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In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.9 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 16.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?

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Answer:

The velocity is [tex]v_b = 20.17 \ m/s[/tex]

Explanation:

From the question we are told that

   The mass of the ball is  [tex]m = 0.245 \ kg[/tex]

   The radius is  [tex]r =  59.8 \  cm  =  0.598 \ m[/tex]

   The force is  [tex]F =  30.9 \ N[/tex]

   The speed of the ball is  [tex]v = 16.0 \ m/s.[/tex]

Generally the kinetic energy at the top of the circle is mathematically represented as

    [tex]K_t  =  \frac{1}{2} *  m  *  v^2[/tex]

=> [tex]K_t  =  \frac{1}{2} *  0.245   *  16.0 ^2[/tex]  

=> [tex]K_t  =  31.36 \ J[/tex]  

Generally the work done by the force applied on the ball from the top to the bottom  is mathematically represented as

       [tex]W =  F *  d[/tex]

Here  d is the length of  a semi - circular arc which is mathematically represented as

       [tex]d =  \pi * r[/tex]

So

      [tex]W =  30.9 *  0.598  [/tex]

      [tex]W = 18.48 \ J [/tex]

Generally the kinetic energy at the bottom is mathematically represented as

      [tex]K_b  =  \frac{1}{2} *  m *  v_b^2[/tex]

=>    [tex]K_b  =  \frac{1}{2} *  0.245  *  v_b^2[/tex]

=>   [tex]K_b  =  0.1225  *  v_b^2[/tex]

From the law of energy conservation

     [tex]K_t +  W  =K_b[/tex]

=>    [tex]31.36+  18.48 = 0.1225  *  v_b^2[/tex]

=>    [tex]v_b = 20.17 \ m/s[/tex]