CIRCULAR
MOTION
NEWTON’S 1ST LAW
• In order to understand how and why objects travel in
circles, we need to look back at Newton’s 1st Law
• Objects in motion (and at a constant velocity) will stay in
motion (at a constant velocity) and objects at rest will stay
at rest unless acted upon by an outside, unbalanced force.
• Remember this is often called the “Law of Inertia”
• Objects naturally want to stay at rest or keep moving at a
constant velocity IN A STRAIGHT LINE
• If they are not moving in a straight line, it’s because
SOMETHING is forcing them to change.
BASIC PROBLEM
• Imagine the motion of a ball on a string being
swung in a horizontal circular motion above a
person’s head.
• Is the ball being swung at a constant speed
accelerating?
• YES!! Velocity is changing. Velocity is a
vector. There is a magnitude and a direction.
The direction is changing so it’s velocity is
changing!
• Acceleration will be caused by
•
•
•
A change in speed
A change in direction
A change in both speed and direction

Look at the below picture (snapshot of a swinging ball –
bird’s eye view) You can see the magnitude of the velocity
vector does not change, but direction does.
WHICH DIRECTION IS
ACCELERATION?
• If you have a toy car that travels only in
a straight line, what would you have to
do to make it travel in a circle?
• You would have to push the car INTO
a circular path. Imagine doing this
with little “nudges.” You would
continuously tap the side of the car
• Since the tapping will always be
directed to the center, this is the
direction of acceleration.
ANSWER: Inwards towards the center.
HOW CAN THIS ACCELERATION
BE CALCULATED?
• We can’t use a =(vf-vt)/t
• Ac = v2
R
• The variable Ac stands for centripetal acceleration.
Centripetal means “center-seeking”.
• The variable v is the velocity of the object that is moving
in the circular path.
• The variable R is the radius of the circular path.
EXAMPLES
1. Find the centripetal acceleration of a cat that travels
around a circular turn of radius 50 m if it travels at a
speed of 25 m/s. (12.5 m/s2 inward towards the center)
Ac =v2/r
Ac = 252/50
Ac = 12.5 m/s2 inward
EXAMPLES
2. Find the velocity of a ball on a string that is being swung in
a horizontal circle above a man’s head. If the circular path
has a radius of 1m and the centripetal acceleration is 100
m/s2, what is the velocity of the ball if the string snaps?
(what direction will the ball travel?)
Ac = v2/r
100 = v2/1
V2 = 100
V = 10 m/s in a straight line tangent to the circle (because of
intertia)
TWO ADDITIONAL CONCEPTS
Period of motion
• We will use the variable T to represent period.
• The period is the length of time that it takes to complete
one full circular revolution.
• A large period refers to a slow moving object
• A small period refers to a fast moving object
• We will use the equation d = rt. Since speed is constant,
this equation is appropriate.
• The distance is the circumference. Since circumference =
C = 2πR
• How fast will it be able to do this? Well, since the speed is
a constant, we will use the variable “v”.
• Therefore, d = rt or 2πR = vT. Rearranging the equation
gives
EXAMPLE
3. A man swings a ball in on a string in a horizontal circle
above his head. If the ball is traveling at a speed of 5 m/s and
the circle has a radius of 50 cm, find the time needed for the
ball to complete one full revolution. [.628 sec]
2r 2 (.50)


 .628 ms
v
5
EXAMPLE
A man holds his son by the arms and starts to spin him
around in a circle of radius 1.5m. If the man spins at a
constant velocity, and if he completes 5 full revolutions in 10
seconds, find
• the period of the boys motion. [2 sec]
• the speed of the little boy’s feet. [4.71 m/s]
• the speed of the little boy’s belt, which is only .75 m from the
center of the circle. [2.36 m/s]
10 sec sec

 2 rev  2 sec
5rev
2r 2 (1.5)
v

 4.71 ms

2
2r 2 (.75)
v

 2.36 ms

2
2ND CONCEPT
Frequency of motion
We could say that a fast moving ball completes a full circle
more frequently than a ball that is moving very slow. The
frequency depends on how fast the object moves around the
circle. We will measure frequency in “revolutions per
second” using the equation.
EXAMPLE
5. On a dare, a softball player pus her forehead on the end of
a bat and begins to spin around in a circle. Amazingly, she
maintains a constant velocity and competes 10 full turns in
12 seconds. Find:
• The period of the motion. [1.2 sec (per full turn)]
• The frequency of the motion. [0.833 turns per second]
12 sec

