Rolling Friction is a frictional force that opposes rolling objects. Rolling friction is applicable where the body moves along its curved surfaces. For example, wheels in vehicles, ball bearings, etc. are examples of rolling friction.
In this article, we will learn about rolling friction, its definition, laws, formulas, causes, coefficient, and the factors that affect it. We will also learn how rolling friction is different from sliding friction.
What is Rolling Friction?
Rolling friction is a type of friction, that opposes the motion of rolling objects. For example, if a ball is rolling on the floor, eventually it will stop. This happens because the ball is going to experience resistance in the direction of its motion. The frictional force acts on the ball, which is opposite to the motion of the ball and slows down the ball. The deformation of surfaces is the reason for rolling friction. Rolling friction is valid in the case for ball bearings-car tires, bowling balls, etc.
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Rolling Friction Examples
Here are some examples of rolling friction
- Wheels of Vehicles: When vehicles move forward, they experience resistance to their motion due to contact with roads.
- Ball Bearings: Ball Bearings uses small metal balls, which creates rolling friction between wheels and axles
- Luggage Wheels: Trolleys use wheels that experience rolling friction when dragged on the surface and help to move heavy luggage with ease.
- Conveyor Belts: Conveyor belts at airports and metro stations are used to scan luggage. In these, rolling friction acts between the belt and the roller.
Laws of Rolling Friction
There are three laws of rolling friction:
Law 1: As the rolling friction becomes smoother, the weight acting on the wheels tends to decrease.
Law 2: Friction expressed as the product of load and constant where constant is raised to a fractional power.
F = kLn
where F is Rolling Friction, K is frictional constant, L is load, n is fractional power.
Law 3: Load is directly proportional to rolling friction force, and the degree of curve is inversely proportional to the rolling friction force.
F = ? ×W/r
Here, F = force of rolling friction
- μ = coefficient of rolling friction
- W = load (weight/mass)
- r = radius of curvature.
Consider the following figure:
There are two formulas of rolling friction which are derived from the laws of rolling friction. The first formula related to rolling friction is given as:
F = KLn
Here,
- F = rolling friction
- K = friction constant
- L = load
- n = fractional power
The second formula of rolling friction is given as
F = ? ×W/r
Here, F = force of rolling friction
- μ = coefficient of rolling friction
- W = load (weight/mass)
- r = radius of curvature.
Cause of Rolling Friction
Rolling friction arises from several factors related to the interaction between a rolling object and the surface over which it rolls. The main causes of rolling friction include:
Deformation of Surfaces: When a wheel or a cylindrical object rolls over a surface, both the object and the surface undergo deformation. The surface deforms slightly under the pressure of the rolling object causing rolling friction to the wheel.
Elastic Hysteresis: Elastic hysteresis is the loss of energy that occurs when an elastic material, such as rubber, undergoes deformation and then returns to its original shape. As the tire or wheel rolls over the surface, it undergoes repeated cycles of compression and expansion, causing energy to be dissipated as heat. This also causes rolling friction.
Internal Friction: Internal friction within the material of the rolling object, such as the rubber in tires, also contributes to rolling resistance. As the tire rolls, internal friction within the tire material generates heat and dissipates energy, increasing the resistance.
Surface Irregularities: Surface irregularities, such as roughness, bumps, and imperfections, on both the rolling object and the surface over which it rolls can increase rolling friction.
Adhesion and Cohesion: Adhesion and cohesion between the rolling object and the surface can also contribute to rolling friction. Molecular interactions between the materials of the rolling object and the surface can create adhesion and cohesion forces, which resist the motion of the rolling object and increase rolling resistance.
Coefficient of Rolling Friction
The coefficient of rolling friction is the ratio of the rolling friction to whole weight of object. The coefficient of rolling friction when given may be formulated mathematically as
F= μ × W
- F= the magnitude of rolling resistance that corresponds to this given force.
- μ = coefficient of road resistance rolling
- W= load(weight)
Factors Affecting Rolling Friction
The following factors affect rolling friction:
- Nature of the surfaces: The more polished the surfaces are, the less turbulent the rolling resistance and vice versa.
- Material of the rolling body: The harder and the more inelastic the material of the rolling thing is, the smaller the rolling friction is. This is simply due to the fact that the body will be less deformed while rolling.
- Load on the rolling body: If pressure increases and load becomes higher the rolling body will deform more and this affects rolling friction.
- Radius of the rolling body: The larger the volume of the body in the rolling motion remains the smaller the rolling resistance. Thus, the body which have bigger radius will not be able to easily deform.
- Speed of the rolling body: With the higher speed of the body that is rolling, an enormous rolling friction comes. This to say that at high speeds slippage between the rolling body and the ground at high speeds will increase due to the rolling friction.
