What is a Frame of Reference?
A frame of reference is a coordinate system that we use to describe the position and motion of objects. Think of it as your “point of view” for observing motion.
Types of Reference Frames
1. Inertial Frame of Reference
An inertial frame is one in which Newton’s laws of motion hold true.
Characteristics:
- The frame is either at rest or moving with constant velocity
- No acceleration of the frame itself
- Objects at rest remain at rest (Newton’s 1st Law applies)
Examples:
- A room at rest on Earth (approximately)
- A train moving at constant speed on straight tracks
- A spaceship far from any gravitational influence
2. Non-Inertial Frame of Reference
A non-inertial frame is an accelerating frame where Newton’s laws don’t directly apply.
Characteristics:
- The frame is accelerating (changing speed or direction)
- Pseudo forces appear to act on objects
- Newton’s laws require modification
Examples:
- A car taking a turn
- An elevator accelerating up or down
- A merry-go-round
Pseudo Forces
In non-inertial frames, we introduce pseudo forces (also called fictitious forces) to apply Newton’s laws.
$$\vec{F}_{pseudo} = -m\vec{a}_{frame}$$where $\vec{a}_{frame}$ is the acceleration of the non-inertial frame.
Example: Elevator
When an elevator accelerates upward with acceleration $a$:
- You feel heavier
- Effective weight = $m(g + a)$
When the elevator accelerates downward:
- You feel lighter
- Effective weight = $m(g - a)$
Interactive Demo: Visualize Motion in Different Frames
Explore how motion appears different from various reference frames.
Choosing the Right Frame
The choice of reference frame can simplify problem-solving:
| Situation | Best Frame to Use |
|---|---|
| Ground-based problems | Earth’s surface (inertial) |
| Problems inside vehicles | Vehicle frame + pseudo forces |
| Relative motion problems | One object as reference |
Practice Problem
A person in a lift feels their weight to be $1.2$ times their actual weight. If $g = 10 \text{ m/s}^2$, find the acceleration of the lift.
Solution:
- Apparent weight = $m(g + a) = 1.2mg$
- Therefore: $g + a = 1.2g$
- $a = 0.2g = 2 \text{ m/s}^2$ (upward)