 1.2 sec
 1.2 sec
rev
10rev
1
1
f  
 .83Hz
 1.2 sec
The Equations of UNIFORM Circular Motion
f 
PERIOD (SEC):
2R 1
T

v
f
FREQUENCY (HZ):
TANGENTIAL VELOCITY (M/S):
1
T
v1
R
2R
v
T
v2
CENTRIPETAL ACCELERATION (M/S2):
v2
ac 
R
4 2 R
2
ac 

4

Rf
2
T
a tangential  0
2
If you see RADIANS  in the Holt Physics book, convert to
REVS by using the following conversion:
1 rad 
1
rev
2
ac
(because the
velocity is
constant.)
HORIZONTAL
CIRCLES
In all examples, the positive direction will be chosen
as “INTO THE CENTER” of the circle.
A BALL ON A STRING
SWINGING IN A
HORIZONTAL CIRCLE
(ON A
FRICTIONLESS TABLE)
T
R
m
T supplies the centripetal force that keeps the mass in its circular orbit
Fc  mac
T  Fc
T  mac
mv2
T
R
A CAR TRAVELING AROUND A
CIRCULAR TRACK (OR CURVE)
m
Ff
R
Friction supplies the centripetal force that keeps the car from slipping off
the track (curve)
Continued on
next page 
FBD for …. A car traveling around a
circular track (or curve)
The frictional force supplies the
centripetal force necessary to
keep the car in the circular path.
FN
The car “wants” to go this
way (because of its inertia,
it wants to continue in the
“straight line” path)
Ff
R
y
x
ac
mg
NEW FORMULAS FOR
FRICTION
Recall: F = ma from Newton’s 2nd Law.
The new formula we’ll be using is Fc = mac (with c meaning
centripetal)
•
Using our previous formula: Ac = v2
R
We will rearrange F = ma to Ac = Fc/m then substitute it
into the above formula.
Result: Fc = mv2/R
The next slide provides a summary of formula
substitutions
FN
FN  mg
F f  FN  mg
Fnet  ma
F f  mac
Ff
R
ac
mg
mv 2
F f  mac  mg  mac  mg 
R
v2
g 
 v  gR
R
This is the maximum velocity that a car can travel around the circular path
without slipping.
You can also use the following formula to find the minimum
coefficient of friction for no slip at a certain velocity.
v2

gR
A CAR TRAVELING
AROUND A CIRCULAR
TRACK WITH A BANKED
TURN
Continued on
next page 
VERTICAL CIRCLES
Vertical circles still use the same formulas as horizontal
circles.
period (sec): T = 2πR = 1
V
f
frequency (Hz): f = 1
T
velocity (m/s): v = 2πR
T
centripetal acceleration (m/s2):
centripetal force (N): Fc = mv2
R
Ac = v2 = 4 π2R
R
T2
HOWEVER….
Vertical circle problems also
include the effects of
GRAVITY!!
INTRO TO VERTICAL CIRCLES
•
All of the problems we looked at so far have dealt with
objects moving in horizontal circles, like a car going around a
turn.
•
In that case, the Centripetal Force is being supplied by the
friction between the tires and the road pushing the car into the
circle. In some cases, like a ball on the end of a string, the
force was from tension in the rope pulling the ball into the
circle. In both cases, SOMETHING has to provide the
centripetal force in order to make the object move in a circle.
FBD:
Since centripetal force must keep the
object IN the circle, we only cared about
FN
the forces that point INTO the circle
notice how weight and normal force
Ff
do not point in?
Thus they were never part
R
of our problems!
ac
mg
Imagine a scenario where a man is in the middle of the circle and
he’s constantly trying to push the car OUT of its circular path.
He’s not very strong, so he can’t make the car leave the circle, but
since he’s trying, we have to consider his pushing force in our
FBD.
FN
Again, the friction is supplying the
centripetal force, BUT the pushing
force is working against friction.
These must be subtracted in order
to find centripetal force.
Fc = Ff – Fpush
Ff
Fpush
mg
So…when multiple forces point into or out of the circle, we need
to find the net force.
The sum of the forces pointing into the circle MINUS the
sum of the forces pointing out of the circle provides
centripetal force.
Fc = Ins - OUTs

