Circular Motion

From ancient times circular trajectories have occupied a special place in our model of the

Universe. Although these orbits have been replaced by the more general elliptical geometry,

circular motion is still an important special motion to understand. The simplicity of the shape

sometimes hides the complexity of the motion.

We start by recognizing that if you choose the simplest origin for the coordinate system, the

center of the circle, that the motion can constrained by the condition r const . This is a true

statement at all times, but it is not a full description of the trajectory. For that we need to know

more about the motion. The general case is that the speed of the particle is not constant.

However, let’s start by considering the special case of uniform circular motion, i.e. the speed of

the particle does not change. It is important to notice that we said the “speed” and not the

“velocity”, for the former is constant, while the latter is not.

The motion made by the hands of a clock is one example of uniform circular motion. If we

consider the particle at the tip of the hand, then that particle repeatedly traces out a circle with a

constant period. Its speed is constant, but it is important to note that the speed does depend of

the radius, so other particles in the clock’s hand will also have constant speeds, but the speeds

will differ from each other depending on the particle’s radius. The quantity that is constant for

all particles is the angular velocity i.e. the rate of angular change of the particles. A diagram

will be helpful.

r

is the angle that the radius vector makes with some reference axis. As the particle rotates

around the center this angle changes. The rate of change of this angle is the angular velocity, .

Mathematically this is written as d

dt

and is constant for uniform circular motion.

Now with a well defined complete description of this special motion we can write down the

trajectory by inspection:

( ) cos( ) sin( )r t r t x r t y where ,x y are orthogonal unit vectors, and sets the

initial position at time 0t . The first thing that you might like to check is that the magnitude of

( )r t is constant. We know how to find the length of the radius vector by ( ) ( ) ( )r t r t r t

which indeed equals the constant r . Since we have an explicit expression for the position as a

function of time, it is easy to find the velocity and acceleration by taking derivatives with respect

to time.

The velocity is ( ) sin( ) cos( )v t r t x r t y . The magnitude of the velocity is the

speed and it is a constant ( )v t r . Notice that as mentioned above, a particle with a larger

radius has a proportionately larger speed.

Taking the derivative we find the acceleration 2 2( ) cos( ) sin( )a t r t x r t y .

Again, this is a vector with a constant magnitude 2( )a t r .

We have discussed the magnitudes of the velocity and acceleration, but not the directions. It is

very important to understand that these are vector quantities. Just because the speed of the

particle is constant does NOT imply that the acceleration is zero.

So what direction is the velocity in? Well, it must be tangent to the circle. If it were not, then

there would be a radial component to the velocity and the radius would change. How can we

check this? We just noted that ( )v t must be perpendicular to ( )r t at all times. Can we check

this? Yes, by taking the dot product and seeing if it is zero at all times. Please find ( ) ( )v t r t and

show that it is indeed always zero. So let’s assume that the rotation is counterclockwise (>0)

and then the diagram looks like this.

The velocity is tangent to the trajectory.

What about the acceleration, what direction is that in? This can be seen by noting the

relationship between the radius and acceleration:

( ) cos( ) sin( )r t r t x r t y

2 2( ) cos( ) sin( )a t r t x r t y

See the relationship? 2( ) ( )a t r t The acceleration is in the opposite direction to the radial

vector. The constant of proportionality is 2 . With this the diagram becomes:

r

v

a

r

v

This is a complete description of the kinematics of uniform circular motion, as we have found

the trajectory, velocity and acceleration as a function of time.

What about the dynamics, in other words what about the forces involved? We have treated the

problem very generally and have not specified any particular forces, however we can already

make a very strong and important statement: whatever the forces, their sum, the net force,

must be equal to the mass times the acceleration or 2( ) ( ) ( )netF t ma t m r t . This is a

very powerful statement, for an object to be in uniform circular motion there must be a very

particular force on the particle. This puts tremendous constraints on particles in circular motion.

Turning the statement around if the net force on a particle is not equal to 2 ( )m r t , then the

particle is not moving in a uniform circular trajectory about the origin.

For some of you this formulation of the force for circular motion will look different from what

you learned in high school. You may have learned that 2

net

vF m

r directed radially inward.

This is simply an equivalent statement because 2 2( )v r so the net force can be rewritten as

2

net

vF m r

r or 2mr r . Whatever form you prefer, there must be a net force on the particle

equal to this magnitude and directed radially inward.

