Question

In her hand, a softball pitcher swings a ball of mass 0.254 kg around a vertical circular path of radius 59.4 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 29.8 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 15.7 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?​

Answer 1

Given

mass m = 0.254 kg

radius of the vertical circle is r = 59.4 cm = 0.594 m

the constanat force F = 29.8 N

the spped of the ball at top of the circle is v2 =15.7 m/s

the spped of the ball at bottom of the circle is v1 = ? m/s

here the total mechanical energy is same that the total energy at bottom and top is same  

let us consider the bottom point as reference where the gravitational potential energy is zero

at botttom point

E1 = p.e1+k.e1 = mgh1 + 0.5*m*v1^2 = 0 + 0.5*m*v1^2

at top point

E2 = p.e2+k.e2 = mgh2 + 0.5*m*v1^2 = mg2r + 0.5*m*v2^2

equating both energies  

E1 = E2

0 + 0.5*m*v1^2 = 2mgr + 0.5*m*v2^2

v1^2 = 4gr + v2^2

substituting the values

v1^2 = 4*9.8*0.594 +15.7^2

solving for v1

v1 = 16.42 m/s

the speed of the ball at the bottom point is 16.42 m/s