Uniform Circular Motion

A particle or an object is said to be in circular motion if it follows a path around the circle or a circular arc. If it has constant speed it is said to be in uniform motion. When both of these conditions are met then the object is in uniform circular motion.

The velocity and the acceleration of the object have constant value but they change direction. The picture below shows the relationship between acceleration and velocity.

The acceleration is always pointing towards the center of the circle. It is called centripetal acceleration. The velocity is a tangent to the circle on point where the object is.

Following the logic from linear motion we can find the equation for centripetal acceleration.

\(a= \frac {v^2}{R}\)

When going around the circle the objects travel the entire circumference. We can use this to find the time needed.

\(T=\frac {2*\pi*R}{v}\)

Each body that has acceleration has some sort of force acting on it. In this case we have centripetal force. We can use Newton's second law to calculate the force.

We have:


Replacing \(a\) we get:

\(F=\frac {m*v^2}{R}\)

Example 1

A military plane is flying with a constant speed of 200 m/s. Then it enters into the loop with a radius of 400 m. What is the value of centripetal acceleration of the plane?


v=200 m/s

R=400 m



\(a=\frac {v^2}{r}=\frac {(200\;m/s)^2}{400\;m}= 100\; m/s^2\)

This value is equal to 100/9.81=10.19g. This is a huge acceleration and if a pilot experiences it, he would die instantly. This is the reason why the special g-suite was invented to keep pilots safe.

Example 2.

The International space station is flying at 520 kilometers above Earth. It needs 90 minutes to make a trip around Earth. Assuming the path is circular calculate:

a. The radius of the circle

b. The total distance traveled in one day

c. The velocity of the station

d. The acceleration of the station

e. The centripetal force acting on the astronaut that has a mass of 80 kg


a. The station is circling at the height of 520 km above Earth. We know that the center of this circle is at Earth's center. The radius of Earth is \(6.37 \times 10^6\; m\). So the radius of the circle is

\(R=\text{radius of earth} + \text{height above earth}\)


b. To calculate the total trips traveled in one day first we need to calculate the number of trips in one day.

\(24\;h \times {60\;min\over 1\;h}= 1440\;minutes\)


Thus, the station completes 16 trips during one day.

Next, we need to calculate distance traveled during one trip. It is equal to the circumference of the circle that the station is on.

          \(d=2\pi R = 2\pi( 6.89\times10^6\;m)= 43,291,146\;m\)

Now we multiply that with number of the trips to get the total distance

\(D=16d=16 \times 43,291,146\;m= 692,658,336\;m\)

c. To calculate the velocity of the station we use the formula: \(v=\frac {d}{T}\)

where \(d\) is the distance traveled during one trip and \(T\) is the time needed for that trip.

\(T=90\;mins \times {60\;s \over 1\;min}=5400\;s\\ \)

Therefore, the speed is

\(v={43,291,146\;m \over 5400\;s}\\ v=8,016.9\;m/s\)

d. To calculate the acceleration of the station we use the formula: \(a=\frac{v^2}{R}\)

\(a=\frac {(8016.9\;m/s)^2}{6.89\times 10^6\;m} = 9.33\;m/s^2\)

e. Given m=80kg, the force can be solved by \(F=ma\)