The Physics of Cycling: Understanding Bicycle Motion and Stability

Bicycles are a fundamental mode of transportation, recreation, and sport, yet the physics behind their motion is often overlooked. Understanding the forces at play when riding a bicycle reveals the intricate balance, stability, and mechanics that make cycling possible. This document explores the fundamental principles behind the physics of cycling. This includes bicycle motion, the role of gravitational, centrifugal, and frictional forces.

We begin by analyzing how a rider maintains stability, examining the effects of leaning and counteracting forces. The study extends to the influence of centrifugal force during turns, the mechanics of braking, and the significance of front and rear wheel dynamics. Additionally, we discuss gear switching and its impact on torque and acceleration, as well as the mathematical principles governing a bicycle’s circular path when turning.

By breaking down these forces and interactions, this document provides a deeper insight into the physics of riding a bicycle, helping both enthusiasts and engineers appreciate the science behind an everyday activity.

Table of Contents

Leaning – Situation Along Primary Axis

Now, we will consider the forces acting on the bike. We will view them from behind along the primary axis.

Figure 1. Ideally positioned bike with the rider perpendicular to the ground.

In an ideally positioned bike, pictured by Figure 1, there is no leaning to the left or to the right.

Gravitational Force Fg

We all know that situation shown in Figure 1 is impossible. The center of mass of the rider will always shift to the left or to the right of a vertical line. Hence, the rider will always be leaning or falling off to either side.

Figure 2 shows the situation where the rider leans to the right. There is a gravitational force Fg acting from the center of mass down perpendicular to the ground. There are two components derived from the gravitational force: Force applied with torque F torque and radial force F radial. Torque force rotates the rider counter-clockwise. Radial force acts along the body of a bike and rider.

Figure 2. Leaning of rider to the right.

We can calculate the torque force and radial force as follows:

\[ F_{radial} = F_g \sin{\theta} \] \[ F_{torque} = F_g \cos{\theta} \]

Radial Force F radial

Next, the radial force is divided into two further components: contact force F contact and force F move of movement to the left. See Figure 3. Contact force Fc is equal to Normal force N. Force of movement F move is parallel to the ground.

Figure 3. Radial force is further divided into two forces: one parallel and one perpendicular to the ground.

We find two forces as follows:

\[ F_{contact} = F_{radial} \sin{\theta} = N \] \[ F_{move} = F_{radial} \cos{\theta} \]

Also, force of friction prevents the bicycle from skidding to the left as long as it is greater than force of movement F move:

\[ F_{friction} = \mu N = \mu F_{radial} \sin{\theta} \] \[ F_{friction} \geq F_{move} \]

When force of friction becomes less than force of movement, the rider and bicycle start skidding to the left.

Centrifugal Force Fc

Now we will consider what happens when the bike is turning to the right as well as leans to the right. When turning, there is centrifugal force Fc that pushes and twists the bicycle to the left as it turns to the right, see Figure 4.

Centrifugal force Fc has components for torque and radial forces:

\[ F_{torque2} = F_{centrifugal} \sin{\theta} \] \[ F_{radial2} = F_{centrifugal} \cos{\theta} \]

Second radial force Fr2 further divides into force along the ground and force perpendicular to the ground:

\[ F_{c move} = F_{radial2} \cos{\theta} \] \[ F_{c contact} = F_{radial2} \sin{\theta} \]

Centrifugal force is equal to the centripetal acceleration times mass:

\[ F_{centrifugal} = m a_{c} = m \frac{v^2}{r} \]

You drive a bike such that torque from centrifugal force equals torque from force of gravity:

\[ F_{torque2} = m \frac{v^2}{r} \sin{\theta} = F_{g} \cos{\theta} \]

This way you do not fall over either side. If you increase speed, it becomes harder to compensate torque from centrifugal force with torque from gravity. As a result, you tend to increase the radius of traveling by straightening the handlebar. As a result, you tend to go off in a straight line.

Reducing speed helps you decrease torque from centrifugal force. It also allows you to turn left or right. The torque from gravitational force compensates torque from centrifugal force during the turn. This is crucial in the physics of cycling.

You also decide intuitively which angle theta to use. You must have force of friction be large enough to counteract force of moving along the ground:

\[ F_{net move} = F_g \sin{\theta} \cos{\theta} + m \frac{v^2}{r} \cos^2{\theta} \] \[ F_{friction} = \mu (F_g \sin^2{\theta} + m \frac{v^2}{r} \cos{\theta} \sin{\theta}) \]

Force of friction F friction must be greater than net force of movement for the bike to not skid.