Turning to nonuniform circular motion, this means that there is a component of the acceleration

parallel to the velocity and therefore the particle speed is not constant. An appropriate diagram

looks like:

r

v

a

The acceleration can be broken up into two components. One component will be radial or

perpendicular to the velocity and the other will be tangential or parallel to the velocity. The

velocity itself must still be tangential or else the radius would not be constant.

The parallel acceleration changes the rotational speed of the particle. In the diagram above the

particle is slowing its rotational speed. What about the perpendicular or radial component of the

acceleration? It must be the same magnitude as before 2

2

perpendicular

va r r r

r . What is

different is that this is no longer a constant. Since the instantaneous and v are changing then

the radial acceleration must also change to maintain circular motion.

Of course this means that the radial component of the force on the particle must change to

produce this result. The classic example of nonuniform circular motion is an object rotating in a

vertical circle in a gravitational field. Two examples are a bucket being swung around vertically

on a rope, or a toy car doing a loop-to-loop. Let’s draw the free-body diagram for an example

such as this, specifically the swinging bucket.

r

v

a a

parallela

perpendiculara

gF

ropeT

There are two forces on the bucket, gravity and the tension due to the rope. We can resolve these

forces into radial and tangential components.

Algebraically ,tangential cos( )gF mg and ,radial sin( )gF mg . The tangential force component

changes the angular velocity, while the radial force component changes the direction of the

velocity just so that a circular trajectory is followed. This means that the total radial force must

have the correct magnitude for circular motion or , ,radialnet radial rope gF T F must equal 2v

m rr

or

equivalently 2mr r . This is a very important point: the sum of the radial forces on an

object in circular motion must be equal to 2v

m rr

.

As the object rotates around the orbit its velocity will change, being smallest at the top and

highest at the bottom. This then demands that the radial force must change as the object is

rotating. This is accomplished by the tension in the rope varying as the bucket goes around. The

tension is highest as the bucket swings through the bottom of its arc and lowest at the top. You

should note that the free-body diagram will look different as the angle changes. If would be

good practice to draw the free-body diagram for the bucket at several positions around the circle.

Then calculate what the tension in the rope must at that time.

I have written a code to simulate the motion of the swinging

bucket. The first parameters that I use are: mass of bucket =

1kg, length of rope = 1 m, initial position was down( 0 ),

initial velocity = 1m/s ( 1 rad/sec ). The image to the right

shows the trajectory in white and the tension vector in blue.

ropeT ,tangentialgF

,radialgF

Here is the angle verses time curve:

This looks a lot like a sine function with a period of ~2 sec and an amplitude of 0.3 radians

(~18°). The angular velocity curve looks similar:

Note the phase difference between the two. When the angle is zero the magnitude of the angular

velocity is greatest. This last plot showed the tension in the rope as a function of angle.

The positive direction is defined as radially inward. If the bucket was just hanging and not

swinging the tension would be 9.8( )mg N . However, due to the need to accelerate the bucket

radially inward to have it execute circular motion, the tension is larger than the resting value.

A number of things change as the bucket is given a larger

initial velocity, 4m/s, see figure on right. Clearly the

amplitude of oscillation is larger (1.3 radians ~78°) and you

can see from the graph below that the period has increased to

2.25 sec.

The angular velocity is beginning to look less sinusoidal and a bit more like a triangle wave.

The most obvious change is that the tension now varies over a much larger range. There are

times ( 0) when the tension is almost triple the weight and other times ( 75 ) when the

tension is less than one third the weight.

Increasing the initial velocity to 6 m/sec yields a completely different type of behavior. Now

clearly no longer is the motion sinusoidal. The period has increased to over 3 seconds

The angular velocity is starting to have inflection points.

But the most important difference is observed in the tension of the rope: it goes negative!

The image on the right

shows a positive

tension, i.e. radially

inward. The image on

the left shows a

negative tension,

pointing radially

outward. This cannot

happen with a rope.

The bucket cannot

have this trajectory, it

would have fallen.

Does this mean that such motion can never occur? No, but

it cannot happen with a rope. You need to use a rod or

other tether that can apply both positive and negative

forces on the bucket. If you use a stiff rod you can find a

starting condition that allows the mass to just reach the

top and almost stop. The trajectory looks like this;

however there are two VERY different situations. The

first is that the mass does not quite make it to the top and

the second is that the mass goes over the top and around

in a complete orbit. The next six plots show the two

different conditions. For the first three plots the initial

velocity was 6.245 m/s, while for the last three graphs the

initial velocity was 6.276 m/s. This small difference

(~1%) in the initial velocities produced a very large change in the trajectories.

