# A thin, uniform rod is bent into a square of side length a. If the total mass is

A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

William Yazzie

Consider the square made of 4 thin uniform rods of length a.
The blue small circles in the middle of each rod represents its center of mass.
If we considere a perpendicular axis of rotation through each center of mass: $$\displaystyle{I}_{{{c}{m}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}$$
To find the moment of inertia of each rod about an axis through the center of the square and perpendicular to its plane (point P in the figure), use the Parallel - Axis Theorem: $$\displaystyle{I}_{{{p}}}={I}_{{{c}{m}}}+{\left({\frac{{{M}}}{{{4}}}}\right)}{\left({\frac{{{a}}}{{{2}}}}\right)}^{{{2}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}+{\frac{{{1}}}{{{4}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}={\frac{{{1}}}{{{12}}}}{M}{a}^{{{2}}}$$
The total moment of inertia is equal to four times the value f or each rod (the figure is symmetric): $$\displaystyle{I}_{{{p}\ to{t}{a}{l}}}={4}{I}_{{{p}}}={4}{\left({\frac{{{1}}}{{{12}}}}\right)}{M}{a}^{{{2}}}={\frac{{{1}}}{{{3}}}}{M}{a}^{{{2}}}$$

###### Have a similar question?
Ralph Lester
The moment of inertia about an axis through the center and perpendicular to the plane of the square is
$$\displaystyle{I}_{{{s}}}={\frac{{{M}{a}^{{{2}}}}}{{{3}}}}$$
Explanation:
From the question we are told that
The length of one side of the square is a
The total mass of the square is M
Generally the mass of one size of the square is mathematically evaluated as
$$\displaystyle{m}_{{{1}}}={\frac{{{M}}}{{{4}}}}$$
Generally the moment of inertia of one side of the square is mathematically represented as
$$\displaystyle{I}_{{{g}}}={\frac{{{1}}}{{{12}}}}\star{m}_{{{1}}}\star{a}^{{{2}}}$$
Generally given that $$\displaystyle{m}_{{{1}}}={m}_{{{2}}}={m}_{{{3}}}={m}_{{{4}}}={m}$$ means that this moment inertia evaluated above apply to every side of the square
Now substituting for $$\displaystyle{m}_{{{1}}}$$
So
$$\displaystyle{I}_{{{g}}}={\frac{{{1}}}{{{12}}}}\star{\frac{{{M}}}{{{4}}}}\star{a}^{{{2}}}$$
Now according to parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as
$$\displaystyle{I}_{{{a}}}={I}_{{{g}}}+{m}{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}$$
$$\displaystyle\Rightarrow{I}_{{{a}}}={I}_{{{g}}}+{\frac{{{M}}}{{{4}}}}\star{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}$$
substituting for $$\displaystyle{I}_{{{g}}}$$
$$\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{1}}}{{{2}}}}\star{\frac{{{M}}}{{{4}}}}\star{a}^{{{2}}}+{\frac{{{M}}}{{{4}}}}\star{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}$$
$$\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{M}{a}^{{{2}}}}}{{{48}}}}+{\frac{{{M}{a}^{{{2}}}}}{{{16}}}}$$
$$\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{M}{a}^{{{2}}}}}{{{12}}}}$$
Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as
$$\displaystyle{I}_{{{s}}}={4}\star{I}_{{{a}}}$$
$$\displaystyle\Rightarrow{I}_{{{s}}}={4}\star{\frac{{{M}{a}^{{{2}}}}}{{{12}}}}$$
$$\displaystyle\Rightarrow{I}_{{{s}}}={\frac{{{M}{a}^{{{2}}}}}{{{3}}}}$$