Question

The moment of inertia of a body rotating about a given axis is $12.0{\text{kg m}}^2$ in S.I units. The value of the moment of inertia in a system of units in which the unit of length is $5\text{cm}$ and the unit of mass is $10\text{g}$ is $x\times {10}^5$. What is the value of $x$?

Answer 1

The value of MOI in new units is given by
$n_2=n_1{\left(\frac{M_1}{M_2}\right)}^a{\left(\frac{L_1}{L_2}\right)}^b{\left(\frac{T_1}{T_2}\right)}^c$

Dimensional formula of moment of inertia $=\left[M^1{L}^2{T}^0\right]$
$\therefore a=1,b=2,c=0$

Given: $n_1=12.0{\text{kg m}}^2$
$M_1=1\text{kg},L_1=1\text{m},T_1=1\text{s}$ [S.I units]
$M_2=10\text{g},L_2=5\text{cm},T_2=1\text{s}$
[unit of time doesn't change]

Therefore, value in new units
$n_2=12.0{\left(\frac{1\text{kg}}{10\text{g}}\right)}^1{\left(\frac{1\text{m}}{5\text{cm}}\right)}^2{\left(\frac{1\text{sec}}{1\text{sec}}\right)}^0$
$=12\times {\left(\frac{1000\text{g}}{10\text{g}}\right)}^1{\left(\frac{100\text{cm}}{5\text{cm}}\right)}^2\times 1$
$=12\times 100\times 400=4.8\times {10}^5$