Physics I
Class 06
Uniform Circular Motion
Rev. 09-Aug-03 GB
06-1
Newton’s Second Law Yet Another Review!
Newton’s Second Law:
 
 F  Fnet

 Fnet

 m a or a 
m
The net force and acceleration are always in the same direction because
m is a positive number.
Acceleration is the rate of any change in the velocity vector – either
magnitude (speed) or direction, or both.
Today’s lecture and activity will stretch our understanding of
acceleration and Newton’s Second Law for a special type of motion.
06-2
Important Facts About
Velocity and Acceleration Vectors
v
a
v
Same direction: speeding up.
a
v
a
Opposite directions: slowing down.
Right angles: changing direction, same speed.
06-3
Circular Motion
 When an object travels in a circle, its velocity is constantly
changing (in direction at least).
 That means the object has a non-zero acceleration even if it
moves at constant speed.
06-4
Uniform Circular Motion
“Uniform” circular motion means that the
object moves in a circle at a constant speed.
Some definitions and equations:
T = period = time to go around exactly once
r = radius of circle
v = speed (scalar, not vector)
circumfere nce 2  r
v

period
T
2r
T
v
06-5
What is the Direction of
Acceleration?
Since the speed is not changing, only the
direction of velocity, acceleration must be always
at right angles to velocity. The acceleration
vector points inward, toward the center of the
circle. This is called centripetal acceleration
from Latin for “to go to or seek the center.”
Like the direction of the velocity vector, the
direction of centripetal acceleration is constantly
changing as the object moves around the circle.
v2
The magnitude of centripetal acceleration is given by a 
.
r
06-6
Centripetal Force
 

Since  F  Fnet  m a , the net force on any object in uniform
circular motion must be given by

v2
Fnet  m a  m
r
and its direction is the same as acceleration: toward the center.
The net force in this case is called the centripetal force. It is not a
separate physical force in its own right, but only a name that we
give to the total or net force on an object in uniform circular
motion. It may, in fact, be the sum of several forces of several
different types.
06-7
Example:
A Rock on a String
Twirl a 1 kg rock attached to a string in a 1 m radius vertical
circle. The speed is 4 m/sec. What forces act on the rock and
what are the directions of those forces?
06-8
Using Newton’s Second Law to
Solve Problems - Review
1.
2.
3.
4.
5.
6.
Identify all forces acting on the object.
Pushes or Pulls
Friction (if specified)
Gravity
Normal (Surface) Forces
Choose a coordinate system.
If you know the direction of acceleration, one
coordinate axis should be in that direction.
Draw a “Free-Body Diagram.”
We know how to do this now.
Express the force vectors in components.
This may require trigonometry.
Use Newton’s Second Law to write one
equation for each direction considered.
We will only consider vertical forces today.
Solve the equation(s).
06-9
Case A:
Rock at the Top of the Circle
The center of the circle is below the rock, so acceleration is down.
1.
2.
3.
4.
5.
Forces: Weight (W) down and Tension (T) down.
Coordinates: +X down. (Why?)
Free-body diagram:
X Components: (W) and (T).
Second Law: (W) + (T) = m a.
 v2

 g
6. Solve: T  m a  W  m a  m g  m
 r

a
T
X
W = mg
T = 1 (16/1–9.8) = 6.2 N.
W = 9.8 N.
06-10
Case B:
Rock at the Bottom of the Circle
The center of the circle is above the rock, so acceleration is up.
1.
2.
3.
4.
5.
Forces: Weight (W) down and Tension (T) up.
Coordinates: +X up. (Why?)
Free-body diagram:
X Components: (–W) and (T).
Second Law: (–W) + (T) = m a.
 v2