Figure B:
VERTICAL CIRCLE
• Look at the FBD above. The ball has 4 key points along its
path.
• At every point along the path, tension points inward.
• At every point along the path, weight (Fg) points downward.
• Remember, Fc = Ins – OUTs
• At the top of the path, there are two “INs”
• At the bottom of the path, there are two “OUTs”
EXAMPLE 1 : A BALL ON A
STRING SWINGING IN
VERTICAL CIRCLES
• What supplies the centripetal force
when the ball is at the top of its path?
• Tension & the balls weight (together
)
• What supplies the centripetal force
when the ball is at the bottom of its
path?
• Tension (with the ball weight acting
against it )
• What supplies the centripetal force
when the ball is at the side (3 O’Clock)
of its circular path?
• Tension alone (b/c the weight is not
pointing into the circle)
v
W
T
R
T
W
Forces pointing
in are positive
EXAMPLE 1 : A BALL ON A
STRING SWINGING IN
VERTICAL CIRCLES
v
W
T
R
Assuming a constant speed
At top:
T
Fc = T + W
mv2 = T + mg
R
At bottom:
W
If moving too fast, the string may break
due to too much tension 
Fc = T – W
If moving too slow, the tension will
2
mv = T – mg
become zero at the top (the ball will fall
R

mv2
mv2
T  mg 
mg 
vmin  gR
R
R
EXAMPLE #2: A BIKE OR
CAR GOING AROUND A
VERTICAL – CIRCLE LOOP
How is this different from a
ball on a string being swung
in vertical circles?
•
It’s the same problem, but
instead of tension
supplying the centripetal
force, the normal force
(from the track) is the
supplier.
Since the normal force is
what the road “feels” (and
thus the force that it pushes
back with), the normal force
is also the “APPARENT
WEIGHT” of the bike/car.
FN  Wapp
A BIKE OR CAR GOING
AROUND A VERTICALCIRCLE LOOP
W
FN
R
Assuming a constant speed:
At top:
Fc  FN  W
Fc  mac
2
mv
 FN  mg
R
FN
v
Forces pointing
in are positive.
At bottom:
Fc  FN  W
Fc  mac
mv 2
 FN  mg
R
W
If moving too slow, the car/bike will lose contact at the top and fall.
Before it separates from the track, FN goes to zero! A certain speed
must be maintained to avoid this.
0
mv2
mv2
FN  mg 
mg 
vcritical  gR
R
R
PRACTICE PROBLEMS
1. A 0.5 kg ball is swung on a 1 m rope in a vertical circle
with a constant velocity of 5 m/s.
a. Find the position of the ball (top, bottom, left, right) when
the tension is the greatest.
b. Find the position of the ball when the tension is the least.
c. Draw a FBD
d. Find maximum tension in the string.
2. A 0.2 kg ball is being swung on a 0.5 m rope in a vertical
circle. The rope can withstand up to 50 N of force.
a. Find the maximum speed of the ball before it breaks the
rope.
b. Find the minimum speed of the ball (critical velocity) so
that it doesn’t fall out of the circle at the top.
3. A 35 kg boy is swinging on a rope 7 m long. He passes
through the lowest position with a speed of 3 m/s. What is
the tension in the rope at that moment?
A 5 kg ball is swung on a 1 m long rope. The rope may be
assumed to be weightless. If the ball travels at a constant
velocity and takes 0.5 seconds to complete a full revolution,
find the tension in the rope at the top and bottom of the
swing.
“WORKED-OUT”
PROBLEMS
T  0.5 sec
At top:

v
Fc  T  W
2R 2 (1m)

 12.566 ms
T
0.5 sec
Fc  mac
mv2
T  mg 
R
5kg  (12.566 ms ) 2
mv 2
T
 mg 
 (5kg)(9.8 sm2 )  740.522 N
R
1m
At bottom:
Fc  T  W
Fc  mac
mv2
T  mg 
R
5kg  (12.566 ms ) 2
mv 2
T
 mg 
 (5kg)(9.8 sm2 )  838.522 N
R
1m
2) For the same 5 kg ball swung on the 1 m long rope, what
is slowest period that will allow the ball to maintain its
circular path?
“WORKED-OUT”
PROBLEMS
vcritical  gR 
2R
T
v

9.8 1m  3.130
m
s2
m
s
2R
2 (1m)
Tcritical 

 2 sec
m
vcritical 3.130 s
A man on a motorcycle (4800 N total weight) attempts to
complete a vertical circular loop of radius 3m. How fast must
he be driving in order to complete the loop (Give your answer in both
m/s & mph)? If he just exceeds this amount, what will be his
apparent weight at the bottom of the loop?
“WORKED-OUT”
PROBLEMS
vcritical  gR 
9.8 3m  5.422
m
s2
m
s
Since 9.8 sm2  32.2 sft2 , there must be 32.2/9.8 ft in 1 m.
Therefore, there are 3.286 ft in 1 m.
 3.286 ft   1mi   3600 sec 
  
5.422 ms  5.422 ms  
  
  12.148mph
1
m
5
,
280
ft
1
h

 

 
His apparent w eight at the bottom will be equal in magnitude to the normal force at the bottom.
N
m 2
( 94800
2 )(5.422 s )
mv 2
Therefore, FN 
 mg  .8 m / s
 4800 N  9,600 N
R
3m
The rider pulled 2 g’s!
(twice his weight)