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Natali [406]
2 years ago
5

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 appl

ies 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?
Physics
1 answer:
lozanna [386]2 years ago
8 0

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

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