Question

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?

Answer 1

Answer:

The length of the rod for the condition on the question to be met is $L = 1.5077 \ m$

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is $M = 3.41 \ kg$

           The mass of each small bodies is  $m = 0.249 \ kg$

           The moment of inertia of the three-body system with respect to the described axis is   $I = 0.929 \ kg \cdot m^2$

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        $I = I_r + 2 I_m$

Where  $I_r$ is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        $I_r = \frac{ML^2 }{12}$

And   $I_m$ the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           $I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}$

Thus  $2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}$

Hence

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

=>   $I = [\frac{M}{12} + \frac{m}{2}] L^2$

substituting vales  we have  

        $0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2$

       $L = \sqrt{\frac{0.929}{0.40867} }$

      $L = 1.5077 \ m$