Next let us consider braking.

Braking Forces – Situation along Side Axis

Every bicycle has two brakes: one for the front wheel and one for the back wheel. The two brakes are not equal and produce different efficiencies. Figure 5 shows bicycle when viewed from side.

Back Wheel Braking

When you apply brakes on the handlebar, you squeeze the braking pads on the corresponding wheel. The pads squeeze the wheel and apply friction to it. Your hand brake applies friction on the wheel. This is the same amount that the wheel will experience by ground friction, see Figure 6.

The amount of friction depends on how hard you squeeze the hand brake. The brake will resist rotational movement of the wheel. When rotational movement is decreased, the wheel resist translational movement along the ground. This, in turn, slows the whole bike down:

Rate of slowing down:

\[ a = \frac{F_{friction back wheel}}{m} \]

The maximum amount of friction by braking for the back wheel is no greater than static friction on the back wheel:

Figure 7. Force of gravity on bike and on each wheel.

Static friction depends on normal force and mu coefficient. When the force of gravity is centered in the middle of two wheels, it gets distributed along two wheels equally. The force per wheel is twice as less as the force of gravity.

\[ F_{static friction} = \mu N = \mu \frac{F_g}{2} \]

Greatest deceleration is achieved when “friction back wheel” by squeeze pads equals static friction:

\[ a_{max} = \frac{F_{friction back wheel}}{m} = \frac{\mu F_g}{2m} \]

Braking back wheel will not affect speed in any way more than what half force of gravity allows. Braking front wheel is different as you lean forward like in Figures 8-10:

Figure 8. When braking is applied to the front wheel, the bike tips over slightly.

Front Wheel Braking

Braking front wheel with pads causes front wheel to slow down. It causes ground friction to decrease translational speed of the bike. It also causes the bike to tip forward, rotating perpendicularly to radial support line. The forward rotation is caused by the force from bike center of mass that is opposite to ground friction. The front wheel is in front of the center of mass. This makes the center of mass move forward when braking is applied. This does not happen when the back wheel is braking. The back wheel is behind the center of mass and so it doesn’t tip over.

The force resulted from front wheel braking propagates along the radial support and also causes torque around radial support. See Figure 9.

Figure 9. Torque and Radial components of the force resulted from applying front brakes (the Center of Gravity Force).

The torque and radial components are calculated as follows:

\[ F_{torque3} = F_{friction front wheel} \sin{\theta} \] \[ F_{radial3} = F_{friction front wheel} \cos{\theta} \]

“Friction front wheel” is the same as the force of applying hand brakes, or the force of braking pads on the wheel in Figure 8. We see that by applying brakes on the front wheel you cause torque that makes the bike tip over. Without gravity, there would be no opposite force to the torque and so the bike would tip over in any case. With gravity, however, the bike is able to counteract the torque. See Figure 10.

Figure 10. Vertically, gravity divides into torque and radial components.

\[ F_{torque4} = F_g \cos{\theta} \] \[ F_{radial4} = F_g \sin{\theta} \] \[ F_{move4} = F_{radial4} \cos{\theta} \] \[ F_{contact4} = F_{radial4} \sin{\theta} \]

So, when you apply brakes on the front wheel, you most likely have Force for torque4 greater than Force for torque3. In other words, backward torque from gravity is greater than forward torque resulted from braking:

\[ F_{torque4} > F_{torque3} \]

This is not true in all situations. Especially on hills, the forward torque from braking can be greater than backward torque from gravity. In this case, your bike can tip over when you apply front brakes. So be careful and keep that in mind! This is important in the physics of cycling. (Note on figure 10 we assume that torque from gravity is greater than torque from braking so the angle theta is same as on other figures as the bike will not tip over).

