizontal circle, making one revolution in 1. 18 s. Find the tension force exerted from the string on the stopper. Formulas you will need to answer this question: ü ac = 4 pi 2 r/T 2 ü Ftension = mac

Step 1: Tool Box Known Variables: m = 13 grams r = 0. 93 meters T = 1. 18 seconds Unknown Variables: ac = ? Fnet = ? (Here's the question one more time!) A 13 -gram rubber stopper is attached to a 0. 93 -meter string. The stopper swung in a horizontal circle, making one revolution in 1. 18 seconds. Find the tension force exerted but the string on the stopper.

Step 2: Substitute known variables and solve for ac ac = 4 pi 2 r/T 2 ac = 4 pi 2(0. 93 m)/(1. 18 s)2 ac = 36. 7 m/1. 39 s ac = 26. 4 (round to two significant figures) ac = 26 m/s 2

Step 3: Subsitute 26 m/s 2 (ac) into the formula FT = mac to find the tension in the string 13 g ---> 0. 013 kg (Answer is in Newtons so grams must be converted to kilograms) FT = mac FT = (0. 013 kg)(26 m/s 2) FT = 0. 34 N The tension on the string is 0. 34 N

Assessment: What have you learned?

Before you start… • This test will be taken individually. • Please write answers on a separate piece of paper • There will be a time limit to complete each question. • You will receive time warnings. • You will be allowed to use your notes. • You may not ask any presenter for additional help.

Question 1 Which formula correctly displays centripetal acceleration? a. �� r/r�� t = �� v/v�� t b. ac = 4 pi 2 r/T 2 c. mc (pi 42) d. 1/r (�� r/�� t) = 1/v (�� v/�� t)

Answer 1 Which formula correctly displays centripetal acceleration? a. �� r/r�� t = �� v/v�� t b. ac = 4 pi 2 r/T 2 c. mc (pi 42) d. 1/r (�� r/�� t) = 1/v (�� v/�� t)

Question 2 Which formula correctly displays Newton's 2 nd law for circular motion? a. Fnet = Tac b. v = �� r/�� t c. Fnet = mac d. a = �� v/�� t

Answer 2 Which formula correctly displays Newton's 2 nd law for circular motion? a. Fnet = Tac b. v = �� r/�� t c. Fnet = mac d. a = �� v/�� t

Question 3 What is the direction of the force that acts on clothes in a spin cycle of a washing machine? What exerts force?

Answer 3 What is the direction of the force that acts on clothes in a spin cycle of a washing machine? What exerts force? Ø The direction of the force that acts on the clothes is towards the center of the washer machine. The walls of the washer machine exert the force on the clothes.

Question 4 A 40. 0 -gram stone is whirled horizontally on the end of a 0. 60 -meter string making a revolution in 2. 28 -seconds. What is the centripetal acceleration? What is the tension in the string?

Answer 4 A 40. 0 -gram stone is whirled horizontally on the end of a 0. 60 -meter string making a revolution in 2. 28 -seconds. What is the centripetal acceleration? What is the tension in the string? Ø FT = 0. 18 N

Question 5 Chris spun a 2 kg ball horizontally on the end of a 0. 8 m chain making a revolution in 5. 4 s. What is the centripetal acceleration? What is the tension in the chain?

Answer 5 Chris spun a 2 kg ball horizontally on the end of a 0. 8 m chain making a revolution in 5. 4 s. What is the centripetal acceleration? What is the tension in the chain? Ø ac = 1. 1 m/s 2 Ø FT = 2. 2 N

Congratulations! You've made it! Thank you for completing this lesson to the best of your ability. We hope this has been beneficial to the class as a whole.