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?

Answers

Answer 1

Answer:

The value is [tex]KE = 259.6 \ J[/tex]

Explanation:

From the question we are told that

     The weight of the horizontal solid disk is  [tex]W = 805 \ N[/tex]

      The radius of the horizontal solid disk is  [tex]r = 1.58 \ m[/tex]

      The force applied by the child is  [tex]F = 49.5 \ N[/tex]

       The time considered is  [tex]t = 2.95 \ s[/tex]

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

          [tex]m_h = \frac{W}{ g}[/tex]

=>       [tex]m_h = \frac{805}{ 9.8 }[/tex]

=>       [tex]m_h = 82.14 \ N[/tex]

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

         [tex]I = \frac{1}{2} * m * r^ 2[/tex]

=>      [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]  

=>      [tex]I = 102.5 \ kg \cdot m^2[/tex]

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

           [tex]T = I * \alpha = F * r[/tex]

=>         [tex]\alpha = \frac{ F * r }{ I }[/tex]

=>         [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]

=>         [tex]\alpha = 0.7628[/tex]

Gnerally from kinematic equation we have that

         [tex]w = w_o + \alpha t[/tex]

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

So

           [tex]w = 0 + 0.7628 * 2.95[/tex]

=>        [tex]w = 2.2503 \ rad/s[/tex]

Generally the kinetic energy is mathematically represented as

        [tex]KE = \frac{1}{2} * I * w^2[/tex]

=>      [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]

=>      [tex]KE = 259.6 \ J[/tex]


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