Difference Between Rolling And Sliding Friction
The differences between rolling friction and sliding friction have been listed in the table below:
Rolling friction
| Sliding friction
|
|---|
The use of a rolling motion allows for the avoidance of the use of the friction.
| Frictional sliding arises when the surfaces are dragged against each other or not.
|
Rolling Friction occurs due to deformation of rolling objects.
| Sliding friction occurs due to interlocking of microscopic surfaces
|
The value of the coefficient of rolling friction is determined by the radius of rolling body, as well as how much it depress, and the hardness of the surface.
| The value of sliding friction coefficient depends on the real surface roughness and might be affected by the temperature to some degree.
|
Coefficient of rolling friction: Fr = μrN
| Coefficient of sliding friction: Fk = μkN
|
Advantages and Disadvantages of Rolling Friction
We know that Friction is a necessary evil. Hence, rolling friction has both advantages and disadvantages. The advantages and disadvantages of rolling friction is tabulated below:
Advantages of Rolling Friction
| Disadvantages of Rolling Friction
|
|---|
It allows for efficient energy transfer between a rolling object and the surface over which it rolls
| It requires energy to overcome, particularly in scenarios where heavy loads or high speeds are involved
|
It reduces wear and tear on both the rolling object and the surface.
| Minimizing rolling friction requires complex design
|
It enables ease of movement for objects such as vehicles, luggage, and machinery.
| In some situations, such as when braking or cornering at high speeds, rolling friction may not provide sufficient grip
|
It improves efficiency in many mechanical systems and devices
| It may not perform well in certain conditions, such as on rough or uneven surfaces, at high temperatures, or in extreme environments.
|
It produces less noise and vibration compared to sliding friction
| It can still contribute to noise pollution and vibration levels, particularly in heavy-duty machinery and transportation systems
|
Numericals on Rolling Friction
Example 1: Considering a static friction, a 100 kg crate push across a horizontal floor is at a constant speed. A coefficient of rolling friction per one the crate and floor unit is 0.02.(157 words) What is the force needed for pushing the truck?
Solution:
Here, μ = 0.02
Fr = μN
The normal force can be calculated as:
N = mg
N = 100 × 9.81
N = 981
Fr = 0.02 × 981
Fr = 19.62N
Example 2: The bike covers twenty meters on the down-slope at a continuous pace of twenty meters per second. The coefficient of rolling resistance of the bicycle tires and the road have 0.01 as the value. What is the magnitude of rolling friction (rostructural material) that acts on the bicycle?
Solution:
Here, μ = 0.01
The normal force is equal to the weight of the bicycle and the rider, which we will assume is 100 kg.
N = mg
N = 100 × 9.81
N = 981
Fr = 0.01 × 981
Fr = 9.81 N
Example 3: A box with 200 kg weight is moving straight out on the floor with the ground speed. The rolling resistance between the crate and the floor is equal to 0.03. Let's see: what is the acceleration?
Solution:
N = mg = 200 × 9.81 = 1962 N
Fr = μ × N = 0.03 × 1962 = 58.86 N
Example 4: A rolling ball having a mass of 5 kg rides on the surface, under a rolling resistance coefficient of 0.02. Analyze the rolling friction force.
Solution:
Given:
Ball weight (W): 5 kg
Coefficient of rolling friction (μ): 0.02
N = W × g = 5 × 9.81 = 49.05 N
Fr = μ × N = 0.02 × 49.05 = 0.981 N
Example 5: Derive the rolling friction force for a rolling object with radius of curvature = 0.5 meters and at 50 kg load by making use of the relevant formula.
Solution:
Given:
Radius of curvature (r): 0.5 meters
Load (W): 50 kg
Fr = μ × W × r
N = W × g = 50 × 9.81 = 490.5N
Fr = μ × W × r = μ × 490.5 × 0.5 = 245.25μ
Example 6: A bicycle wheels along the road 30 meters in distance with a coefficient of the rolling resistance of 0.015. If the total weight of the bicycle and rider amounts to 150kg, what is the maximum value of rolling friction?
Solution:
N = mg = 150 × 9.81 = 1471.5 N
Fr = μ × N = 0.015 × 1471.5 = 22.0725 N
Rolling Friction Practice Problems
1. Considering a static friction, a 1000 kg crate push across a horizontal floor is at a constant speed. A coefficient of rolling friction per one the crate and floor unit is 0.06. What is the force needed for pushing the truck?
2. The bike covers twenty meters on the down-slope at a continuous pace of twenty meters per second. The coefficient of rolling resistance of the bicycle tires and the road have 0.09 as the value. What is the magnitude of rolling friction (restructure material) that acts on the bicycle?
3. A box with 2900 kg weight is moving straight out on the floor with the ground speed. The rolling resistance between the crate and the floor is equal to 0.09. what is the acceleration?
4. Derive the rolling friction force for a rolling object with radius of curvature = 60 meters and at 5000 kg load by making use of the relevant formula.
5. Derive the rolling friction force for a rolling object with radius of curvature = 0.8 meters and at 5 kg load by making use of the relevant formula.
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