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A kart consists of several elements of a certain weight. This results in so-called motion inertia, i.e. the force opposing acceleration. It is an important factor and deserves further analysis

WHAT IS IT?

HOW DOES IT WORK?

IN PRACTICE

**All** objects that have a **mass** observe the **laws** of **physics** and, in **particular**, the equation of **motion**. The axiom, in itself, can be **found** on the first **pages** of any **basic physics** book, but its practical **implications** are not as easy to analyse, **especially** when, as in karting, they can **become crucial** to achieving top performance.

Let’s start with the **theory**: excluding all **forms** of **friction** that can exists in reality, the motion **equations** can be **simplified** according to the following formulas:

**Rotating motion: T = I * w**

where * T* is

are represented by tyres, wheel rims, brake disc, chain, plate, etc.).

**Linear motion: F = m * a**

where * F* is the

**Basically**: the more a vehicle **weighs** and has heavy rotating masses, the **greater** the torque and the force **needed** to accelerate.

WEIGHT IS THE FACTOR THAT DETERMES MOTION INERZIA, BOTH FOR LINEAR AND ROTATING MOTION

WHAT IS IT?

HOW DOES IT WORK?

IN PRACTICE

SIMPLE MASSES ROTATING MASSES DISC BRAKE TYRES

SIMPLE MASSES

ROTATING MASSES

DISC BRAKE

TYRES

**Focussing** on the specific **components** of a kart, we can say that inertia is **determined** by **simple masses** (such as the driver’s weight, the weight of the engine and the chassis, with tank and fuel), which **determine** the **parameter** ” m “, and rotating masses (tyres, wheel rims, disc brake, chain, plate, etc.) that determine the **instant inertia** “I”. **Regarding** the masses “m”, the situation is **simple**, since they are determined by the overall **weight** of the kart (and engineered) plus the **driver.** The moment of inertia, on the other hand, does not **depend** solely on the weight of the rotating **objects,** but also by the **square** of the **distance** of the masses from the **centre** of rotation (radius r).

WHAT IS IT?

HOW DOES IT WORK?

IN PRACTICE

For example, let’s take a **disc that rotates**: its moment of inertia is given by the formula:

*I = m * R ^{2 }*

However, if we consider the brake disc, which is actually a

As a result, the **greater** the outer radius of the disc or plate, the **greater** the moment of inertia ** I**, i.e. the

effect compared to the **total weight** of the disc.

Let’s look at an **example** to gain a better **understanding**.

On the one hand, a cast **iron disc** weighing **1 kg**, with a **9 cm** outer radius and a** 7 cm** inner radius. **Reasonable** values for a **traditional** rear brake disc.

On the other hand, a disc with a **mass** weighing about half of the former, i.e.** 0.6 kg**, perhaps made of **Ergal** (aluminium alloy), with a **7 cm** inner diameter (i.e. with a disc plate of the same width as the former), but with an **11 cm** outer diameter. The respective moments of **inertia **will be:

**I _{ghisa} = 1 * (92^{2} – 7^{2}) = 32 kg m^{2 }**

THE HEAVIER THE KART AND THE MORE ROTATING MASSES IT HAS, THE GREATER THE INERTIA OPPOSED TO ACCELERATION AND, THEREFORE, THE GREATER THE POWER REQUIRED FROM THE ENGINE

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