Circular Motion

Circular Motion, as the title suggests, is the section where we look at things traveling in a circle. To figure out the speed or acceleration of these kinds of objects we use the standard velocity and acceleration formulas, but instead replace parts it with the formula for the circumference of a circle (2×piPi is that special number used to calculate measurements of circles. It is 3·141592...×radiusThe length from the center of the circle to the edge.).

There are three formulas we will use in this section:

As in previous sections we would select the formula that has all the figures we know, or want to know.

By the way, the r stands for the radius of the circle, and t stands for the time to make one full circuit.

And something else to remember: acceleration is always towards the center of the circular path, whereas velocity acts at right angles to the circle. More about that later.

For our first example we are going to take a coin, tie it to a piece of string, and whirl it around above our heads. If the string is 15cm long and it takes 2·0 seconds to go around once. How fast is the coin traveling?

Below is a diagram of the above question.

For this question we will need to use the third formula written above, and jam all the information we know into it. But first, we need to convert the 15cm into the proper SI unitsA list of these can be found in the Before We Begin section..

v = (2×3·142×r)÷t
v = (2×3·142×0·15)÷2
v = 0·47ms^-1 (2 s.f.Don't know what this is? Significant figures are explained in the Before We Begin section.)

Its that simple.

Now for an example that includes acceleration. Do you remember that on the acceleration page we said that "acceleration is a change in velocity, so it can come with a direction. If an object is traveling around a corner at a constant speed, it would be "accelerating" because even though the speed is the same, the velocity is changing"? This is a section where this rule will apply. Below is a diagram of the directions of the velocity and acceleration on a circular path.

Basically, these next examples involve objects traveling at a constant speed, but because they are going around in circles, their direction is changing, hence they are "accelerating". Using the formulas at the top of the page, we will find the centripetalA word that means "movement toward the center". acceleration.

Instead a coin on a string, lets find something a bit more applicable and take a field trip to the local playground. Do you remember those roundabout things that you would spin around on during primary school? Lets find one and take some measurements.

This particular roundabout is 3 meters across, and we'll spin it so that it makes 12 revolutions in a minute. What is the centripetal acceleration, assuming there is no friction on the roundabout?

There's a problem here: We don't know any values that we can put in any formula!

This is easily fixed. First, if a circle is 3 meters across, and the radius is half that length, what's the radius?

If you are reaching for a calculator, STOP! You're doing Year 12 physics and you need to use a calculator to figure out what half of three is? The shame.

If you still went through with using a calculator, you should now know that 3÷2 equals 1·5. Now you'd better remember that, you never know when that fact might save your life...

Second problem we need to fix. We found out that the roundabout is making 12 revolutions a minute, but we need to know how long it takes to make one revolution. This is calculated by taking the amount of time taken and dividing by the amount of revolutions made. In this case:

60÷12 = t
t = 5·0

So we know the radius and the time for one revolution. Now we'll just type that into our recently bought computer that can do all the working for us and get... an error.

Oops... I forgot that the second formula needs the velocity, not the radius. It looks like we'll need to use the third formula first. But because the computer is being stroppy, we'll have to do it by hand.

v = (2×3·142×r)÷t
v = (2×3·142×1·5)÷2
v = 9·4÷2
v = 4·7ms^-1 (2 s.f.)

At this point because we know the velocity we could use either formula, but we'll use the second formula this time.

a = (2×3·142×v)÷t
a = (2×3·142×4·7)÷2
a = 29·6÷2
a = 15ms^-2 (2 s.f.)

And because the centripetal acceleration is always to the center of the circle, we wont need to calculate any angles.

Now we have an opportunity to test the first formula. This time we're going on a very expensive school trip to London. Before I tell you what our final destination is, see if you can figure it out: to test the formula we'll need something that travels around in a circle. So, are we going to:

a) Buckingham Palace (For the "rotating" of the guards)?
b) Charing Cross Station (For getting around in the underground)?
c) Royal Albert Hall (Because it's round)?
d) The London Eye (Because it travels around in a circle)?
e) The M25 (A large, orbital motorway)?
f) Big Ben (The clock hands travel around in a circle)?

Answer: D! That wasn't too hard, was it?

Here's some information about the eye.

• It takes 30 minutes for one full trip, so it travels at around ·24ms^-1 (2 s.f.).
• Each of the 32 pods on the wheel holds up to 25 people.
• It is 135 meters tall.
• It was officially opened by Tony Blair on December 31, 1999.

That's enough information for now, because we can find out the centripetal acceleration if, say, it was knocked over and spinning around while on the ground...

All we need to do is extract the information from that list that tells us what the speed and radius is.

...it travels at around ·24ms^-1 (2 s.f.)
It is 135 meters tall

However, we've reached the problem we encountered in the previous example. We don't know the radius!

Ok everyone, thinking caps on. How do we get the radius?

135÷2 = r
r = 67·5m (3 s.f.)

And jamming all this into the first formula gives us...

a = v²÷r
a = ·24²÷67·5
a = ·0576÷67·5
a = ·00085ms^-2 (2 s.f.)

So because the wheel is moving so slowly, the centripetal acceleration is hardly anything!