Let’s examine net contact force when torques from forward move from braking and torque from center of gravity are equal. In this case, there is equilibrium in net torque and net ground contact force:

\[ F_{torque3} = F_{friction front wheel} \sin{\theta} = F_{torque 4} = F_g \cos{\theta} \] \[ F_{friction front wheel} = \frac{F_g \cos{\theta}}{\sin{\theta}} \] \[ F_{contact3} = F_{radial 3} \sin{\theta} = F_{fritction front wheel} \cos{\theta} \sin{\theta} = F_g \cos^2{\theta}\] \[ F_{contact4} = F_{radial 4} \sin{\theta} = F_g \sin^2{\theta} \] \[ F_{net contact} = F_{contact 3} + F_{contact 4} = F_g \]

We have to consider one more fact, The force resulted from applying front brakes was divided into two components on Figure 9: torque component and radial component. The radial component is further divided into force of movement and ground contact force (Figure 11):

We have that net contact is still the force of gravity, hence the equilibrium is reached at that point. Friction of the front wheel is Fg times cot theta. It is greater than force of gravity for smaller angle theta and is zero for angle at 90 degrees. Since the bike has frames usually at lower degrees theta, the front wheel friction is greater than gravity, hence braking force is greater for the front wheel than the back wheel.

Therefore, it is much easier to brake with front brakes than with back brakes when driving on a bike!

Gear Switching

On most modern bikes there are gears present so you can switch from low gear to high gears when driving.

You start with low gears numbered 1, 2 ,3 etc. Low gears have larger radiuses and so their torque will be greater at the point of force application:

\[ \textbf{Torque } T = F r \]

You could pedal with less effort on lower gear. However, the speed limit is small. Since the radius is large, faster wheel rotations will correspond to faster chain movement along the rim of a lower gear. To maintain high speed on a lower gear you will need to pedal very fast.

The problem with maintaining speed is less so when riding on a higher gear. Higher gears will have smaller radiuses so their rotations will correspond to smaller speeds and slower pedaling. However, to apply same acceleration you must apply greater force through smaller radius.

\[ v_{tan} = \omega r \]

Here tangential velocity is proportionate to radius of the gear. Higher radius will correspond to greater tangential velocity. Smaller radius will decrease the tangential velocity.

Determining Circular Path of a Turning Bicycle

It is known that a bicycle travels through a circular path when turning left or right. However, determining the exact path is more difficult. The key is that you must find the point where lines from each wheel axes intersect:

Figure 13. Circular Path of a turning bicycle.

The point of intersection of two axes lines will be the cetner of a circular path. This makes sense as in case of a 90 degree angle, the radius of a circular path will be the distance between two wheels. In case of a 0 degree angle. The circular path will extend indefinitely.

Findings Summary

We have explored the following findings regarding dynamics of riding a bicycle:

Finding	Example
Key factors determining whether you fall off to the side: angle of leaning, speed of movement and radius of turning.	The torque from gravitational force is greater for larger lean angle. So, to counteract it, you must travel at higher speed and at smaller radius.
Braking with the back wheel begins with squeezing the braking pads on the wheel. Slowing rotational movement of the wheel will slow translational movement of the bike.	In case when center of gravity lies in the middle of the bike, both front wheel and back wheel get half of net gravitational force acting on them. This determines max static friction force for braking.
Braking with front wheel will result in higher braking force than that of a back wheel.	Braking with front wheel will tip the bike over to the front and apply additional pressure on the wheel.
The larger the radius of a gear chain the easier it is to accelerate. However, the harder it is to maintain high speeds.	Rotational force T depends on radius r and applied force F. The larger the radius the smaller F will result in same torque T.
Center of the circular path is in the intersection of two lines from wheel axes.	When the front wheel is tilted at 90 degrees relative to the back wheel, the circular path will have radius equal to distance between front and back wheels.

Conclusion

The physics of riding a bicycle is a complex interplay of forces, balance, and motion dynamics. Throughout this document, we have explored how gravitational, centrifugal, frictional, and torque forces contribute to the stability, maneuverability, and efficiency of a bicycle. From understanding the effects of leaning and turning to analyzing braking mechanics and gear switching, we have seen how these fundamental principles dictate a rider’s control over the bicycle.

Maintaining balance requires continuous adjustments to counteract gravitational and centrifugal forces, ensuring that torque and friction forces are properly balanced. The mechanics of braking highlight the differences between front and rear wheel braking, demonstrating the importance of weight distribution and normal force. Additionally, the physics behind gear ratios emphasizes how torque and speed are optimized for different riding conditions.

By applying these principles, cyclists can improve their riding techniques, engineers can design more efficient bicycles, and scientists can further explore the mechanics of two-wheeled motion. Whether for transportation, sport, or recreation, understanding the physics of cycling enhances both performance and safety, making it not just a practical skill but also a fascinating study of applied physics. If you are interested in more of this type of articles, read Physics of Running!