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 [tex]I{xx}= 0.119 kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The density `of wooden handle is [tex]\rho_w = 860 kg/m^3[/tex]
The density `of soft-metal head is [tex]\rho_s =8000kg/m^3[/tex]
Generally the mass of the wooden can be mathematically obtained with this formula
[tex]m_w = \rho_w A_w l_w[/tex]
Where [tex]A_w[/tex] is mass of wooden handle which is mathematically obtain with the formula
[tex]A_w = \frac{\pi}{4} d^2_w[/tex]
Where [tex]d_w[/tex] is the diameter of the wooden handle which from the diagram is
[tex]27mm = \frac{27}{1000} = 0.027m[/tex]
So [tex]A_w = \frac{\pi}{4} * 0.027^2[/tex]
[tex]l_w[/tex] is the length of the the wooden handle which is given in the diagram as [tex]l_w = 315mm = \frac{315}{1000} = 0.315m[/tex]
Substituting these value into the formula for mass
[tex]m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315[/tex]
[tex]= 0.155kg[/tex]
Generally the mass of the soft-metal head can be mathematically obtained with this formula
[tex]m_s = \rho_s A_s l_s[/tex]
Where [tex]A_s[/tex] is mass of soft-metal head which is mathematically obtain with the formula
[tex]A_s = \frac{\pi}{4} d^2_s[/tex]
Where [tex]d_s[/tex] is the diameter of the soft-metal head which from the diagram is
[tex]36mm = \frac{36}{1000} = 0.036m[/tex]
So [tex]A_s = \frac{\pi}{4} * 0.036^2[/tex]
[tex]l_s[/tex] is the length of the the soft-metal head which is given in the diagram
as [tex]l_s = 90mm = \frac{90}{1000} = 0.090m[/tex]
Substituting these value into the formula for mass
[tex]m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090[/tex]
[tex]=0.733kg[/tex]
Generally the mass moment of inertia about x-axis for the wooden handle is
[tex](I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ][/tex]
Substituting values
[tex](I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ][/tex]
[tex]=5.12*10^{-3}kg \cdot m^2[/tex]
Generally the mass moment of inertia about x-axis for the soft-metal head is
[tex](I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2][/tex]
Where b is the distance from the centroid to the axis of the head which is mathematically given as
[tex]b=l_w +\frac{d_s}{2}[/tex]
Substituting values
[tex]b = 0.315 + \frac{0.036}{2}[/tex]
[tex]= 0.336m[/tex]
Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head
[tex](I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2][/tex]
[tex]=0.113 kg \cdot m^2[/tex]
Generally the mass moment of inertia about x-axis is mathematically represented as
[tex]I_{xx} = (I_{xx})_w + (I_{xx})_s[/tex]
[tex]= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2][/tex]
Substituting values
[tex]I_{xx} = 5.12*10^{-3} +0.113[/tex]
[tex]I{xx}= 0.119 kg \cdot m^2[/tex]
