Tuesday, October 3, 2017

Chapter 6 Part 1 Circular Motion, Centripetal Acceleration and Force, Centrifugal

Prof. Clements Notes/Hints for Physics Courses
Chapter 6 Part 1 Open Stax College Physics or Most Introductory Physics Courses
Rotation Notation, Centripetal Acceleration and Force, Coriolis "Force", Centrifugal "Force"

Terms you should know at the end of this unit: arc, central angle, angular velocity, centripetal, centrifugal, Coriolis

Angle of Rotation and Angular Velocity
Think about some objects that you have seen rotate: e.g. bicycle wheel, Ferris Wheel, merry-go-round, etc. The spinning motion of these objects leads to a different set of characteristics that describe the rotation of the objects: angle of rotation, angular velocity, angular acceleration, etc. The objects may also have a linear motion while they are spinning (e.g. bicycle wheel) but I will generally only discuss the rotation where the axis of rotation is fixed.

The objects don’t travel in a straight line so you can’t use our 4 kinematic equations found in Chapter 2. But have no fear, you will find that there exists 4 kinematic rotations for rotation that are extremely similar to the familiar kinematic equations you have mastered. And, the solution of problems for rotation will be accomplished in a very similar manner with those equations.

Consider the sector of a circle formed by the arc and the two radius lines from the ends of the arc to the center of the circle.
Δθ = Δ S / r or S = r * θ The angle formed between the two radius lines is the central angle, θ. The central angle MUST be measured in radians for this formula to apply. A reminder: 2π radians = 360 degrees. r is the radius of the circle. S is the length of the arc opposite the central angle. S and r must have the same units.

You should know how to convert degrees to radians and radians to degrees using the conversion factor 2π radians = 360 degrees
 e.g. 45 degrees * 2π radians / 360 degrees = 0.785 radians

What is the arc length if the radius is 1.6 meters and the central angle is 45 degrees?
Answer: S = rθ        45 degrees must be converted to radians, as shown above.
S = 1.6 meters * 0.785 radians = 1.26 meters

The angular velocity is found by dividing the angle covered during the motion by the time.

ω = Δθ / Δt Angular velocity has units of radians/second

As the object moves around the circle it has a velocity tangent to the circle. The magnitude of the velocity is related to the angular velocity V = r * ω The units for ω must be radians/second.

e.g. If ω is 3 radians/second and r is 1.2 meters then V is 3.6 meters/second.


Centripetal Acceleration
You should notice that the direction of the velocity changes as the object moves in a circle. At any instant the velocity vector is drawn "tangent" to the circular path of the object. The velocity vector is perpendicular to the radius line that extends from the object to the center of the circle. The magnitude of the velocity is constant but the direction of the velocity changes. An acceleration is needed to change velocity (speed or direction). The name of the acceleration for uniform circular motion is centripetal acceleration. Centripetal acceleration is directed towards the center of the circular path of the object which is moving in circular motion.

The centripetal acceleration is calculated from ac = V2 / r where r is the radius of the circle and V is the magnitude of the velocity. Sometimes it is useful to use the rotation values to calculate the centripetal acceleration. You can substitute V = r * ω for V in the ac = V2 / r expression. When you simplify you find that
ac = r * ω2 . Take note that the acceleration is not tangent to the circle.

What happens to the value of the centripetal acceleration if the radius is held constant but the velocity (m/s) increases by a factor of 3 (e.g. from 8 m/s to 24 m/s)?
Answer: Think about the equation: ac = V2 / r . ac is proportional to the square of the velocity.
The value of the original ac is ac = (1 * V)2 / r or ac = 1 * V2 / r . If the velocity increases by a factor of 3 the value of ac is now ac = (3*V)2 / r or ac = 9 *V2 / r . ac increases by 32 or 9 times the original value for this example.


Centripetal Force
As you learned in Chapter 4, net external force causes acceleration. The centripetal acceleration, ac, is caused by the centripetal force, Fc. It is true that Fc = m * ac 
 or Fc = mV2 / r . Note that the centripetal force is perpendicular to the velocity. 
 Fc only changes the direction of the velocity, it does not change the magnitude (speed) of the velocity.

Examples of centripetal force include
1. Rubber stopper swung in a horizontal circle on the end of a string
The string tension supplies the centripetal force.

2. Car driving around a flat curve
Friction of the road on the tires supplies the centripetal force.

3. A satellite moving in a circular orbit around the Earth
The force of gravity of the Earth on the satellite supplies the centripetal force.

