Question

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?

Short answer

1. Percentage of decrease in rotational inertia when the innermost particle is removed (l_i) is 7.1%.
2. Percentage of decrease in rotational inertia when the innermost particle is removed (l_i) is 64%.

Step-by-step solution

Understanding the given information

1. The mass of each particle is, m=0.0100kg.
2. The length of the rod is, L = 6.00m.

Concept and Formula used for the given question

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

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

(a) Calculation for the percentage decrease in the rotational inertia of the assembly around the rotation axis when that removed particle is (a) the innermost one

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%.

(b) Calculation for the percentage decrease in the rotational inertia of the assembly around the rotation axis when that removed particle is (b) the outermost one

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%.