CLASSICAL-MECHANICS · Interactive Practice | Unit 1 · Video 6

The Hidden Grammar of Physics: How M, L, and T Encode Every Quantity

IKey Formulas

Concept	Dimensional Formula	Notes
Velocity	$[v] = L \cdot T^{-1}$	distance / time
Force	$[F] = M \cdot L \cdot T^{-2}$	mass × acceleration
Work / Energy / Torque	$[E] = M \cdot L^{2} \cdot T^{-2}$	the shared "fingerprint"
Power	$[P] = M \cdot L^{2} \cdot T^{-3}$	energy / time

The Big Idea: Every derived quantity is a product of powers of $M$ , $L$ , $T$ . Sharing dimensions signals a relationship — but not necessarily equivalence.

IIVisualization 1 — The Dimensional Builder

Which powers of $M$ , $L$ , and $T$ produce the energy fingerprint — and which quantities share it?

Power of M 1 Power of L 2 Power of T -2

💡 Challenge: from the energy fingerprint $M \cdot L^{2} \cdot T^{-2}$ , drop the $T$ power once more to $M \cdot L^{2} \cdot T^{-3}$ — the dimensions of power.

IIIVisualization 2 — Building Quantities from the Ground Up

Each quantity inherits its dimensions from the one before it, one base dimension at a time.

Quantity Step 1

IVVisualization 3 — Reverse-Engineering Constants

A constant's dimensions are forced on it by the law it lives in — isolate it, substitute, simplify.

Constant Step 1

Defining law

F = \dfrac{G\, m_1 m_2}{r^{2}}

Isolate the constant

G = \dfrac{F\, r^{2}}{m_1 m_2}

Substitute dimensions

[G] = \dfrac{(M \cdot L \cdot T^{-2})\, L^{2}}{M^{2}}

Simplify

[G] = L^{3} \cdot M^{-1} \cdot T^{-2}

Defining law

E = h f

Isolate the constant

h = \dfrac{E}{f}

Substitute dimensions

[h] = \dfrac{M \cdot L^{2} \cdot T^{-2}}{T^{-1}}

Simplify

[h] = M \cdot L^{2} \cdot T^{-1}

Defining law

F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^{2}}{r^{2}}

Isolate the constant

\dfrac{1}{4\pi\varepsilon_0} = \dfrac{F\, r^{2}}{e^{2}}

Substitute dimensions

\left[\tfrac{1}{4\pi\varepsilon_0}\right] e^{2} = (M \cdot L \cdot T^{-2})\, L^{2}

Simplify

\left[\tfrac{1}{4\pi\varepsilon_0}\right] = M \cdot L^{3} \cdot T^{-2}\ (\text{per charge}^{2})

💡 A constant's numerical value changes between unit systems, but its dimensional formula never does.

VQuiz Questions

Question 1

Velocity is distance divided by time. What is the dimensional formula of velocity?

✅ Correct! Distance (L) over time (T) gives L · T⁻¹.

❌ Not quite. Dividing by time gives a negative power of T, and no mass is involved.

Show solution

Solution:

Velocity is distance over time: $[v] = \frac{L}{T} = L \cdot T^{-1}$

Dividing by time means a negative power of $T$ . There is no mass involved, so $M$ does not appear.

Question 2

Kinetic energy is $\tfrac{1}{2}mv^{2}$ . A common mistake is to forget that the velocity term is squared. What is the correct dimensional formula of kinetic energy?

✅ Correct! Squaring velocity gives L² · T⁻², times mass = M · L² · T⁻².

❌ Not quite. That is the dimension of force. You forgot to square the velocity.

❌ Not quite. Remember velocity is squared, so both its L and T powers double.

Show solution

Solution:

The factor $\tfrac{1}{2}$ is a pure number and carries no dimension. So: $[KE] = M \cdot (L \cdot T^{-1})^{2} = M \cdot L^{2} \cdot T^{-2}$

Squaring the velocity squares both the L and T powers. This is the same "fingerprint" shared by work and torque.

Question 3

Newton's law of gravitation is $F = \dfrac{G\, m_1 m_2}{r^{2}}$ . Isolating $G$ gives $G = \dfrac{F\, r^{2}}{m_1 m_2}$ .

Using $[F] = M \cdot L \cdot T^{-2}$ , what is the dimensional formula of $G$ ?

✅ Correct! The two masses in the denominator give M⁻¹, and L¹·L² = L³.

❌ Not quite. Watch the mass: dividing by m₁m₂ (= M²) against the M¹ from force gives M⁻¹, not M¹.

❌ Not quite. Combine L¹ (from force) with L² (from r²) to get L³, and track the mass powers carefully.

Show solution

Solution:

Substitute the dimensions into $G = \dfrac{F\, r^{2}}{m_1 m_2}$ : $[G] = \frac{(M \cdot L \cdot T^{-2}) \cdot L^{2}}{M^{2}}$

Combine the L powers: $L^{1} \cdot L^{2} = L^{3}$ . The mass powers: $\dfrac{M^{1}}{M^{2}} = M^{-1}$ .

$[G] = L^{3} \cdot M^{-1} \cdot T^{-2}$

Question 4

Work and torque both have the dimensional formula $M \cdot L^{2} \cdot T^{-2}$ .

True or False: Because they share the same dimensions, work and torque are physically the same quantity.

✅ Correct! Same dimensions ≠ same thing. Work is a dot product; torque is a cross product.

❌ Not quite. Shared dimensions reveal a relationship, but work and torque are genuinely different quantities.

Show solution

Solution:

False. Sharing dimensions signals a relationship, not equivalence.

Work is a dot product ( $\vec{F} \cdot \vec{d}$ ) — a scalar measure of energy transfer.
Torque is a cross product ( $\vec{r} \times \vec{F}$ ) — a vector describing rotational tendency.

They carry the identical dimensional fingerprint $M \cdot L^{2} \cdot T^{-2}$ , yet they are fundamentally different physical quantities. Dimensional identity is a clue about structure — it does not mean two things are the same.

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