Inertial and Non-Inertial Frames — JEE Advanced Physics

Key Concepts

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Examples

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Introduction to Inertial and Non-Inertial Frames

Welcome to Inertial and Non-Inertial Frames. Content to be added.

Non-Inertial Frames and Pseudo Forces

Inertial vs Non-Inertial Frames

Newton's Laws are not universally applicable in all reference frames!

Inertial Frame: A reference frame where Newton's First Law holds - objects with zero net force remain at constant velocity (including rest).

Non-Inertial Frame: An accelerating reference frame where Newton's Laws in standard form do NOT hold.

Block observed from different frames

Example: A block at rest on the floor of an accelerating train:

Ground observer (inertial): Block at rest, forces balanced ✓
Train observer (non-inertial): Block accelerates backward, but forces still balanced? ✗

The Pseudo Force

To use Newton's Laws in a non-inertial frame, we introduce a pseudo force (fictitious force):

$\vec{F}_{pseudo} = -m\vec{a}_{frame}$

where $\vec{a}_{frame}$ is the acceleration of the non-inertial frame relative to an inertial frame.

Key points:

Acts on every object of mass $m$ in the frame
Directed opposite to frame's acceleration
Not a real physical interaction
Does NOT have an action-reaction pair

FBD from inertial frame

FBD from non-inertial frame (with pseudo force)

Is the pseudo force always directed opposite to the acceleration of the non-inertial frame?

Modified Newton's Second Law

In a non-inertial frame:

$\vec{F}_{real} + \vec{F}_{pseudo} = m\vec{a}_{relative}$

where $\vec{a}_{relative}$ is acceleration relative to the non-inertial frame.

This allows us to analyze motion within the accelerating frame by treating pseudo force as another force!

Example 1: Pendulum in Accelerating Car

A pendulum hangs in a car accelerating at $a_{car}$ . Find the angle $\theta$ from vertical at equilibrium (relative to car).

Pendulum in accelerating car

Solution (car's frame):

Pseudo force: $F_{pseudo} = ma_{car}$ (backward)

Equilibrium in car frame: $a_{relative} = 0$

Forces on bob: Weight $mg$ (down), Tension $T$ (along string), Pseudo force $ma_{car}$ (backward)

Resolve tension: $T_x = T\sin\theta$ , $T_y = T\cos\theta$

Horizontal: $T\sin\theta = ma_{car}$ ... (1)

Vertical: $T\cos\theta = mg$ ... (2)

Divide (1) by (2):

$\tan\theta = \dfrac{a_{car}}{g} \implies \theta = \arctan\left(\dfrac{a_{car}}{g}\right)$

If the car accelerates faster, would the pendulum angle $\theta$ increase or decrease?

Example 2: Apparent Weight in Elevator

A 70 kg person in an elevator accelerating upward at 3 m/s². Find scale reading.

Person in elevator

FBD of person

Method 1 (Inertial frame - ground):

$N - W = ma \implies N = mg + ma = 700 + 210 = 910 \, \text{N}$

Method 2 (Non-inertial frame - elevator):

Pseudo force: $F_{pseudo} = 210$ N (downward)

Equilibrium in elevator: $N - W - F_{pseudo} = 0$

$N = W + F_{pseudo} = 700 + 210 = 910 \, \text{N}$

Scale reading: $910/10 = 91$ kg (feels heavier!)

Atwood Machine - Detailed Analysis

Masses $m_1 = 5$ kg and $m_2 = 7$ kg on ideal Atwood machine.

Atwood machine with 5kg and 7kg

FBDs:

FBD of 5kg mass

FBD of 7kg mass

Solution: $m_2 > m_1$ , so $m_2$ goes down, $m_1$ goes up

For $m_1$ : $T - 50 = 5a$ ... (1)

For $m_2$ : $70 - T = 7a$ ... (2)

Add: $20 = 12a \implies a = \dfrac{5}{3} \approx 1.67 \, \text{m/s}^2$

From (1): $T = 50 + 5(5/3) = \dfrac{175}{3} \approx 58.33 \, \text{N}$