Uniform Circular Motion Circular Motion Lab Results Part

Uniform Circular Motion

Circular Motion Lab Results Part 1: Radius’ effect on velocity Mass, centripetal force kept constant Part 2: Mass’ effect on velocity Radius, centripetal force kept constant Part 3: Force’s effect on velocity Radius, mass kept constant

Circular Motion Lab Results If the sum of the forces on an object is pointed toward a common center of rotation, perpendicular to its velocity, it will travel in a circle at a constant speed. V m ∑F r

Warm-up Question (1/10) What is the average speed of the Moon in its orbit given the information below? Answer:

Example Problem #2 What is the average speed of the Earth in its orbit around the Sun given the information below? Answer:

Vector nature of circular motion Draw a circle on your whiteboard. Determine a direction of circular motion (clockwise or counterclockwise) Draw 2 dots, labelling them A and B—B will be further along the motion than A Draw the instantaneous velocity vectors at each of the points Draw a resultant vector showing the change in velocity between A and B

Terminology reminders: Centripetal Force: The net force, acting towards the center of rotation, the allows an object to travel in uniform circular motion. Centripetal Acceleration: The acceleration of the object, caused by the centripetal force, directed towards the center of the circular path.

Applying Newton’s 2 nd Law

Question 7. 1 Tetherball a) toward the top of the pole In the game of tetherball, b) toward the ground the struck ball whirls c) along the horizontal component of the tension force around a pole. In what direction does the net d) along the vertical component of the tension force on the ball point? e) tangential to the circle T W

Question 7. 1 Tetherball In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) tangential to the circle The vertical component of the tension balances the weight. The horizontal component of tension W T T provides the centripetal force that points toward the center of the circle. W

Question 7. 2 a Around the Curve I You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor

Question 7. 2 a Around the Curve I You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door. a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor

Question 7. 2 b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above

Question 7. 2 b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.

Question 7. 2 c Around the Curve III You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? a) car’s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above

Question 7. 2 c Around the Curve III You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? a) car’s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line! Follow-up: What could be done to the road or car to prevent skidding?

Question 7. 3 Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow? d a b c e

Question 7. 3 Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow? d a e b c Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires! Follow-up: What physical force provides the centripetal force?

Question 7. 4 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of a) T 2 = ¼T 1 b) T 2 = ½T 1 circle 2 is twice that of circle 1. If the period c) T 2 = T 1 of motion is the same for both rocks, what is d) T 2 = 2 T 1 the tension in cord 2 compared to cord 1? e) T 2 = 4 T 1

Question 7. 4 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of a) T 2 = ¼T 1 b) T 2 = ½T 1 circle 2 is twice that of circle 1. If the period c) T 2 = T 1 of motion is the same for both rocks, what is d) T 2 = 2 T 1 the tension in cord 2 compared to cord 1? e) T 2 = 4 T 1 The centripetal force in this case is given by the tension, so T = mv 2/r. For the same period, we find that v 2 = 2 v 1 (and this term is squared). However, for the denominator, we see that r 2 = 2 r 1 which gives us the relation T 2 = 2 T 1.

Question 7. 5 Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? a b c d e

Question 7. 5 Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? a b c d e The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down?

Practice Problem #3 An astronaut rotates at the end of a test machine whose arm has a length of 10. 0 m. If the maximum acceleration she reaches must not exceed 5 g, what is the maximum number of revolutions per minute of the arm?

Practice Problem #4 A neutron star has a radius of 50. 0 km and completes 1 rotation every 25. 0 ms. What is the centripetal acceleration experienced at the equator? Hypothetically, what is the centripetal force acting on a 50. 0 kg object at the surface of the neutron star? How many times its weight is this?

Question 7. 6 a Going in Circles I You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by a) N remains equal to mg b) N is smaller than mg your seat is equal to your weight mg. c) N is larger than mg How does N change at the top of the d) none of the above Ferris wheel when you are in motion?

