Question

A ball of mass,m, is attached to a string of length,l . It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion.Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, theball's speeds are vt and vb, and the corresponding tensions in the string are Tt and Tb. Tt and Tb have magnitudes Tt and Tb . Find Tb-Tt, the difference betweenthe magnitude of the tension in the string at the bottom relative to that at the top of the circle.
Express the difference in tension in terms of m and g. The quantities vt and vb should not appear in your final answer.

Answer 1

At the top of the circle, the weight mg of the object and the tension Tt in the string are both acting down. Therefore, taking the upward direction aspositive:
- Tt - mg = - m vt^2 / L ...(1)

At the bottom, the weight acts downwards, and the tension upwards:
Tb - mg = m vb^2 / L ...(2)

If you try swinging something on a string in a vertical circle, you will discover that the speed at the top is less than the speed at the bottom. If thespeed at the bottom is insufficiently great, the string will go slack before the top, and the object will not complete the circle.

Let the kinetic energy of the object at the top be kt and at the bottom be kb. Taking the potential energy at the bottom to be 0, then the potential energyat the top is 2Lmg.

Equating total energy at the top and bottom:
kb = kt + 2Lmg
m vb^2 / 2 = m vt^2 / 2 + 2Lmg
vb^2 - vt^2 = 4Lg ...(3)

Adding (1) &(2):
Tb - Tt = m(vb^2 - vt^2) / L

Substituting for vb^2 - vt^2 from (3):
Tb - Tt = 4mg.

Answer 2

Tb-Tt = 6mg