Question

A child of mass 46.2 kg sits on the edge of a merry-go-round with radius 1.9 m and moment of inertia 130.09 kg m2 . The merrygo-round rotates with an angular velocity of 2.4 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.779 m from the center. Now what is the angular velocity of the merry-go-round

Answer 1

Answer:

The angular velocity is $w_f = 4.503 \ rad/s$

Explanation:

From the question we are told that

   The mass of the child is  $m_c = 46.2 \ kg$

    The radius of the merry go round is  $r = 1.9 \ m$

     The moment of inertia of the merry go round is $I_m = 130.09 \ kg \cdot m^2$

      The angular velocity of the merry-go round is  $w = 2.4 \ rad/s$

       The position of the child from the center of the merry-go-round is  $x = 0.779 \ m$

According to the law of angular momentum conservation

    The initial angular momentum  =  final  angular momentum

So  

       $L_i = L_f$

=>     $I_i w_i = I_fw_f$

Now   $I_i$ is the initial moment of inertia of the system which is mathematically represented as

          $I_i = I_m + I_{b_1}$

Where  $I_{b_i}$ is the initial moment of inertia of the boy which is mathematically evaluated as

      $I_{b_i} = m_c * r$

substituting values

      $I_{b_i} = 46.2 * 1.9^2$

      $I_{b_i} = 166.8 \ kg \cdot m^2$

Thus

   $I_i =130.09 + 166.8$        

   $I_i = 296.9 \ kg \cdot m^2$      

Thus  

     $I_i * w_i =L_i= 296.9 * 2.4$

       $L_i = 712.5 \ kg \cdot m^2/s$

Now  

     $I_f = I_m + I_{b_f }$

Where  $I_{b_f}$ is the final  moment of inertia of the boy which is mathematically evaluated as

         $I_{b_f} = m_c * x$

substituting values

         $I_{b_f} = 46.2 * 0.779^2$

         $I_{b_f} = 28.03 kg \cdot m^2$

Thus

      $I_f = 130.09 + 28.03$

      $I_f = 158.12 \ kg \ m^2$

Thus

     $L_f = 158.12 * w_f$

Hence

      $712.5 = 158.12 * w_f$

       $w_f = 4.503 \ rad/s$