Aball whirls around in a vertical circle at the end of a string. the other end of the string is fixed at the center of the circle. assume that the total energy of the ball-earth system remains constant.
(a) what is the tension in the string at the bottom? (use the following as necessary: m for mass of the ball, g for gravitational acceleration, vb for velocity at the bottom, andr for radius of the circle.)
tb =
(b) what is the tension in the string at the top? (use the following as necessary: m, g, vt for velocity at the top, and r.)
tt =
(c) how much greater is the tension at the bottom? (use the following as necessary: m, g.)
tb = tt +

a) Tb = m (g + vb² / R) , b)  Tt = m (g - vt² /R),    c) Tb = Ty + (vb² -vt²) /R

Explanation:

a) For this exercise we will use Newton's second law, at the bottom

       Tb - W = m a

Where the acceleration is centripetal

      a = v² / R

     Tb = W + m vb² / r

    Tb = m (g + vb² / R)

b) the tension at the top

     -Tt -w = m a

      Tt = W - m vt² / R

      Tt = m (g - vt² / R)

c) write the two equations and solve system

     Tb = m (g + vb² / R)

      Tt = m (g - vt² / R)            (-1)

We multiply by (1) and add

      Tb -Ty = vb² /R  -vt² /R

      Tb = Ty + (vb² -vt²) / R

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s