Question

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child applies a force 49.5 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?

Answer 1

Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

     The weight of the horizontal solid disk is  $W = 805 \ N$

      The radius of the horizontal solid disk is  $r = 1.58 \ m$

      The force applied by the child is  $F = 49.5 \ N$

       The time considered is  $t = 2.95 \ s$

Generally the mass of the  horizontal solid disk is mathematically represented as

          $m_h = \frac{W}{ g}$

=>       $m_h = \frac{805}{ 9.8 }$

=>       $m_h = 82.14 \ N$

Generally the moment of inertia  of the horizontal solid disk is mathematically represented as  

         $I = \frac{1}{2} * m * r^ 2$

=>      $I = \frac{1}{2} * 82.14 * 1.58^ 2$  

=>      $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

           $T = I * \alpha = F * r$

=>         $\alpha = \frac{ F * r }{ I }$

=>         $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=>         $\alpha = 0.7628$

Gnerally from kinematic equation we have that

         $w = w_o + \alpha t$

Here  $w_o$ is the initial angular velocity velocity of the horizontal solid disk  which is  $w_o = 0\ rad/s$

So

           $w = 0 + 0.7628 * 2.95$

=>        $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

        $KE = \frac{1}{2} * I * w^2$

=>      $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=>      $KE = 259.6 \ J$