4. Clothes in a washer during the spin cycle
The wall of the rotating tub supplies an inward force on the clothes.
Why does some water leave the clothes during the spin cycle?
And, why does the tub of the washer have to spin so fast?
Answer: The water molecules need an inward centripetal force to move in a circle during the spin cycle. The adhesion force between the water molecules and the clothes cannot provide enough centripetal force when the velocity of spin is high. Recall: Fc = mV2 / r . The water tends to move in a straight line instead of a circle. The wall of the rotating tub in the washer constrains the water somewhat but the water flows along the wall of the tub and out through the holes on the wall of the tub. The clothes are too big (we hope) to fit through the holes in the wall of the tub and don’t leave the tub (except for an occasional sock !?).
The tub has to spin fast to make the required value of Fc larger than the available force from the adhesion of the water to the clothes. The water will remain on the clothes and move in a circle until the spin speed reaches a point where Fc = mV2 /r is larger than the adhesion force.

You should work through the examples of centripetal acceleration for a computer hard drive and the friction needed for a car on a curve shown in the image below.


Fictitious Forces and Non-inertial Frames: The Coriolis "Force"
If you make observations in a non-inertial reference frame you will think that fictitious forces exist. That is to say, you will observe that objects don’t obey Newton’s laws of motion unless you invent a force that is not really present. In a rotating system this fictitious force is sometimes identified as an outward, centrifugal force.

e.g. The driver of a car making a sharp right turn will think there is a force pushing the driver towards the door near the left arm. Actually, your body is trying to move in a straight line and the car is turning into your body.

e.g. All parts of the earth make one rotation in 24 hours. The Earth rotates towards the east (recall where the Sun rises!) . But, objects located closer to the Earth’s equator are farther from the rotation axis of the Earth that runs from the North Pole to the South Pole through the center of the Earth. Objects closer to the equator (as compared to objects in Nebraska, or Canada etc.) have a larger value for r in the equation V = r * ω and thus have a larger velocity in meters per second towards the east.

Imagine you are at the Earth's equator. You are shooting a long range artillery weapon that is aimed directly towards the north. Before the weapon is fired the artillery shell and weapon are moving toward the east due to the rotation of the Earth. The artillery shell will keep its large eastward velocity due to the rotation of the Earth as it is shot out the barrel of the weapon. As the shell moves towards the north, it is traveling above land that has a smaller velocity (meters/second) towards the east (recall V = r * ω ). The observer on Earth will see the artillery shell land east of the target because the shell is moving at a faster speed towards the east than is the land it flies over. The observer will invent a “Coriolis” force directed towards the east to explain why the shell hit east of the target. The “Coriolis” "force" is not a real force. The deviation of the shell toward the east is just a consequence of the rotating Earth. This same effect is responsible for the direction of rotation of high and low pressure weather systems.

Centrifugal "Force"
e.g. Imagine rolling a ball across a merry-go-round in the playground while you riding the merry-go-round. If friction is not a big effect the ball will travel in a straight line as viewed by an observer on a tree branch above the merry-go-round. But, the observer on the merry-go-round will see the ball move in a curve such that the ball moves towards the outer edge of the merry-go-round without passing through the center. The observer will claim that there is an outward centrifugal "force" that pushes the ball away from the center. The centrifugal “force” does not exist.
YouTube videos that explain the misconception of centrifugal "force."
Or you may do your own search for videos that state that centrifugal "force" is fictitious.

Here are quantities that are uniform (constant) in uniform circular motion: magnitude of velocity, magnitude of centripetal acceleration, and magnitude of centripetal force are constant. The full vectors for these items have a changing direction but have a constant length.
Other constant items are radius and angular velocity. (The angular velocity is a different sort of vector that will be discussed briefly in the unit on angular momentum.)

What force is necessary to keep the earth moving in a (nearly) circular orbit around the sun?
Answer: The force of gravity on the Earth due to the attraction to the Sun. This force is directed towards the Sun.


e.g. A Ferris wheel has a radius of 8 meters and rotates once in 15 seconds. What force does the seat of the Ferris wheel apply to a passenger who has a mass of 80 kg and is currently at the highest point in the ride?



Copyright© 2017 by Greg Clements Permission is granted to reproduce this document as long as 1) this copyright notice is included, 2) no charge of any kind is made, and, 3) the use is for an educational purpose. Editing of the document to suit your own class style and purposes is allowed.

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