With the lower initial velocity the angle vs. time curve shows a very asymmetric shape. The flat

tops are where the mass remains near the top of the trajectory for a long time. You can see that

the angular displacement is bounded by . The angular velocity crosses zero and changes sign

as the particle approaches the top and reverses direction to fall back down over the same side.

The tension still displays the negative regions and so this behaviour cannot be observed by using

a rope.

Turning to the next initial condition with a 1% larger starting velocity, below are the results from

that simulation. The first thing to notice is that the angle is no longer bounded by , but will

now ever increase as the mass rotates counterclockwise around and around. The low slope

regions are where the mass is near the top and the angle changes very slowly. This is easily seen

in the next graph of angular velocity vs. time.

In the angular velocity vs. time curve you can see where it reaches a low value and stays there

for some time. However in contrast to before the velocity never reaches zero or changes sign.

Although the tension graph looks quite different from before, actually little has changed, only the

range of the abscissa. The tension values are essentially identical for the two situations.

These two cases are important to understand. They show a large change in behavior with a very

small change in initial conditions. This is not the typical circumstance that we have investigated

before in class, but we will see it again by the end of the term.

There is just one last condition that we need to discuss. In the last two situations we still had a

negative tension. However, we know that if we swing a bucket fast enough we can get it to go

around. There is a critical initial condition that must be met for this to occur. You can see it

from the free-body diagram of the bucket at the top of the orbit. The ropeT must be zero or

greater, the limiting condition is that the tension is zero.

Therefore the minimum radially inward force would equal mg , but we know that this must equal

22v

m mrr

so this set a minimum velocity for the particle to have at the top. To obtain this

velocity at the top, the mass must have sufficient initial velocity. Starting the mass at the bottom,

the critical initial value for 5g

r or 5 7 /v gr m s . (We will later see how to derive this

analytically when we examine energy conservation.) This is when the tension will be zero and

the top. For larger initial angular velocities the tension will be greater than zero. But this is fine

for a rope. Below are the orbital graphs for this condition.

The displacement angle is steadily increasing and the angular velocity shows oscillations as the

mass accelerates towards the bottom and decelerates towards the top.

ropeT

gF

The tension vs. angle graph shows that indeed the tension is always positive. When the bucket is

at the top of the orbit the tension goes to zero, there is a slack rope. However, when the mass

rounds the bottom the tension is six times the weight of the bucket.

Circular motion is very common. It is important that you understand how it arises and what is

special about the forces that produce it.

Top 180°

Important Note

You may have noticed that in these notes the words “centripetal force” or “centripetal

acceleration” have not yet appeared. This is because I really don’t want you to use them. The

forces that produce circular motion are real forces, i.e. gravitational, spring, tension, contact,

magnetic. There is no force in nature named the centripetal force, no particle generates it, nor

any field carries it. This is just a name given to the radial component of the net sum of the

forces on an object that is moving in a circle, therefore its magnitude must be 2

2 or v

m mrr

.

Many students want to put a “centripetal force” on the mass to make it move in a circle. Please

do not do this! Put a gravitational force or a tension or a normal force, put a real force on the

object. If you have to use the word “centripetal” make sure you know what you are doing and

just don’t let me hear it.

Appendix, proof of acceleration components for nonuniform circular motion

Let ( )t describe the angular variation of the particle with time. In the case of uniform circular

motion this would be ( )t t with being a constant. We want to consider the more general

case with an arbitrary ( )t . Now ( )d

tdt

. The position of the particle is given by

( ) cos( ( )) sin( ( ))r t r t x r t y Differentiating this with respect to time yields the velocity

( ) ( sin( ( )) cos( ( )) ) ( )d

v t r t x t y r t vdt

Taking another time derivative

22

2( ) ( sin( ( )) cos( ( )) ) (cos( ( )) sin( ( )) )

d da t r t x t y r t x t y

dtdt

or

2( )d

a t r v r rdt

. It is customary to denote the angular acceleration

( )( )

d tt

dt

And so the total acceleration equals

2( ) ( ) ( )a t r t v r t r

This is a simple expression to understand, the first term gives the acceleration parallel to the

velocity and the second term give the radially inward acceleration needed to move in a circle.

Defining the parallel and perpendicular accelerations yields the total acceleration

a a a as the sum of the parallel a r v and perpendicular 2a r r accelerations.