In this post we’ll look at circular motion in situations involving multiple forces.
- Concepts you need to know
- Circular Motion with 1 force
- Horizontal Circles
- Stone on a string
- Airplanes
- Cars on a curved road
- Vertical Circles
- Stone on a string
- Rollercoasters
- Ferris Wheels
- Cars on bumps
Before solving problems, you must keep a few concepts in mind.
Concepts you need to know:
- IF an object is moving in a circle at a constant speed, THEN there is a constant centripetal acceleration, & thus constant centripetal force
- Centripetal force = NET force acting on an object towards the centre of a circle
- If the object only moves in a circle and does not accelerate in any other way, the sum of forces in any direction EXCEPT centripetally MUST be 0
Really internalise that:
ΣF = Fc
IF an object is in uniform circular motion (at a constant angular & linear speed) & it does not accelerate in any other direction.
Below, we’ll see how these rules apply in different situations. Once you understand these, you can apply them to find many values in a problem!
Circular Motion with 1 Force
The simplest example: the only force acting on an object is the centripetal force. For example, a satellite orbits a planet at a constant speed: no thrusters, just a natural circular orbit.
In this case, the only force acting on the satellite is the gravitational force. Thus,
Fc = Fg
mv2/r = GMm/r2
You can solve a bunch of problems with this equality! Now let’s move on to something a bit more complicated…
Circular Motion with Multiple Forces
For the following sections, remember that ΣF = Fc
A) Horizontal Circles
Rock on a string:
Let’s start simple: imagine a rock tied to a string, held up by a fixed point.
There are only 2 forces here: the weight W of the rock, & the tension T of the string. Both are balanced, so the rock does not accelerate.
Now, I start swinging the rock. It starts to rise, forming an angle θ between the string and the vertical.
Ignoring air resistance, the forces acting on it are:
- Tension of the string, towards the attachment point
- Weight, downwards
Since the rock is moving at a constant speed in a circle, the net force acting on it MUST be the centripetal force. Here’s a resolution of the forces:
The weight is countered by the vertical component of tension.
In this situation, Fc is the horizontal component of tension.
Thus, the only net force acting on the rock is T sin θ, which acts as the Fc !
*If you realise, this limits the angle θ to be above 0. If the circle is perfectly horizontal, there won’t be any vertical component to oppose the stone’s weight, & it would start to fall back down!
Let’s expand the rock situation to more complicated ones. Below are 2 different vehicles – both travelling in a horizontal circle. We’ll break down what the forces are & what contributes to centripetal acceleration.
| Airplane banking |  Forces acting on the plane:Weight (W = mg) vertically downwards Lift (L) at an angle θ to the vertical W is balanced by the vertical component of L, so L = mg cos θ The only force acting in the horizontal direction is the horizontal component of L Net force = Fc Fc = L sin θ = mg tan θ |
| Car turning on a slanted road (negligible friction) |  acting on the car:Weight (W = mg) vertically downwards Reaction force (R) perpendicular to the road W is balanced by the vertical component of R, so R = mg cos θ The only force acting in the horizontal direction is the horizontal component of R Net force = Fc Fc = R sin θ = mg tan θ |
| Car turning on a slanted road (accounting friction) | Forces acting on the car: Weight (W = mg) vertically downwards Reaction force (R) perpendicular to the road Friction (F) parallel to the road* *Friction may act in either direction, depending on the speed of the car: if it goes too fast, it would slip ‘outwards’ from the circle, so friction acts ‘inwards’ opposing that. If it goes too slow, it would slip ‘inwards’ to the circle’s centre, so friction would act ‘outwards’. Situation 1: Friction inwards  The only forces acting in the horizontal direction are the horizontal component of R + the horizontal component of friction Net force = Fc Fc = R sin θ + F cos θ Situation 2: Friction outwards  The only forces acting in the horizontal direction are the horizontal component of R + the horizontal component of friction Net force = Fc Fc = R sin θ – F cos θ |
B) Vertical Circles
This is where things get interesting.
Travelling in a vertical circle, you must remember that the direction of gravity is ALWAYS vertically downwards. However, the direction of the centripetal force will always be towards the centre of the circle. This means that at different points in the circle, each individual force may change in magnitude, but the net force will always remain constant & point towards the circle’s centre.
