Question

A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answer 1

Answer:

The moment of inertia of the system is  $I = 400.5 \ kg \cdot m^2$

Explanation:

From the question we are told that

    The mass of the platform is  $m = 133\ kg$

     The  radius of the  platform is  r = 1.95 m

     The mass of the person is $m_p = 62.7 \ kg$

     The position of the person from the center is  $d = 1.19 \ m$

       The mass of the dog is $m_D = 28.5 \ kg$

     The position of the dog from the center is  $D = 1.45 \ m$

   

Generally the moment of inertia of the platform with respect to its axis is  mathematically represented as

       $I_p = \frac{m r^2}{2}$

The  moment of inertia of the person with respect to the axis is mathematically represented as

        $I_z = m_p* d^2$

The  moment of inertia of the dog with respect to the axis is mathematically represented as

       $I_D = m_d * D^2$

So the moment of inertia of the system about the axis  is mathematically evaluated as

        $I = I_p + I_z + I_D$

=>      $I = \frac{mr^2}{2} + m_p * d^2 + m_d * D^2$

substituting values  

            $I = \frac{(133) * (1.95)^2}{2} + (62.7) * (1.19)^2 + (28.5) * (1.45)^2$

          $I = 400.5 \ kg \cdot m^2$