Conservation of Energy softball pitcher

AI Thread Summary
A softball pitcher rotates a 0.250 kg ball in a vertical circular path with a radius of 0.4 m, applying a force of 33.0 N. The ball's speed at the top of the circle is 15.0 m/s, and the problem asks for its speed upon release at the bottom. The conservation of energy approach is discussed, equating kinetic and potential energy, while an alternative method using work and kinematics is also suggested. The calculations indicate a potential release speed of approximately 23.595 m/s. The discussion emphasizes the importance of considering forces and energy transformations in solving the problem.
nickhassan38
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Homework Statement


A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.4 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 15.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?


Homework Equations


KE = 1/2mv^2
PE = mgh

The Attempt at a Solution


1/2mv^2 = 1/2mv^2 + mgh
225.125 = .125v^2 +26.4
198.725 = .125v^2
7.43
 
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nickhassan38 said:

Homework Statement


A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.4 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 15.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?

I think you may want to approach this problem from an approach using work, rather than potential energy. Consider that the ball is on a circular arc from the top of the pitch to the bottom, where it is released. The force applied by the pitcher to the ball is always aligned with its instantaneous velocity (that is, the ball is undergoing a tangential acceleration). How would you work out the change in velocity along the path?

(This problem can also be done entirely with kinematics.)
 
23.595?
 
nickhassan38 said:
23.595?

Would you like to show us how you found that?
 
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