Let’s start with that rock again, now swung in a vertical circle:
| At the top of the circle | T & W act in the same direction: vertically downwards, towards the centre of the circle. Fc = T + W |
| At the lowest point of the circle | T & W act in opposite directions. Fc = T – W |
| How forces change throughout the motion | Since Fc must remain constant to maintain a constant speed, & W remains constant due to gravity, it is only the value of T that changes. At the top of the circle, T is at its minimum. At the bottom of the circle, T is at its maximum. You can feel this yourself; the string will pull on your hand stronger at the bottom of the circle. |
Let’s apply this idea to some situations:
Rollercoaster Loops
| Forces acting on the rollercoaster car | Weight (W = mg) always vertically downwards. Reaction force between wheels & track (R), perpendicular to the track, changes in direction & magnitude. Net force between these must be a centripetal force (Fc = mv2/r) |
| At the top of the loop | R & W act in the same direction: vertically downwards, towards the centre of the circle. Fc = R + W |
| At the bottom of the loop | R & W act in opposite directions. Fc = R – W |
These are the same forces acting on a passenger inside the car: they ‘feel’ lighter at the top of the loop because the reaction force between them & the seat is at a minimum. They ‘feel’ heavier at the bottom of the loop because the reaction force between them & the seat is at a maximum.
Ferris Wheels
Now this is a little different: the ‘cars’ in a ferris wheel maintain the same orientation (they do not turn upside-down like a roller coaster). We shall consider the forces acting on a single passenger – try think about what you would ‘feel’ riding a ferris wheel.
| Forces acting on a passenger | Weight (W = mg) always vertically downwards. Reaction force between passenger & seat (R), always vertically upwards, but changes in magnitude. R & W are always in opposite directions, but Fc always acts towards the centre of the ferris wheel. Since magnitude of Fc remains constant, & W remains constant, R must change. |
| At the top of the wheel | Fc acts vertically downwards. Fc = W – R R is at its minimum |
| At the bottom of the wheel | Fc acts vertically upwards. Fc = R – W R is at its maximum |
I’ve only been on a ferris wheel twice, so I had to verify whether you really do feel lighter at the top of a ferris wheel – apparently, you do! https://www.wired.com/story/why-do-you-feel-lighter-at-the-top-of-a-ferris-wheel/
Cars on Bumps
We rarely think about it, but cars actually undergo circular motion as they drive over bumps or hills. For these analysis we’ll just look at the forces acting as the car cross the top of the bump (as cars don’t tend to go underneath them unless something is very wrong).
| Forces acting on the car |  Weight (W = mg), always vertically downwards Reaction force from bump (R), perpendicular to the surface of the bump The net force = centripetal force |
| At the top of the car | Fc = mg – R |
Now, one final situation…
People on the Earth
The Earth is rotating at a constant rate, which means that everything on its surface undergoes uniform circular motion (approximately). Let’s look at 2 people standing on scales in separate locations: Alexa at the North Pole, & Brendan at the equator.
| Alexa, at the North Pole | The only forces acting on Alexa are: Weight (W = mg) vertically downwards, towards the centre of the Earth Reaction force between Alexa & the scale (R), opposite in direction to W. At the poles, there is practically no rotation, so Alexa is NOT in uniform circular motion. Thus, the net force acting on Alexa is 0. mg – R = 0 R = mg This R is what is measured by the weighing scale, & what Alexa “feels” as their weight. This is the apparent weight. At the poles, apparent weight = mg |
| Brendan, at the equator | The only forces acting on Brendan are: weight (W = mg) vertically downwards, towards the centre of the Earth Reaction force between Brendan & the scale (R), opposite in direction to W. At the equator, every object is undergoing uniform circular motion! Thus, the net force acting on Brendan is the centripetal force, towards the centre of the Earth. Fc = mg – R R = mg – Fc This R is what is measured by the weighing scale, & what Brendan “feels” as their weight. This is the apparent weight. At the equator, apparent weight = mg – Fc Thus, we ‘feel’ LIGHTER at the equator! This holds true for any locations except the poles: the closer to the equator you go, the ‘lighter’ your apparent weight is. |
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