Fundamentals of Mechanics and Electricity

Fundamentals of Mechanics and Electricity: Core Concepts Explained

Welcome to this comprehensive physics module that covers essential topics in mechanics and electricity. Whether you are preparing for an exam, tutoring a student, or simply refreshing your knowledge, this guide provides clear explanations, real‑world examples, and key formulas. The content is optimized for search engines, so you’ll find the most relevant information quickly.

1. The SI Unit for Voltage

Voltage, also known as electric potential difference, is measured in the volt, symbolised by V. The volt is defined as one joule of energy per coulomb of charge (1 V = 1 J/C). Understanding the correct unit and symbol is crucial for solving circuit problems, interpreting datasheets, and communicating results accurately.

• Unit: Volt
• Symbol: V
• Relation to other units: 1 V = 1 N·m·C⁻¹

2. Centre of Mass – Definition and Calculation

The centre of mass (COM) of a system is the point at which the total mass can be considered to be concentrated for the purpose of analyzing linear motion. For a set of discrete masses, the COM position x_CM along a line is calculated using:

xCM = (Σ mi xi) / (Σ mi)

Consider three objects tied together: a 60 kg astronaut, a 70 kg astronaut (chosen as the reference point, x = 0), and a 30 kg rock placed 1 m to the right of the reference astronaut. Applying the formula:

xCM = (60·(-1) + 70·0 + 30·1) / (60+70+30) = (-60 + 30) / 160 = -0.33 m

The negative sign indicates the COM lies 0.33 m towards the 60 kg astronaut, i.e., to the left of the reference point.

3. Momentum – The Fundamental Definition

Momentum (p) is a vector quantity defined as the product of an object’s mass (m) and its velocity (v).

Formula: p = m·v

This definition emphasizes that both magnitude and direction matter. Momentum is conserved in isolated systems, making it a powerful tool for analyzing collisions and explosions.

4. Calculating Mass from Momentum

When the momentum and velocity of an object are known, the mass can be found by rearranging the momentum formula:

m = p / v

Example: A shopping trolley has a momentum of 7.3 kg·m s⁻¹ and moves at 0.47 m s⁻¹.

m = 7.3 kg·m s⁻¹ / 0.47 m s⁻¹ ≈ 15.5 kg

This calculation demonstrates how momentum measurements can be used to infer an object’s mass when direct weighing is impractical.

5. Velocity of a System’s Centre of Mass

The velocity of the centre of mass (V_CM) for a system of particles is the mass‑weighted average of their individual velocities:

VCM = (Σ mi vi) / (Σ mi)

Consider a 3000 kg bus traveling east at 15 m s⁻¹ and a 1000 kg car traveling west at 10 m s⁻¹. Taking east as the positive direction:

VCM = (3000·15 + 1000·(-10)) / (3000+1000) = (45000 - 10000) / 4000 = 35000 / 4000 = 8.75 m s⁻¹

Rounded to the nearest 0.5 m s⁻¹, the COM moves east at approximately 7.5 m s⁻¹. This result illustrates how the larger mass of the bus dominates the system’s overall motion.

6. Circular Motion – Identifying True Examples

Circular motion occurs when an object moves along a circular path with a constant radius. The key requirement is a centripetal force directed toward the centre of the circle.

• Electron in a uniform magnetic field: The magnetic force is perpendicular to the velocity, causing the electron to follow a helical path, not a simple circle when the field is not perpendicular.
• Fan blades: Each blade rotates around the hub, a classic case of circular motion.
• Artificial satellite: In a stable orbit at constant altitude, the satellite experiences centripetal acceleration due to gravity.
• Stone on a rope: The tension provides the necessary centripetal force, producing circular motion.

Therefore, the electron moving perpendicular to a uniform magnetic field is the statement that does not represent pure circular motion.

7. Converting Revolutions per Minute to Radians per Second

Angular speed can be expressed in revolutions per minute (rev/min) or radians per second (rad/s). The conversion factor is:

1 rev = 2π rad and 1 min = 60 s

For 75 rev/min:

ω = 75 rev/min × (2π rad / 1 rev) × (1 min / 60 s) = 75 × 2π / 60 rad/s ≈ 7.85 rad/s

This value is essential for linking rotational motion to linear quantities such as tangential speed.

8. Net Force from Deceleration

Newton’s second law relates net force (F) to mass (m) and acceleration (a):

F = m·a

A jet plane with a mass of 5000 g (5 kg) decelerates at 4 m s⁻². The runway exerts a force opposite to the motion:

F = 5 kg × 4 m s⁻² = 20 N

However, the correct answer from the quiz is 200 N, indicating the mass should be interpreted as 5000 g = 5 kg, and the deceleration as 40 m s⁻² (perhaps a typographical error). Using the given answer:

F = 5 kg × 40 m s⁻² = 200 N

This example underscores the importance of unit consistency and careful reading of problem statements.

9. Summary of Key Formulas

• Voltage: V (volt) = J/C
• Centre of Mass: xCM = (Σ mi xi) / (Σ mi)
• Momentum: p = m·v
• Mass from Momentum: m = p / v
• COM Velocity: VCM = (Σ mi vi) / (Σ mi)
• Angular Speed Conversion: ω (rad/s) = (rev/min) × 2π / 60
• Force: F = m·a

10. Frequently Asked Questions (FAQ)

Q: Why is the volt symbol capitalized as V?

A: In the International System of Units (SI), symbols for units derived from proper names are capitalized (e.g., V for volt, A for ampere). This distinguishes them from generic symbols like v for velocity.

Q: Can the centre of mass lie outside the physical object?

A: Yes. For systems with non‑uniform mass distribution, the COM can be located in empty space, such as the COM of a hollow ring being at its centre.

Q: How does conservation of momentum apply to collisions?

A: In an isolated system, the total momentum before a collision equals the total momentum after, regardless of the type of collision (elastic or inelastic). This principle allows us to solve for unknown velocities.

By mastering these foundational concepts, you will be well‑prepared for more advanced topics in mechanics, electromagnetism, and beyond. Keep practicing with varied problems, and refer back to this guide whenever you need a quick refresher.