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In her hand a softball pitcher swings a ball of mass 0.251 kg around a vertical
circular path of radius 59.5 cm before releasing it from her hand. The pitcher
maintains a component of force on the ball of constant magnitude 30.2 N in the
direction of motion around the complete path. The speed of the ball at the top of the
circle is 14.4 m/s. If she releases the ball at the bottom of the circle, what is its
speed upon release?
The problem can be solved easily by using the work energy rule i.e.
the work done on an object = gain in its kinetic energy.
Here the force component in tangential direction is parallel to direction of motion and
hence angle q between the small displacement ds and the force component is zero.
Hence the work done is given by
 
dW  F  ds  Fds cos   Fds
But the ball is moving on a semicircular distance from the top to the bottom and
hence the distance s is changes from 0 to R. So the total work done is given by
W   Fds  F  ds  FR  30.2  3.14  0.595  56.42 J
(This result can be obtained directly using Fs because the force is always in the
direction of motion.)
If the final velocity at the end of path be v then increase in the kinetic energy is(1/2)m(v2 – u2)
Hence using the work energy rule we have
(1/2)0.251(v2 – 14.42) = 56.42
v2 = 656.92
v = 25.63 m/s