Chapter 8: Motion in Circles

8.1 Circular Motion

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 angular displacement

angular speed

axis

centrifugal force

centripetal acceleration

centripetal force

circumference

ellipse

gravitational constant

law of universal gravitation

linear speed

orbit

radian

revolve

rotate

satellite

Inv 8.1 Motion in Circles

Investigation Key Question:

How do we describe circular motion?

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 Motion in Circles

Earth revolves around the Sun once each year while it rotates around its north-south axis once each day.

8.1 Motion in Circles

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 Circular Motion 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 The relationship between linear and angular speed The circumference is the distance

around a circle. The circumference depends on the

radius of the circle.

8.1 The relationship between linear and angular speed The linear speed (v) of a point at the

edge of a turning circle is the circumference divided by the time it takes to make one full turn.

The linear speed of a point on a wheel depends on the radius, r, which is the distance from the center of rotation.

8.1 The relationship between linear and angular speed

C = 2π rRadius (m)

Circumference(m)

v = d t

Distance (m)

Speed(m/sec)

Time (sec)

2πr

8.1 The relationship between linear and angular speed

v = rRadius (m)

Linear speed(m/sec)

Angular speed(rad/sec)

*Angular speed is represented with a lowercase Greek omega (ω).

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. Use v = ωr

4. Solve: For Siv: v = (1 rad/s)(4 m) v = 4 m/s. For Holly: v = (1 rad/s)(2 m) v = 2 m/s.

Calculate linear from angular speed

Two children are spinning around on a merry-go-round. 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 The units of radians per second One radian is the angle

you get when you rotate the radius of a circle a distance on the circumference equal to the length of the radius.

One radian is approximately 57.3 degrees, so a radian is a larger unit of angle measure than a degree.

8.1 The units of radians per second Angular speed

naturally comes out in units of radians per second.

For the purpose of angular speed, the radian is a better unit for angles.

Radians are better for angular speed because a radian is a ratio of two lengths.

8.1 Angular Speed

= t

Angle turned (rad)

Time taken (sec)

Angular speed (rad/sec)

1. You are asked for the angular speed.

2. You are given turns and time.

3. There are 2π radians in one full turn. Use: ω = θ ÷ t

4. Solve: ω = (6 × 2π) ÷ (2 s) = 18.8 rad/s

Calculating angular speedin rad/s

A bicycle wheel makes six turns in2 seconds. What is its angular speed in radians per second?

8.1 Relating angular speed, linear speed anddisplacement As a wheel rotates, the point touching the ground passes around its circumference.

When the wheel has turned one full rotation, it has moved forward a distance equal to its circumference.

Therefore, the linear speed of a wheel is its angular speed multiplied by its radius.

1. You are asked for the angular speed in rpm.

2. You are given the linear speed and radius of the wheel.

3. Use: v = ωr, 1 rotation = 2π radians

4. Solve: ω = v ÷ r = (11 m/s) ÷ (0.35 m) = 31.4 rad/s. Convert to rpm: 31.4 rad x 60 s x 1 rotation = 300

rpm1 s 1 min 2 π rad

Calculating angular speedfrom linear speed

A bicycle has wheels that are 70 cm in diameter (35 cm radius). The bicycle is moving forward with a linear speed of 11 m/s. Assume the bicycle wheels are not slipping and calculate the angular speed of the wheels in rpm.

Chapter 8: Motion in Circles

8.1 Circular Motion

8.2 Centripetal Force

8.3 Universal Gravitation and Orbital Motion

Inv 8.2 Centripetal ForceInvestigation Key Question:

Why does a roller coaster stay on a track upside down on a loop?

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.

8.2 Calculating centripetal force

The magnitude of the centripetal force needed to move an object in a circle depends on the object’s mass and speed, and on the radius of the circle.

8.2 Centripetal Force

Fc = mv2

r

Linear speed(m/sec)

Radius of path(m)

Centripetalforce (N)

Mass (kg)

1. You are asked to find the centripetal force.

2. You are given the radius, mass, and linear speed.

3. Use: Fc = mv2 ÷ r

4. Solve: Fc = (50 kg)(6 m/s)2 ÷ (3 m) = 600 N

Calculating 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?

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

ac = v2

r

Speed(m/sec)

Radius of path(m)

Centripetalacceleration (m/sec2)

1. You are asked for centripetal acceleration and a comparison with g (9.8 m/s2).

2. You are given the linear speed and radius of the motion.

3. Use: ac = v2 ÷ r

4. 4. Solve: ac = (10 m/s)2 ÷ (50 m) = 2 m/s2

5. The centripetal acceleration is about 20%, or 1/5 that of gravity.

Calculating centripetal acceleration

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

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 often incorrectly identified as 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.

8.2 Centrifugal Force

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.

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.

Chapter 8: Motion in Circles

8.1 Circular Motion

8.2 Centripetal Force

8.3 Universal Gravitation and Orbital Motion

Inv 8.3 Universal Gravitation and Orbital MotionInvestigation Key

Question:

How strong is gravity in other places in the universe?

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

F = m1m2

r2

Mass 1Mass 2

Distance between

masses (m)

Force (N)

1. You are asked to find a person’s weight on the Moon.

2. You are given the radius and the masses.

3. Use: Fg = Gm1m2 ÷ r 2

4. Solve:

Calculating the weight of a person on the moon

The mass of the Moon is 7.36 × 1022 kg. The radius of the moon is 1.74 × 106 m. Use the equation of universal gravitation to calculate the weight of a 90-kg astronaut on the Moon’s surface.

8.3 Orbital Motion A satellite is an object

that is bound by gravity to another object such as a planet or star.

An orbit is the path followed by a satellite. The orbits of many natural and man-made satellites are circular, or nearly circular.

8.3 Orbital Motion The motion of a satellite

is closely related to projectile motion.

If an object is launched above Earth’s surface at a slow speed, it will follow a parabolic path and fall back to Earth.

At a launch speed of about 8 kilometers per second, the curve of a projectile’s path matches the curvature of the planet.

8.3 Satellite Motion

The first artificial satellite, Sputnik I, which translates as “traveling companion,” was launched by the former Soviet Union on October 4, 1957.

For a satellite in a circular orbit, the force of Earth’s gravity pulling on the satellite equals the centripetal force required to keep it in its orbit.

8.3 Orbit Equation The relationship between a satellite’s

orbital radius, r, and its orbital velocity, v is found by combining the equations for centripetal and gravitational force.

8.3 Geostationary orbits

To keep up with Earth’s rotation, a geostationary satellite must travel the entire circumference of its orbit (2π r) in 24 hours, or 86,400 seconds.

To stay in orbit, the satellite’s radius and velocity must also satisfy the orbit equation.

All geostationary satellites must orbit directly above the equator.

This means that the geostationary “belt” is the prime real estate of the satellite world.

There have been international disputes over the right to the prime geostationary slots, and there have even been cases where satellites in adjacent slots have interfered with each other.

Use of HEO