Q. Transformation of kinetic energy to potential energy?

A tiny ball at the end of string is swung in a circular vertical loop. The speed of rotation is reducing to the point that the tension on the string at the top of the loop drops to zero, analyze the system for this condition. The forces on the ball consist of its weight the centrifugal force because of motion along a circular path and the tension from the string which provides the centripetal force.

Ft = (mv2)/l- mg - T_s,

Where F_t is the total of the forces m is the mass of the ball and l is the length of the string and T_s is the tension. by the pinnacle of the loop the forces are in

Equilibrium thus F_t = 0 Let v_t be the velocity of the ball at the top. If the tension falls to zero there we have

mvt²/l= mg,                             so,

vt2= gl                                     and,

vt =√gl.

The kinetic energy of a ball at the top of the loop is

mv _t ²/2.

At the bottom of the loop the kinetic energy is enhanced by the potential energy at the top. From this we are able to determine the velocity at the bottom

mv_b²/2 = (mv_t²/2)+ 2mgl

Solving for v_b:

                                                v_b²= v_t²+ 4gl

v_b²= gl + 4gl = 5gl

v_b =√5gl.

The total force on the ball at the bottom is the sum of the centrifugal force and its weight. This concludes the tension on the string.

Fb = (mv_b²/l) + mg

Fb = (5mgl/l) + mg

Fb = 6mg.

Consequently when the rotation rate is such that the ball experiences no vertical forces at the top of the loop it experiences 6 g's at the bottom.