Uniform Circular Motion
Centripetal force and acceleration
• What force must be applied to Helen to keep her
moving in a circle?
• How does it depend on the Helen’s radius r ?
 How does it depend
r
on Helen’s velocity v
 How does it depend
m
v
on Helen’s mass m?
On the next pass,
however, Helen failed
to clear the
mountains.
Uniform Circular Motion
Zitzewitz Section 6.2


Circular motion is the name for an object moving in a circle
at constant speed, such as a stone being swung in circles
at the end of a string or a fixed horse on a merry-go-round.
Are these objects accelerating if they are moving in a circle
at constant speed?
∆𝑣
∆𝑡

What is the definition of acceleration? 𝑎𝑎𝑣𝑔 =

But v is a vector quantity and that means speed and
direction are both parts of it.
We said the speed of a horse on a merry go round isn’t
changing, but what about its direction?
Since its direction is changing, therefore it is accelerating!!


Uniform Circular Motion
Uniform circular motion is defined as an
object moving at constant speed (note I
didn’t say constant velocity!) around a
circle with fixed radius.
 As the object moves around the circle,
the length of its position vector (the
radius) doesn’t change, but the
direction does. Same thing with its
velocity vector (constant speed).

Uniform Circular Motion
Δr is the resultant displacement
between r1 and r2
The velocity vector is
perpendicular to the
displacement vectors (r) and
tangent to the circle.
Uniform
Circular Motion
Δv
• Now let’s look what direction the object’s acceleration is in.
The diagram above shows two velocity vectors separated
by time Δt.
• Using vector addition, v1 + Δv = v2
• Recall that aaverage = Δv/Δt so that Δv vector when divided
by Δt gives the acceleration.
• Note that acceleration a, points the same direction as Δv,
towards the center of the circle = “centripetal
acceleration”
Centripetal Acceleration
• The velocity tangent line is
always a 90 degree angle to the
radius, so this angle is
preserved.
∆𝑟
∆𝑣
• So =
𝑟
𝑣
• Now if you divide both sides by
∆𝑟
∆𝑣
Δt:
=
𝑟∆𝑡
𝑣∆𝑡
• Since by definition: 𝑣 =
𝑎=
•
𝑣
𝑟
=
∆𝑣
∆𝑡
𝑎
𝑣
𝑜𝑟 𝒂 =
𝒗𝟐
𝒓
∆𝑟
∆𝑡
and
Centripetal Acceleration
How can we measure the speed of an object
moving in a circle?
 Measure T (period) which is the time for the
object to make one complete revolution around
the center.
 One complete revolution = circumference = 2𝜋𝑟



So 𝑣 =
∆𝑥
∆𝑡
=
And then 𝒂 =
2𝜋𝑟
𝑇𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑
𝒗𝟐
𝒓
=
𝟒𝝅𝟐 𝒓
𝑻𝟐
Centripetal Force
Because there is always an acceleration, a,
towards the center, there is also a force
directed inwards towards the center.
 Newton’s 2nd law for circular motion:
 ΣFnet = mac where ac = centripetal
acceleration. This force is called
centripetal force.
 Note that the centripetal acceleration
always goes on the right side of the
equation! (not as a force on the left)

UCM Example
Zitzewitz p.155 #2
A rubber stopper is being swung in a
horizontal circle on a string.
 m=13g, length of string = l = 0.93m,
T=1.18 sec, find centripetal force



𝒂𝒄 =
𝒗𝟐
𝒓
=
𝟒𝝅𝟐 𝒓
𝑻𝟐
=
𝟒𝝅𝟐
𝟏.𝟏𝟖𝟐
𝑭 = 𝒎𝒂𝒄 = . 𝟎𝟏𝟑 𝒌𝒈
. 𝟗𝟑 = 𝟐𝟔 𝒎/𝒔𝟐
𝒎
𝟐𝟔 𝟐
𝒔
= 𝟎. 𝟑𝟒 𝑵
Centrifugal Force (not real!)
• Let’s say you are in a car and you
go around this corner to the left
too fast. The passenger on the
right side might get thrown
against the right (or outer) door.
• Is there an outward force on the
passenger? Some people say it is
centrifugal force, pushing you
outward, but they are wrong –
there is no such thing, it is
fictitious.
Centrifugal force
The person being “forced” against
the outside door is simply following
Newton’s 1st law, saying an object
will continue moving with the same
velocity in the same direction until a
force acts to change it. The person
has no force acting to change his
motion until the seat belt and/or
door he’s leaning against do that
job.
Common misconception!
Vertical UCM
Vertical UCM




