Question

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. (Hint: Use the parallel-axis theorem.)

Answer 1

Answer:

The  moment of inertia about an axis through the center and perpendicular to the plane of the square is

    $I_s = \frac{Ma^2}{3}$

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

    $m_1 = \frac{M}{4}$

Generally the moment of inertia of one side of the square is mathematically represented as

        $I_g = \frac{1}{12} * m_1 * a^2$

Generally given that $m_1 = m_2 = m_3 = m_4 = m$ it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  $m_1$

  So

       $I _g= \frac{1}{12} * \frac{M}{4} * a^2$

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

      $I_a = I_g + m [\frac{q}{2} ]^2$

=>    $I_a = I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2$

substituting for $I_g$

=>    $I_a = \frac{1}{12} * \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2$

=>    $I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}$

=>    $I_a = \frac{Ma^2}{12}$

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

      $I_s = 4 * I_a$

=>   $I_s = 4 * \frac{Ma^2}{12}$

=>   $I_s = \frac{Ma^2}{3}$