Question

If ${\text{I}}$ is the moment of inertia and $\omega$ the angular velocity, what is the dimensional formula of rotational kinetic energy, $\dfrac{1}{2}{\text{I}}{\omega ^2}$ ?
A. $[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 1}}]$
B. $[{{\text{M}}^2}{{\text{L}}^{ - 1}}{{\text{T}}^{ - 2}}]$
C. $[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$
D. $[{{\text{M}}^2}{{\text{L}}^{ - 1}}{{\text{T}}^{ - 1}}]$

Answer 1

To examine the dimensional formula of any two multiplied quantities first we have to know the dimensional formula of the individual quantities. As in this question, rotational kinetic energy is the multiplication of moment of inertia and square of angular velocity.

Complete step by step answer:
To find: dimensional formula of rotational kinetic energy, $\dfrac{1}{2}{\text{I}}{\omega ^2}$.
Dimensions of moment of inertia, ${\text{I}}\,{\text{ = }}[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^0}]$
Dimensions of angular velocity, $\omega \,{\text{ = }}[{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^{ - 1}}]$
Therefore, dimensions of
${\omega ^2}\,{\text{ = }}{[{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^{ - 1}}]^2}\, = \,{\text{ }}[{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^{ - 2}}]$
Now, the dimensions of rotational kinetic energy,
$\dfrac{1}{2}{\text{I}}{\omega ^2}\, = \,{\text{ }}[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^0}]\,[{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^{ - 2}}] \\\ \Rightarrow \dfrac{1}{2}{\text{I}}{\omega ^2}= \,[{{\text{M}}^{0 + 1}}{{\text{L}}^{0 + 2}}{{\text{T}}^{ - 2 + 0}}] \\\ \therefore \dfrac{1}{2}{\text{I}}{\omega ^2}= \,[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$
Hence, required dimensions are $[{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$.

Therefore, option C is correct.

Note: Analyse the given terms carefully and try to find their dimensions from any equation you remember. Try to solve units and dimensions questions in S.I. units to avoid any confusion in between the questions. BODMAS rule should be followed very carefully.