Newton’s 1st : An object will travel in a straight line
unless a force acts on it to change that.
In other words, in the absence of a modifying force, it
is not natural for an object to travel in a circular path.
So for a ball swung in vertical circles on a string, the
modifying force is the tension in the string that forces
it to move in a circle. It pulls the ball off its straight
path onto a circular path.
Centripetal force is therefore a pulling force which acts
towards the center of the circle.
Centrifugal force – not real

B
Centrifugal force, an outward pushing
force is a “phantom” force. Here’s
proof.
A
C
If you cut the string when the ball is
at A, the balls flies off in a direction
of B, not C. If there had been an
outward force, it would have gone
towards C (or between B and C) but
it doesn’t.
Vertical UCM
But, you say, when you whirl a ball around
vertically on a string, you can feel the string/ball
“pulling” on your hand outwards.
 What you ARE feeling is the resistance of the ball
to moving in a circular path. Remember at every
point, it wants to just go straight. To keep the
ball going in a circle, the person must constantly
PULL the ball towards the center.
 This creates the tension in the string which is the
centripetal force.

Demonstration – water in cup
on platform
Note that the water in the cup always stays
parallel to the platform, not to the ground like
most water does – why?
 This is because I’m creating a pseudo-gravity
greater than the Earth’s gravity.
 This pseudo gravity is always perpendicular to
the surface of the tray. What causes this?
 It’s the centripetal force being exerted by the
tension in the string.

Continued…



The centripetal force is directed to the center of the
circle at all times. Therefore the tray constantly
pushes on the cup towards the center of the circle at
all times.
Fg
Fc
ocean water
The water level in the ocean and the water level in
the cup are both perpendicular to the force.
But wait, those arrows look like the go in opposite
directions…..
Continued…
ocean water
Fg
Earth
Fc
Vertical circle
So really, they are both going in the same direction,
towards the center of rotation!!
Now let’s consider the top and
bottom of a vertical circle
top
top
T
mg
T
mg
Top
y
T
mg
At the top, we know net force
must act towards the center of the circle.
 The net force is the centripetal force.
 You never label Fc in a free body diagram, it comes
in as the mac part.
To calculate the minimum

𝐹 = 𝑇 + 𝑚𝑔 = 𝑚𝑎𝑐
velocity so the string won’t

𝑇 + 𝑚𝑔 =
So
𝑻=
𝒎𝒗𝟐
𝒓
𝑚𝑣 2
𝑟
sag, set T=0 and solve for v.
− 𝒎𝒈 that – sign is why tension is less
at the top of the swing
Bottom
• Now
y
𝑭 = 𝑻 − 𝒎𝒈 = 𝒎𝒂𝒄
𝑻 − 𝒎𝒈 =
• So
𝑻=
𝒎𝒗𝟐
𝒓
𝒎𝒗𝟐
𝒓
+ 𝒎𝒈
• Unlike the tension up at the top, the
tension at the bottom is now the weight
PLUS the centripetal force, so it feels
like “it’s pulling harder” or “it weighs
more”.
T
mg
Rotational Inertia Video from
Space

Ok, this isn’t exactly UCM, but I couldn’t
resist this really cool 1 minute video
anyway!

http://www.youtube.com/watch?v=fPI-

Here’s some Professor Lewin from MIT with 5’ video
on what centripetal acceleration looks like in space.

http://www.youtube.com/watch?v=O2SL5MBTK-Q