Question

Three pairs of balls are connected by very light rods as shown inthe figure.

Rank in order, from smallest to largest, the moments ofinertia I₁,I₂ ,and I₃ about axes through the centers of therods

Answer 1

Concepts and reason

The concept required to solve this problem is moment of inertia of system of rigid bodies.

Initially, write an expression for the moment of inertia of rigid bodies. Later, draw a free body diagram showing the balls with the axis of rotation. Finally, calculate the moment of inertia of different balls and then rank them.

Fundamentals

The expression for the moment of inertia is as follows:

$I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$

Here, ${m_{\rm{i}}}$ is the mass and ${r_{\rm{i}}}$ is the distance from the axis of rotation.

Also, the expression for the moment of inertia is as follows:

$\begin{array}{c}\\I = \int {dl} \\\\ = \int {rdm} \\\end{array}$

Here, I is the moment of inertia, dm is the elemental mass distant from the axis of rotation.

The diagram considering the first system of two equivalent masses is as follows:

Now, calculate the net moment of inertia of the above system about an axis passing through the center of rod using the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

Substitute 2 for i in the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

$\begin{array}{c}\\{I_1} = \sum\limits_2 {{m_{\rm{i}}}r_{\rm{i}}^2} \\\\ = {m_1}r_1^2 + {m_2}r_2^2\\\end{array}$

Here, ${m_1}$ and ${m_2}$ are the masses of the balls 1 and 2 respectively, ${r_1}$ and ${r_2}$ is the distance of the ball 1 and 2 from the axis of rotation.

Substitute m for ${m_1}$ , m for ${m_2}$ , and $\frac{R}{2}$ for ${r_1}$ and ${r_2}$ in the equation ${I_1} = {m_1}r_1^2 + {m_2}r_2^2$ .

$\begin{array}{c}\\{I_1} = m{\left( {\frac{R}{2}} \right)^2} + m{\left( {\frac{R}{2}} \right)^2}\\\\ = 2m{\left( {\frac{R}{2}} \right)^2}\\\\ = \frac{{m{R^2}}}{2}\\\end{array}$

Thus, the inertia of the system having first two masses is,

${I_1} = \frac{{m{R^2}}}{2}$

The diagram considering the second system of two equivalent masses is as follows:

Now, calculate the net moment of inertia of the above system about an axis passing through the center of rod using the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

Substitute 2 for i in the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

$\begin{array}{c}\\{I_2} = \sum\limits_2 {{m_{\rm{i}}}r_{\rm{i}}^2} \\\\ = {m_1}r_1^2 + {m_2}r_2^2\\\end{array}$

Here, ${m_1}$ and ${m_2}$ are the masses of the balls 1 and 2 respectively, ${r_1}$ and ${r_2}$ is the distance of the ball 1 and 2 from the axis of rotation.

Substitute 2m for ${m_1}$ , m for ${m_2}$ , and $\frac{R}{4}$ for ${r_1}$ and ${r_2}$ in the equation ${I_2} = {m_1}r_1^2 + {m_2}r_2^2$ .

$\begin{array}{c}\\{I_2} = 2m{\left( {\frac{R}{4}} \right)^2}\\\\ = \frac{{m{R^2}}}{8}\\\end{array}$

Thus, the inertia of the system having second two masses is,

${I_2} = \frac{{m{R^2}}}{8}$

The diagram considering the third system of two equivalent masses is as follows:

Now, calculate the net moment of inertia of the above system about an axis passing through the center of rod using the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

Substitute 2 for i in the equation $I = \sum\limits_{\rm{i}} {{m_{\rm{i}}}r_{\rm{i}}^2}$ .

$\begin{array}{c}\\{I_3} = \sum\limits_2 {{m_{\rm{i}}}r_{\rm{i}}^2} \\\\ = {m_1}r_1^2 + {m_2}r_2^2\\\end{array}$

Here, ${m_1}$ and ${m_2}$ are the masses of the balls 1 and 2 respectively, ${r_1}$ and ${r_2}$ is the distance of the ball 1 and 2 from the axis of rotation.

Substitute $\frac{m}{2}$ for ${m_1}$ , $\frac{m}{2}$ for ${m_2}$ , and $R$ for ${r_1}$ and ${r_2}$ in the equation ${I_3} = {m_1}r_1^2 + {m_2}r_2^2$ .

$\begin{array}{c}\\{I_3} = \left( {\frac{m}{2}} \right){\left( R \right)^2} + \left( {\frac{m}{2}} \right){\left( R \right)^2}\\\\ = 2\left( {\frac{m}{2}} \right){\left( R \right)^2}\\\\ = m{R^2}\\\end{array}$

Thus, the inertia of the system having third two masses is,

${I_3} = m{R^2}$

Now, ranking the moment of inertia from smallest to largest is as follows:

${I_2} < {I_1} < {I_3}$

Ans:

The ranking of the moment of inertia from smallest to largest is ${I_2} < {I_1} < {I_3}$ .