Foundations of Physics CPO Science Foundations of Physics Chapter 9 Unit 3, Chapter 8

Unit 3: Motion and Forces in 2 and 3 Dimensions Chapter 8 Using Vectors: Forces and Motion 8.1 Motion in Circles 8.2 Centripetal Force 8.3 Universal Gravitation and Orbital Motion

Chapter 8 Objectives Calculate angular speed in radians per second. Calculate linear speed from angular speed and vice-versa. Describe and calculate centripetal forces and accelerations. Describe the relationship between the force of gravity and the masses and distance between objects. Calculate the force of gravity when given masses and distance between two objects. Describe why satellites remain in orbit around a planet.

Chapter 8 Vocabulary Terms rotate revolve axis law of universal gravitation circumference linear speed angular speed centrifugal force radian orbit centripetal force centripetal acceleration ellipse satellite angular displacement gravitational constant

8.1 Vectors and Direction Key Question: How do we describe circular motion? *Students read Section 8.1 AFTER Investigation 8.1

8.1 Motion in Circles We say an object rotates about its axis when the axis is part of the moving object. A child revolves on a merry-go-round because he is external to the merry-go-round's axis.

8.1 Angular Speed Angular speed is the rate at which an object rotates or revolves. There are two ways to measure angular speed number of turns per unit of time (rotations/minute) change in angle per unit of time (deg/sec or rad/sec)

8.1 Angular Speed For the purpose of angular speed, the radian is a better unit for angles. One radian is approx. 57.3 degrees. Radians are better for angular speed because a radian is a ratio of two lengths.

w = q t 8.1 Angular Speed Angle turned (rad) Angular speed (rad/sec) Time taken (sec)

8.1 Calculate angular speed A bicycle wheel makes six turns in 2 seconds. What is its angular speed in radians per second? 1) You are asked for the angular speed. 2) You are given turns and time. 3) There are 2π radians in one full turn. ω = θ/t 4) Solve: ω = (6 × 2π) ÷ (2 sec) = 18.8 rad/sec

8.1 Linear and Angular Speed A wheel rolling along the ground has both a linear speed and an angular speed. A point at the edge of a wheel moves one circumference in each turn of the circle.

8.1 Linear and Angular Speed Radius (m) Circumference (m) C = 2 P r Distance (m) 2 P r Speed (m/sec) v = d t Time (sec)

8.1 Linear and Angular Speed Radius (m) Linear speed (m/sec) v = w r Angular speed (rad/sec) *This formula is used in automobile speedometers based on a tire's radius.

8.1 Calculate linear from angular speed Two children are spinning around on a merry-go-round. 1) You are asked for the children’s linear speeds. 2) You are given the angular speed of the merry-go-round and radius to each child. 3) The relationship that applies is v = ωr. 4) Solve: For Siv: v = (1 rad/sec)(4 m) = 4 m/sec. For Holly: v = (1 rad/sec)(2 m) = 2 m/sec. Siv is standing 4 meters from the axis of rotation and Holly is standing 2 meters from the axis. Calculate each child’s linear speed when the angular speed of the merry go-round is 1 rad/sec.

8.1 Linear and Angular Speed and Displacement

8.1 Calculate angular from linear speed A bicycle has wheels that are 70 cm in diameter (35 cm radius). 1) You are asked for the angular speed in RPM. 2) You are given the linear speed and radius of the wheel. 3) v = ωr, 1 rotation = 2π radians 4) Solve: ω = v ÷ r = (11 m/sec) ÷ (.35 m) = 31.4 rad/sec. Convert to rpm using dimensional analysis: ω = 31.4 rad/sec x 60 sec/min x 1 rotation/2π radians = 300 rpm The bicycle is moving forward with a linear speed of 11 m/sec. Assume the bicycle wheels are not slipping and calculate the angular speed of the wheels in RPM.

*Students read Section 8.2 AFTER Investigation 8.2 8.2 Centripetal Force Key Question: Why does a roller coaster stay on a track upside down on a loop? *Students read Section 8.2 AFTER Investigation 8.2

8.2 Centripetal Force We usually think of acceleration as a change in speed. Because velocity includes both speed and direction, acceleration can also be a change in the direction of motion.

