Question

What is the moment of inertia I of this assembly about the axis through which it is pivoted?Express the moment of inertia in terms of mr, m1, m2, and x. Keep in mind that the length of the rod is 2x, not x.

What is the moment of inertia I of this assembly about the axis through which it is pivoted?

Express the moment of inertia in terms of mr, m1, m2, and x. Keep in mind that the length of the rod is 2x, not x.

Answer 1

Concepts and reason

The concept used in calculating Moment of inertia of discrete and continuous mass distribution.

First, find the moment of inertia of the given system about the axis. The sum of these moments of inertia will be the total moment of inertia of the system.

Fundamentals

The moment of inertia is always defined with respect to the axis of rotation.

Moment of inertia of point particle with mass $M$ is given by the product of its mass and the square of its distance from the axis.

$I = M{R^2}$

Here, $R$ is the distance of the point from the axis of rotation.

Moment of inertia for continuous mass systems is found out by integrating the mass:

$I = \int\limits_0^M {{R^2}dm}$

If the moment of inertia for $n$ components of a system is ${I_1},{I_2},{I_3} \cdots {I_n}$ respectively then the total moment of inertia of the system is given by their sum:

$I = {I_1} + {I_2} + {I_3} \cdots + {I_n}$

Moment of inertia of a rod of mass $M$ and length $L$ about its center is given by:

$I = \frac{{M{L^2}}}{{12}}$

Substitute ${m_r}$ for $M$ and $2x$ for$L$:

${I_r} = \frac{{{m_r}{{\left( {2x} \right)}^2}}}{{12}}$

$= \frac{{{m_r}{x^2}}}{3}$ ...$\left( 1 \right)$

Moment of inertia of mass ${m_1}$, distance $x$ from the axis:

${I_1} = {m_1}{x^2}$ ...$\left( 2 \right)$

Moment of inertia of mass ${m_2}$, distance $x$ from the axis:

${I_2} = {m_2}{x^2}$ ...$\left( 3 \right)$

The total moment of inertia of the system is given by:

${I_{tot}} = {I_r} + {I_1} + {I_2}$

Substitute from $\left( 1 \right),\left( 2 \right),\left( 3 \right)$to the above relation:

$\begin{array}{l}\\{I_{tot}} = \frac{{{m_r}{x^2}}}{3} + {m_1}{x^2} + {m_2}{x^2}\\\\ = \left( {\frac{{{m_r}}}{3} + {m_1} + {m_2}} \right){x^2}\\\end{array}$

Ans:

The moment of inertia of the assembly is $\left( {\frac{{{m_r}}}{3} + {m_1} + {m_2}} \right){x^2}$.