Question 7. 6 a Going in Circles I You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by a) N remains equal to mg b) N is smaller than mg your seat is equal to your weight mg. c) N is larger than mg How does N change at the top of the d) none of the above Ferris wheel when you are in motion? You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg?

Question 7. 6 b Going in Circles II A skier goes over a small round hill a) Fc = N + mg with radius R. Because she is in b) Fc = mg – N circular motion, there has to be a c) Fc = T + N – mg centripetal force. At the top of the d) Fc = N hill, what is Fc of the skier equal to? e) Fc = mg v R

Question 7. 6 b Going in Circles II A skier goes over a small round hill a) Fc = N + mg with radius R. Because she is in b) Fc = mg – N circular motion, there has to be a c) Fc = T + N – mg centripetal force. At the top of the d) Fc = N hill, what is Fc of the skier equal to? Fc points toward the center of the circle (i. e. , downward in this case). The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is Fc = mg – N. e) Fc = mg v mg N R Follow-up: What happens when the skier goes into a small dip?

Question 7. 6 c Going in Circles III You swing a ball at the end of string a) Fc = T – mg in a vertical circle. Because the ball b) Fc = T + N – mg is in circular motion there has to be a c) Fc = T + mg centripetal force. At the top of the d) Fc = T ball’s path, what is Fc equal to? v e) Fc = mg top R

Question 7. 6 c Going in Circles III You swing a ball at the end of string a) Fc = T – mg in a vertical circle. Because the ball b) Fc = T + N – mg is in circular motion there has to be a c) Fc = T + mg centripetal force. At the top of the ball’s path, what is Fc equal to? Fc points toward the center of the circle (i. e. , downward in this case). The weight vector points down and the d) Fc = T e) Fc = mg v mg T tension (exerted by the string) also points down. The magnitude of the net force, therefore, is Fc = T+ mg. Follow-up: What is Fc at the bottom of the ball’s path? R

Vertical Uniform Circular Motion (UCM) implies that the tangential speed of a object moving in a circular path around a center will be constant. Therefore, there is a constant net force (the Centripetal Force) acting toward the center of the circular path. Vertically, we must now always consider the weight of the object In order to keep the net force constant so that the speed remains constant, what must happen to the tension (or normal, or whatever additional force is acting to give UCM)

At the very TOP of a circular path Imagine a stopper at the end of a string that is being whirled in a vertical circle. What forces are acting on the stopper at the very top of its path? Draw a force diagram to illustrate the forces acting on the stopper In a 2 nd color, draw in the vector of the net force Write a vector equation that can be used to determine the magnitude of the tension in the string.

At the very BOTTOM of a circular path Imagine a stopper at the end of a string that is being whirled in a vertical circle. What forces are acting on the stopper at the very bottom of its path? Draw a force diagram to illustrate the forces acting on the stopper In a 2 nd color, draw in the vector of the net force Write a vector equation that can be used to determine the magnitude of the tension in the string.

A stunt pilot in an air show performs a loop-the-loop in a vertical circle of radius 3. 21 x 3 10 m. During this performance the pilot whose weight is 806 N, maintains a constant speed of 2. 25 x 102 m/s. What is the magnitude of the normal force acting on the pilot at the top of his path? What is the magnitude of the normal force acting on the pilot at the bottom of the path?

Practice problem #2 Keys, m= 150. 0 g, are attached to a string and swung around in a vertical circle with radius of 35. 0 cm. What is the minimum speed at which the keys must be moving in order to remain in uniform circular motion? What is the tension in the string when the keys, at this speed, are at the bottom of their circular path?

Practice Problem #3 A student is to swing a bucket of water in a vertical circle without spilling any If the distance from his shoulder to the center of mass of the bucket of water is 1. 2 m, what is the minimum speed required to keep the water from coming out of the bucket at the top of the swing? If the combined mass of the water and bucket is 7. 5 kg, what is the force the student must supply at the bottom of the swing in order to keep the system in uniform circular motion?