Figure 10-35shows three 0.0100kg particles that have been glued to a rod of length L=6.00cm and negligible mass. The assembly can rotate around a perpendicular axis through point Oat the left end. If we remove one particle (that is, 33%of the mass), by what percentage does the rotational inertia of the assembly around the rotation axis decrease when that removed particle is (a) the innermost one and (b) the outermost one?

By using the concept of moment of inertia of the discrete particle system you can find percentage decrease in rotational inertia for the given conditions. The formula of above is given as follows

Moment of inertia for discrete particle system

$\mathit{l}\mathbf{=}\mathbf{\sum }\mathit{m}{\mathit{r}}^{\mathbf{2}}$

For a system with discrete particles, total moment of inertia is given by

$l=\sum m{r}^2$

As we have three particles, moment of inertia of total system

$l_T=m_l{r}_1^2+m_2{r}_2^2+m_3{r}_3^2$

Since $m_1=m_2=m_3=m$

$l_T=m\left(r_1^2+r_2^2+r_3^2\right)$ …(1)

Distance of the innermost particle from the axis r₁ = d

Distance of the innermost particle from the axis r₂ = 2d

Distance of the innermost particle from the axis r₃ = 3d

Hence equation 1 become,

$\begin{array}{rcl}{l}_T& =& m\left(d^2+{\left(2d\right)}^2+{\left(3d\right)}^2\right)\\ & =& m{d}^2\left(1+4+9\right)\\ & =& m\left(d^2+4d^2+9d^2\right)\end{array}$

$l_T=14m{d}^2$ …(2)

To find the percentage of decrease in rotational inertia after removing innermost particle,

$\begin{array}{rcl}{l}_i& =& m\left(r_2^2+r_3^2\right)\\ l_T& =& m\left(4d^2+9d^2\right)\\ l_i& =& 13m{d}^2\end{array}$

Percentage of decrease in rotational inertia

$=\frac{l_T-l_i}{l_T}\times 100$

Percentage of decrease in rotational inertia

$=\frac{14m{d}^2-13m{d}^2}{14m{d}^2}\times 100$

Percentage of decrease in rotational inertia
$$\begin{gathered} =\frac{1}{14}\times 100 \\ =7.14\% \end{gathered}$$

Hence the percentage of decrease in rotational inertia is, 7.14% or 7.1%.

To find the percentage of decrease in rotational inertia after removing outermost particle,

$\begin{array}{rcl}{l}_o& =& m\left(r_1^2+r_2^2\right)\\ & =& m\left(d^2+4d^2\right)\\ & =& 5m{d}^2\end{array}$

Percentage of decrease in rotational inertia

$=\frac{l_T-l_o}{l_T}\times 100$

Percentage of decrease in rotational inertia

$=\frac{14m{d}^2-5m{d}^2}{14m{d}^2}\times 100$

Percentage of decrease in rotational inertia

$$\begin{gathered} =\frac{9}{14}\times 100 \\ =64.3\% \end{gathered}$$

Hence the percentage of decrease in rotational inertia is, 64.3% or 64%.