Question

If we choose velocity $V$, acceleration $A$ and force as the fundamental quantities, then the angular momentum in terms of $V$, $A$ and $F$ would be:
A. $\left[ {F{A^{ - 1}}V} \right]$
B. $\left[ {F{V^3}{A^{ - 2}}} \right]$
C. $\left[ {F{V^2}{A^{ - 3}}} \right]$
D. $\left[ {M{L^2}{T^{ - 1}}} \right]$

Answer 1

Hint:Here we have to find the dimensions of force, velocity, acceleration and angular momentum one by one to get the answer. We need to put the dimensions in the angular momentum formula.

Complete step by step answer:
Force: A force is a push or pull on an object that results from the contact of the object with another object. There is a force on each of the objects if there is an interaction with two objects. The unit of force is Newton. The dimension of force is $\left[ {ML{T^{ - 2}}} \right]$

Velocity: Velocity is the rate at which there is a change in place. The displacement or position transition (a vector quantity) per time ratio is the average velocity. The formula for velocity is $m{s^{ - 1}}$. The dimension of velocity is $\left[ {{M^ \circ }L{T^{ - 1}}} \right]$.

Acceleration: Acceleration is the rate at which an object's velocity varies in response to time. Accelerations (in that they have magnitude and direction) are vector quantities. The unit of acceleration is $m{s^{ - 2}}$. The dimension of acceleration is $\left[ {L{T^{ - 2}}} \right]$.

Angular momentum: In mechanics, angular momentum is the rotational counterpart to linear momentum (rarely, momentum or rotational momentum). In mechanics, it is an important quantity since it is a conserved quantity; a closed system 's cumulative angular momentum stays unchanged. The unit of angular momentum is $kg{m^2}{s^{ - 1}}$. The dimension of angular momentum is $\left[ {M{L^2}{T^{ - 1}}} \right]$. The formula for angular momentum is:
$
L = F \times V \times A \\
\left[ {\overrightarrow L } \right] = {\left[ F \right]^a} \times {\left[ V \right]^b} \times {\left[ A \right]^c} \\
\Rightarrow \left[ {M{L^2}{T^{ - 1}}} \right] = {\left[ {ML{T^{ - 2}}} \right]^a} \times {\left[ {L{T^{ - 1}}} \right]^b} \times {\left[ {L{T^{ - 2}}} \right]^c} \\
\Rightarrow \left[ {M{L^2}{T^{ - 1}}} \right] = \left[ {{M^a}{L^{\left( {a + b + c} \right)}}{T^{\left( { - 2a - b - 2c} \right)}}} \right] \\
$
By equating the power we get:
$
a = 1 \\
\Rightarrow a + b + c = 2 \\
\Rightarrow 2a + b + 2c = 1 \\
$
By solving these equations we get:
$
a = 1 \\
\Rightarrow b = 3 \\
\Rightarrow c = - 2 \\
$
So, the dimension of angular momentum is:
$\therefore\left[ L \right] = \left[ {F{V^3}{A^{ - 2}}} \right]$

Hence, option B is correct.

Note:Here we have to be careful while writing the dimensions. If we write a wrong value in the power the whole answer would be wrong.Moreover, the study of the relationship between physical quantities with the help of dimensions and units of measurement is termed as dimensional analysis. Dimensional analysis is essential because it keeps the units the same, helping us perform mathematical calculations smoothly.