 https://www.youtube.com/watch?v=URn-wonYoPQ
Uniform Circular Motion
An object moves at uniform
speed in a circle of constant
radius.

Uniform circular motion is
accelerated motion. Why?
What do the words centripetal and centrifugal mean?
Discuss with 1-2 people near you.
 Centripetal (from Latin centrum "center" and petere
"to seek") means moving or tending to move toward a
center
 Centrifugal (from Latin centrum, meaning "center",
and fugere, meaning "to flee") means moving or
tending to move away from a center.
Centripetal forces or Centrifugal forces?
One of these exists, one does not. Which is which?
Choose and defend your answer on your whiteboard in
a short paragraph.
Centrifugal Force
It’s a myth!
We need to go back to Newton’s
Laws to properly explain the
feeling you get on a merry-goround or in a turning car.
When a car accelerates
You, as a passenger, feel as if you
are flung backward.
Your inertia (mass) resists
acceleration.
You are NOT flung backward.
Your body wants to remain at
rest as the car accelerates
forward.
When a car decelerates
You, as a passenger, feel as if you are
flung forward.
Your inertia (mass) resists the
negative acceleration.
You are NOT flung forward. Your
body wants to remain in motion at
constant velocity as the car
accelerates backwards.
When a car turns
 You feel as if you are flung to
the outside. Your inertia resists
acceleration.
 You are not flung out, your
body simply wants to keep
moving in straight line motion!
As a general rule
 Whenever you feel you are flung in a
certain direction, you can bet the
acceleration is pointing in the opposite
direction.
 Remember the elevator problems?
When you feel you are flying up,
acceleration of the elevator is down.
When you feel you are sinking down,
acceleration is up.
Acceleration in Uniform Circular
Motion
The acceleration responsible
for uniform circular motion is
referred to as centripetal
acceleration.
Centripetal Accelerationv
•𝑎𝑐 =
2
v
𝑟
ac: centripetal
acceleration in m/s2
v: tangential speed in m/s
r: radius in meters
a
c
v ac
Centripetal acceleration always points
toward center of circle!
ac
v
A rotating object has a linear speed of 1.5 m/s. It undergoes a
centripetal acceleration of 3.6 m/s2. What is the radius of the
mass's circular motion?
2
v
ac 
r
𝑟=
𝑚 2
(1.5 )
𝑠
𝑚
3.6 2
𝑠
r = 0.63 m
2
v
r
ac
 Get a sheet of paper out, talk with a partner and write
down your thoughts on the problem below. – 2.5
minutes
Force in Uniform Circular Motion
A force responsible for uniform
circular motion is referred to as a
centripetal force.
Centripetal force is simply mass
times centripetal acceleration.
Fc = mac
Centripetal Force
• Fc = m ac
•𝐹𝑐 = 𝑚
v
Fc
2
v
𝑟
centripetal force in N
v: tangential speed in
m/s
r: radius in meters
v
Fc
Fc
v
Always toward
center of circle!
More on Centripetal Force
 Centripetal force is not a unique type
of force.
 Centripetal forces always arise from other
forces.
 You can always identify the real force
which is causing the centripetal
acceleration.
 Nearly any kind of force can act as a
centripetal force, it just has to push or
pull towards the center
Friction as centripetal force
As a car makes a turn,
the force of friction
acting upon the turned
wheels of the car
provide the centripetal
force required for
circular motion.
A car is traveling at a constant speed and makes a turn with a radius of 50.0 m.
Its speed is 15.0 m/s. Find the minimum coefficient of friction needed
to keep the car traveling along the path.
The frictional force must equal the centripetal force.
The centripetal force is given by:
mv 2
FC 
r
We know that the centripetal force is the frictional force of
the tires and road. We also know that the frictional force is:
f s  s N
Assume the road is flat, so n = mg
Plug the formula for Frictional force in the place of FC in the original formula
and solve for the coefficient of friction:
v2
s 
gr
mv 2
 s mg 
r
2
m
1

 s  15.0 
s  
m 

9.8
50.0 m

2
s 


0.459
Normal force as centripetal force
An automobile turning
on a banked curve uses
the normal force to
provide the necessary
centripetal force.
Tension as centripetal force
As a bucket of water
is tied to a string and
spun in a circle, the
force of tension
acting upon the
bucket provides the
centripetal force
required for circular
motion.
1.2 kg stone is attached to a 1.3 m line and swung in a circle. If it
has a linear speed of 13 m/s, what is the centripetal force?
Fc = mac
mv 2
FC 
r
 m
1.2 kg 13 
s