 g
6. Solve: T  m a  W  m a  m g  m
 r

T = 1 (16/1+9.8) = 25.8 N.
W = 9.8 N.
X
a
T
W = mg
06-11
Normal Force - A Concept
We Will Use in Today’s Activity
W
N
Floor
W
N
Elevator Cab
“Normal” force is the force generated by a solid
object to keep other objects from penetrating into it.
As the name implies, the direction of this force is at
right angles (“normal”) to the surface. The physical
cause of this force is the stretching of chemical
bonds, much like the stretching of a lattice of springs.
Normal force is often equal and opposite to weight,
but not always. Consider an elevator cab. How does
the normal force compare to weight if the cab is
moving at a constant velocity? Accelerating upward?
Accelerating downward?
06-12
Discussion
Swinging a Water Bucket Overhead
Take a sturdy water bucket ½ full of water and swing
it in a circle over your head. What will the water do
if
1.
2.
3.
you swing it quickly?
you swing it slowly?
you stop it when it is directly overhead?
Can we relate the concepts and math formulas we
have talked about today to the transition between
situation #1 and situation #2?
06-13
Class #6
Problems of the Day
_____1. A stone is twirled at constant speed in a vertical circle at the end of a string.
Which of the following is a correct statement of the separate physical forces
(and their directions) acting on the stone when it is at its highest point?
A)
B)
C)
D)
E)
F)
There are two forces: tension (down) and weight (down).
There are two forces: tension (up) and weight (down).
There are three forces: tension (down), weight (down), and centripetal force (down)
There are three forces: tension (up), weight (down), and centripetal force (down)
More than one of the above is correct.
None of the above is correct.
06-14
Answer to Problem 1 for Class #6
The answer is A.
Strings cannot push, they can only pull.
Weight (gravity) is always down.
Centripetal force is not a separate physical force, it is only a name
we give the net force on any object in uniform circular motion. The
net force can be the sum of more than one different physical force.
In this case, it is the sum of tension from the string plus weight.
06-15
Class #6
Problems of the Day
2. You take your little sister to the park and go for a ride on the park’s
carousel (“Merry-Go-Round”). You warn her to hang on tightly to the
carousel horse so that she does not slide off. (You stand next to her on the
platform just to make sure.) The horse is 4 m from the axis of rotation of
the carousel and the carousel takes 8 seconds to make one revolution at
full speed. Your sister’s mass is 20 kg. What is the centripetal force
needed to keep her on the horse?
06-16
Answer to Problem 2 for Class #6
The first step is to determine the speed of the carousel at the location of your sister.
The radius is 4 m, so the circumference at that point is 8  m or 25.1 m. The time for
one revolution is 8 s, so the speed is 8  / 8 =  or 3.14 m/s.
Then the centripetal force is m v2 / r = 20 * 2 / 4 = 49.3 N.
Because your sister is moving in a circle at a constant speed, we know the net force on
her. However, we don’t know most of the individual physical forces that make up this
total. These include her weight (which we know = 196 N down), the normal force
from sitting on the horse, the friction from sitting on the horse, and the force of her
hands holding on.
06-17
Class #6
Take-Away Concepts
1.
2.
Acceleration (or net force) at a right angle to velocity causes a change
of direction but not a change of speed.
As an object moves around a circle at a constant speed, it accelerates
toward the center with magnitude given by
v2
a
r


3. By Newton’s Second Law, Fnet  m a , the magnitude of the net force
on such an object must be given by

v2
Fnet  m a  m
r
4.
This net force is called centripetal force. It is not a separate physical
force but a name that we give the net force in this situation.
06-18
Activity #6
Ferris Wheel Thrill Ride
Objectives of the Activity:
1.
2.
3.
More experience with VideoPoint.
Investigate uniform circular motion –
Is the acceleration really directed toward the center?
What is the subjective experience of traveling in a rapid
vertical circular motion?
06-19
Class #6 Optional Material
“Centrifugal” Force?
Turn this way.
Feel a force
this way.
Centrifugal is from Latin for “to flee from the center.”
Is centrifugal force a “real” force?
06-20
Accelerated Frames of
Reference
Newton’s Second Law applies to an inertial reference frame, meaning a
reference system for measuring position and time that is not accelerating.
If we wish to use Newton’s Second Law in an accelerating reference frame,
we need to add extra terms to the equation that can be considered as forces
operating on every object that we track using the accelerating reference
frame. These are commonly called inertial forces.
Why would we do a crazy thing like using an accelerated reference frame
instead of an inertial reference frame? In a way, it is built into human
nature to view ourselves as sitting still while the rest of the universe zips by.
It is often a convenient way to calculate things as long as we are careful.
06-21
The Inertial Forces
 
  
 
 

F  m A  F  m a 0  m     R   2 m   V  m   R

#1
#2
#3
#4
Don’t worry about understanding the details of this equation. The important
thing is that each term represents a different type of inertial force.
1.
2.
3.
4.
This term is due to linear (in a straight line) acceleration of the reference
frame. For example, in a car if you slam on the brakes (acceleration to the
rear), it feels like everything in the car is thrown forward.
Centrifugal force – this term is due to rotation of the reference frame.
Coriolis force – this term is due to moving in a rotating reference frame.
If you ever tried to walk down the aisle of a train as it rounded a curve,
you experienced this force.
This term is due to acceleration of the rotation of a reference frame. If you
are riding on a merry-go-round, you need to hold on tighter as it starts up.
06-22