Learn the idea: a short explanation, then a check.
We all know what a circle is, but what physics principles govern an object moving in a circular path?
How do we analyze and describe its motion as it travels in a circle? In this series of lessons on kinematics and dynamics, we will explore:
What circular motion is and how it is defined
How circular motion is measured mathematically
Real-world applications and complex examples
Centripetal force and the underlying causes of circular motion
Before you start, it would be helpful to have reviewed linear kinematics and the concept of force
Understanding Circular Motion
In this lesson we will examine the principles behind uniform circular motion. In other words, situations when the object undergoing circular motion is traveling at a constant speed. The path has a constant radius (r) and a Period (T). However, the velocity is not constant.
Although the magnitude of the velocity (which is the speed) is constant, its direction is constantly changing, since it is a vector.
This leads to an important point. Since the velocity is changing, it must be accelerating. Since it is accelerating, it must be experiencing a net force. It is this force, the centripetal force, that causes the object to undergoing circular motion.
The following video explores this in greater detail, including the formulas involved in the analysis of circular motion
In a rush? Need only a quick review? Watch this Otherwise continue on
📝Practise
Worked examples first, then problems to try, then worksheets.
Check Your Understanding: Circular Motion
1. An object is moving at a constant speed in a horizontal circle. Which of the following statements is true regarding its velocity and acceleration?
Velocity is a vector, so it changes whenever its direction changes, even if the speed stays the same. In a circle the direction of the velocity changes continuously, so the object is accelerating. This centripetal acceleration points towards the centre of the circle, at right angles to the velocity, and its size is a = v2/r.
Velocity is a vector, so a change in direction is a change in velocity even when the speed is constant. That rules out A and B: the velocity is not constant, and a changing velocity means there is an acceleration. In uniform circular motion that acceleration points towards the centre of the circle (it is the centripetal acceleration), not along the path. A tangential acceleration would change the speed, which does not happen here, so D is wrong. The size of the acceleration is constant (v2/r), but its direction keeps changing, so it is not constant either.
2. If a car travels around a circular track at a constant speed, what happens to the required centripetal force if the car’s speed is doubled while the radius stays the same?
The centripetal force needed is Fc = mv2/r, so it is proportional to the square of the speed. If v doubles, v2 becomes (2v)2 = 4v2. The mass and radius are unchanged, so the required force is 4 times as large.
The centripetal force needed is Fc = mv2/r, so it depends directly on the speed, which rules out A and C (a higher speed needs a larger force). The speed is squared, so doubling v multiplies v2 by 22 = 4. The force increases by a factor of 4, not 2, so B is wrong.
3. A ball attached to a string is swung in a clockwise horizontal circle. If the string breaks suddenly when the ball is at the top of the circle (the 12 o’clock position, viewed from above), in which direction will the ball travel immediately after the break?
In a clockwise circle the ball is moving to the right as it passes the top, so its velocity there is along the tangent, to the right. The tension in the string was the force supplying the centripetal force. When the string breaks there is no net force on the ball, so by Newton’s first law it keeps moving with the velocity it had: in a straight line along the tangent.
When the string breaks, the inward tension force disappears and there is no net force on the ball. By Newton’s first law it then moves in a straight line at constant velocity, which is along the tangent at the point of release (to the right at 12 o’clock for clockwise motion). There is no outward force pushing the ball away from the centre, so B is wrong. The inward force is exactly what was removed, so C is wrong. An object cannot keep following a curve once no force is acting on it, so D is wrong.
We are now ready to try some sample problems Below are sample problems with videos that explain how to solve them. It is suggested you try the problems beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Practice Problems: Circular Motion
Use v = 2πr / T, ac = v2 / r and Fc = mv2 / r. The centripetal force and acceleration point towards the centre of the circle, and the velocity is tangent to the path.
1. An ant is riding on the tip of a 10 cm long second hand of a clock. Find the linear velocity of the ant.
Known
r = 10 cm = 0.10 m
T = 60 s (the second hand makes one rotation in 60 s)
Unknown
v = ?
Formula
v = 2πr / T
Substitute and solve:v = (2 × π × 0.10) / 60 = 0.628 / 60 = 0.0105 m/s
Answer: 0.0105 m/s (1.05 cm/s), directed along the tangent to the circle
2. Calculate the centripetal force required to keep a 5 kg mass revolving in a circular path of radius 125 cm with a period of two seconds.
Known
m = 5 kg
r = 125 cm = 1.25 m
T = 2 s
Unknown
Fc = ?
Formula
v = 2πr / T
Fc = mv2 / r
Step 1 (find the speed):v = (2 × π × 1.25) / 2 = 1.25π = 3.927 m/s
Step 2 (find the force):Fc = (5 × 3.9272) / 1.25 = 77.1 / 1.25 = 61.7 N
Answer: 61.7 N, directed towards the centre of the circle
3. An astronaut is on a spinning space station so that he experiences artificial gravity, just like in the film 2001: A Space Odyssey. If the radius of the station is 1000 m, what must be the tangential velocity to simulate Earth’s gravity (g = 9.8 m/s2)?
Known
r = 1000 m
ac = g = 9.8 m/s2
Unknown
v = ?
Formula
ac = v2 / r
so v = √(acr)
Substitute and solve:v = √(9.8 × 1000) = √9800 = 99.0 m/s
Answer: 99 m/s
4. As you go around a corner at 72 km/h, you experience a force of 1120 N from the seat of your car. If your mass is 70 kg, what is the radius of the curve?
5. A centripetal force of 8.94 N is required to keep an object in a circular orbit. What is the mass of the object if it is travelling with a constant speed of 8.10 m/s and a radius of 5.29 m?
Known
Fc = 8.94 N
v = 8.10 m/s
r = 5.29 m
Unknown
m = ?
Formula
Fc = mv2 / r
so m = Fcr / v2
Substitute and solve:m = (8.94 × 5.29) / 8.102 = 47.29 / 65.61 = 0.721 kg
Answer: 0.721 kg
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.
Check in - Circular Motion : If you are using the video in a flipped classroom, this presentation can be used a check understanding after the students have watched the video for homework