1.3 m
2

156 N
Suppose another stone of mass M was swung in a circle of
radius R with a linear speed of S by a force of F. What
would be the resulting force if the radius was increased by
a factor of 3?
𝐹𝑐 =
𝑚𝑣 2
𝑟
𝐹 =
?? =
=
𝑀𝑆 2
𝑅
𝑀𝑆 2
3𝑅
𝐹
3
A child twirls a yo yo. If angle of the cord with the
T
vertical is 30.0, find ac. Notice that our tension0 is at an
angle…that cues us in that we need
to be thinking
about x and y
mg
T cos  mgcomponents!!
0 T 
cos
 Fy = 0
0
mg
The horizontal component of T is the centripetal force.
FC  T sin 
Fc
Plug into equation for T:
FC 
mg
sin 
cos
We also know that:
maC  mg tan 
m
aC  9.8 2 tan 30.0
s
FC  mg tan 
FC  maC
aC  g tan 
aC 
m
5.66 2
s
Centripetal Force can do no work
A centripetal force alters the
direction of the object without
altering its speed.
Since speed remains constant,
kinetic energy remains constant, and
work is zero.
Gravity as centripetal force
As the moon orbits the
Earth, the force of gravity
acting upon the moon
provides the centripetal
force required for circular
motion.
Newton’s Universal Gravitation
Gm1m2
F
2
r
G = 6.67 x 10-11 Nm2/kg2 (Universal gravitational constant)
Acceleration and distance
Kepler’s Laws
1. Planets orbit the sun in elliptical orbits.
2. Planets orbiting the sun carve out equal
area triangles in equal times.
3. The planet’s year is related to its distance
from the sun in a predictable way.
Kepler’s
Laws
Satellites
From geometry, we can calculate orbital
speed at any altitude.
Orbital Motion
• Gmems /
2
r =m
2
v /
r=
• The mass of the orbiting body
does not affect the orbital motion!
e
Consider the see saw
Consider the see saw
Consider the see saw
Consider the see saw
The weight of each child is a
downward force that causes the
see saw to twist.
The force is more effective at
causing the twist if it is greater
OR if it is further from the point
of rotation.
Consider the see saw
The twisting force, coupled
with the distance from the
point of rotation is called a
torque.
What is Torque?
Torque is a “twist” (whereas
force is a push or pull).
Torque is called “moment” by
engineers.
The larger the torque, the more
easily it causes a system to twist.
Torque
Consider a beam connected to a wall by a hinge.
Now consider a
force F on the
beam that is
applied a distance r
from the hinge.
Hinge (rotates)
r
Direction of
rotation
F
What happens? A rotation occurs due to the
combination of r and F. In this case, the
direction is clockwise.
Torque
If we know the angle the force acts at, we
can calculate torque!
Hinge (rotates)
r
 = F r sin 
 is torque
is force
Direction of
rotation
r is “moment arm”
 is angle between F and r

F
Torque equation: simplified
If  is 90o…
 = F r
 is torque
F is force
r is “moment arm
Hinge: rotates
r
Direction of
rotation
 F
We use torque every day
 Consider the door to the classroom. We use torque to
open it.
 Identify the following:
 The point of rotation.
 The moment arm.
 The point of application of force.
Question
Why is the doorknob far from the
hinges of the door? Why is it not in
the middle of the door? Or near the
hinges?
Torque Units
 What are the SI units for torque?
 mN or Nm.
 Can you substitute Joule for Nm?
 No. Even though a Joule is a Nm, it is a
scalar. Torque is a vector and cannot be
ascribed energy units.
Now consider a balanced
situation
40 kg
40 kg
If the weights are equal, and the moment arms are equal,
then the clockwise and counterclockwise torques are equal
and no net rotation will occur. The kids can balance!
Now consider a balanced
situation
40 kg
40 kg
 ccw = cw
 This is called rotational equilibrium!