Question

Problem 24.3 The assembly is made from a steel hemisphere, rho st = 7. 80 Mg/m3 , and an aluminum cylinder, rho al = 2. 70 Mg/m3 . If the height of the cylinder is h = 180 mm, determine the location z of the mass center of the assembly.

Answer 1

Answer:

0.12 m    

Explanation:

The mass of the steel hemisphere is:

$m_h=\rho_s\times \frac{2}{3}\pi r^2\\m_h = 7.8\times 10^3 \times \frac{2}{3}\pi (0.16)^2\\m_h=66.9 kg$

Mass of the aluminium cylinder is:

$m_c=\rho_a\times \pi r^2 h\\m_c=2.7\times 10^3 \times \pi (0.08)^2(0.18) \\m_c =9.7 kg$

mass center of steel hemisphere from the bottom:

$z_1=r-3r/8\\z_1=0.16-(3\times 0.16/8) =0.1 m$

mass center of aluminium cylinder from the bottom:

$z_2=r+h/2 \\z_2=0.16+0.18/2=0.25 m$

center of mass is

$Z=\frac{m_hz_1+z_2m_c}{m_s+m_c}\\Z=\frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76} = 0.12 m$

Answer 2

The center of mass will be "0.12 m".

Given:

• $\rho_s = 7.8 \ Mg/m^3$
• $\rho_a = 2.70 \ Mg/m^3$
• Height of cylinder, $h = 180 \ mm$

Now,

The mass of the steel hemisphere will be:

→ $m_h = \rho_s\times \frac{2}{3} \pi r^2$

By putting the values, we get

         $= 7.8\times 10^3\times \frac{2}{3} \pi (0.16)^2$

         $= 66.9 \ kg$

and,

The mass of aluminum cylinder will be:

→ $m_c = \rho_a \times \pi r^2 h$

        $= 2.7\times 10^3\times \pi (0.08)^2 (0.18)$

        $= 9.7 \ kg$

Now,

The mass center of steel hemisphere will be:

→ $z_1 = r - \frac{3r}{8}$

       $= 0.16-(3\times \frac{0.16}{8} )$

       $= 0.1 \ m$

The mass center of aluminum hemisphere will be:

→ $z_2 = r+\frac{h}{2}$

       $= 0.16+\frac{0.18}{2}$

       $= 0.25 \ m$

hence,

The center of mass will be:

→ $Z = \frac{m_h z_1 +z_2 m_c}{m_s+m_c}$

By putting the values, we get

      $= \frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76}$

      $= 0.12 \ m$

Thus the above answer is right.

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