Question

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?

Answer 1

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

   The mass of the ball is  $m = 0.245 \ kg$

   The radius is  $r = 59.8 \ cm = 0.598 \ m$

   The force is  $F = 30.9 \ N$

   The speed of the ball is  $v = 16.0 \ m/s.$

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

    $K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$  

=> $K_t = 31.36 \ J$  

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

       $W = F * d$

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

       $d = \pi * r$

So

      $W = 30.9 * 0.598$

      $W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

      $K_b = \frac{1}{2} * m * v_b^2$

=>    $K_b = \frac{1}{2} * 0.245 * v_b^2$

=>   $K_b = 0.1225 * v_b^2$

From the law of energy conservation

     $K_t + W =K_b$

=>    $31.36+ 18.48 = 0.1225 * v_b^2$

=>    $v_b = 20.17 \ m/s$