8.2 Centripetal Force Any force that causes an object to move in a circle is called a centripetal force. A centripetal force is always perpendicular to an object’s motion, toward the center of the circle.

Fc = mv2 r 8.2 Centripetal Force Mass (kg) Linear speed (m/sec) force (N) Fc = mv2 r Radius of path (m)

8.2 Calculate centripetal force A 50-kilogram passenger on an amusement park ride stands with his back against the wall of a cylindrical room with radius of 3 m. What is the centripetal force of the wall pressing into his back when the room spins and he is moving at 6 m/sec? 1) You are asked to find the centripetal force. 2) You are given the radius, mass, and linear speed. 3) The formula that applies is Fc = mv2 ÷ r. 4) Solve: Fc = (50 kg)(6 m/sec)2 ÷ (3 m) = 600 N

8.2 Centripetal Acceleration Acceleration is the rate at which an object’s velocity changes as the result of a force. Centripetal acceleration is the acceleration of an object moving in a circle due to the centripetal force.

8.2 Centripetal Acceleration Speed (m/sec) Centripetal acceleration (m/sec2) ac = v2 r Radius of path (m)

8.2 Calculate centripetal acceleration 1) You are asked for centripetal acceleration and a comparison with g (9.8 m/sec2). 2) You are given the linear speed and radius of the motion. 3) ac = v2 ÷ r 4) Solve: ac = (10 m/sec)2 ÷ (50 m) = 2 m/sec2 The centripetal acceleration is about 20% or 1/5 that of gravity. A motorcycle drives around a bend with a 50-meter radius at 10 m/sec. Find the motor cycle’s centripetal acceleration and compare it with g, the acceleration of gravity.

8.2 Centrifugal Force We call an object’s tendency to resist a change in its motion its inertia. An object moving in a circle is constantly changing its direction of motion. Although the centripetal force pushes you toward the center of the circular path... ...it seems as if there also is a force pushing you to the outside. This apparent outward force is called centrifugal force.

8.2 Centrifugal Force Centrifugal force is not a true force exerted on your body. It is simply your tendency to move in a straight line due to inertia. This is easy to observe by twirling a small object at the end of a string. When the string is released, the object flies off in a straight line tangent to the circle.

8.3 Universal Gravitation and Orbital Motion Key Question: How strong is gravity in other places in the universe? *Students read Section 8.3 AFTER Investigation 8.3

8.3 Universal Gravitation and Orbital Motion Sir Isaac Newton first deduced that the force responsible for making objects fall on Earth is the same force that keeps the moon in orbit. This idea is known as the law of universal gravitation. Gravitational force exists between all objects that have mass. The strength of the gravitational force depends on the mass of the objects and the distance between them.

8.3 Law of Universal Gravitation Mass 1 Mass 2 Force (N) F = m1m2 r2 Distance between masses (m)

8.3 Calculate gravitational force The mass of the moon is 7.36 × 1022 kg. The radius of the moon is 1.74 × 106 m. 1) You are asked to find a person’s weight on the moon. 2) You are given the radius and the masses. 3) The formula that applies is Fg = Gm1m2 ÷ r2 4) Solve: Fg = (6.67 x 10-11 N.m2/kg2) x (90 kg) (7.36 x 1022 kg) / (1.74 x 106 m)2 = 146 N By comparison, on Earth the astronaut’s weight would be 90 kg x 9.8 m/s2 or 882 N. The force of gravity on the moon is approximately one-sixth what it is on Earth. Use the equation of universal gravitation to calculate the weight of a 90 kg astronaut on the surface of the moon.

8.3 Orbital Motion A satellite is an object that is bound by gravity to another object such as a planet or star. If a satellite is launched above Earth at more than 8 kilometers per second, the orbit will be a noncircular ellipse. A satellite in an elliptical orbit does not move at a constant speed.

Application: Satellite Motion