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52739
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?
Answer:
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
R
R
0
0
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
or
or
(1/2)0.251(v2 – 14.42) = 56.42
v2 = 656.92
v = 25.63 m/s