Question

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of the soft-metal head is 8000 kg/m3. The longitudinal axis of the cylindrical head is normal to the x-axis. Assume that the handle does not penetrate the head.

Answer 1

Complete Question

Diagram for this  shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

        The density `of wooden handle is  $\rho_w = 860 kg/m^3$

        The density `of soft-metal head  is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

          $m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is  mathematically obtain with the formula

             $A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter  of the wooden handle which from the diagram is

       $27mm = \frac{27}{1000} = 0.027m$

So  $A_w = \frac{\pi}{4} * 0.027^2$

      $l_w$ is the length of the the wooden handle which is given in the diagram as   $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

      $m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

            $= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

           $m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is  mathematically obtain with the formula

            $A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter  of the soft-metal head which from the diagram is            

       $36mm = \frac{36}{1000} = 0.036m$

So  $A_s = \frac{\pi}{4} * 0.036^2$

 $l_s$ is the length of the the soft-metal head which is given in the diagram

     as   $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass  

                  $m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

                       $=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

                  $(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$  

Substituting values

                   $(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

                              $=5.12*10^{-3}kg \cdot m^2$  

Generally the mass moment of inertia about x-axis for the soft-metal head is

    $(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

                   $b=l_w +\frac{d_s}{2}$

Substituting values

                 $b = 0.315 + \frac{0.036}{2}$

                    $= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

                            $(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

                                      $=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

         $I_{xx} = (I_{xx})_w + (I_{xx})_s$

                $= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

        $I_{xx} = 5.12*10^{-3} +0.113$

               $I{xx}= 0.119 kg \cdot m^2$