Animated Solution for Physics - Physics and Measurement: If time ($t$), velocity ($v$) and angular momentum ($l$) are taken as the fundamental units, then the dimension of mass ($m$) in terms of $t$, $v$ and $l$ is

Imagine we are rewriting the rules of physics. Instead of the traditional mass, length, and time being our fundamental building blocks, we are given a new set: time ($t$), velocity ($v$), and angular momentum ($l$). Our mission is to figure out how to construct the dimension of mass ($m$) using only these new tools.

In dimensional analysis, when we want to express one quantity in terms of others, we assume a proportional relationship with unknown powers. Let's assume that mass is proportional to time to the power $a$, velocity to the power $b$, and angular momentum to the power $c$. We can write this mathematically as:

Or, introducing a dimensionless constant $k$:

To solve for these unknown powers, we need the standard dimensional formulas for all the quantities involved in terms of the classic $\text{M}$, $\text{L}$, and $\text{T}$:

Mass ($m$): $[{\text{M}}^1{\text{L}}^0{\text{T}}^0]$ Time ($t$): $[{\text{M}}^0{\text{L}}^0{\text{T}}^1]$ Velocity ($v$): $[{\text{M}}^0{\text{L}}^1{\text{T}}^{-1}]$ Angular Momentum ($l$): Remember, angular momentum is $mvr$, so its dimension is $[{\text{M}}^1][{\text{L}}^1{\text{T}}^{-1}][{\text{L}}^1]=[{\text{M}}^1{\text{L}}^2{\text{T}}^{-1}]$

Now, let's substitute these standard dimensions into our assumed equation. We are essentially translating our new rulebook back into the classic language to see how they match up.

Next, we need to group the terms on the right side. We collect all the $\text{M}$'s, $\text{L}$'s, and $\text{T}$'s together by adding their exponents.

According to the principle of dimensional homogeneity, for an equation to be physically valid, the dimensions on both sides must be identical. This means the power of $\text{M}$ on the left must equal the power of $\text{M}$ on the right, and similarly for $\text{L}$ and $\text{T}$. Let's set up our equations:

Comparing powers of $\text{M}$: So, we immediately find that $c=1$.

Comparing powers of $\text{L}$: Substituting our known value of $c=1$:

Comparing powers of $\text{T}$: Now, substitute the values of $b=-2$ and $c=1$ that we just found:

We have successfully cracked the code! The powers are $a=-1$, $b=-2$, and $c=1$. Substituting these back into our initial assumption, we get the final dimensional formula for mass in this new system:

Therefore, the dimension of mass is $[t^{-1}{v}^{-2}{l}^1]$. This elegant method allows us to translate between any arbitrary systems of units!