Question

A small body of mass 'm' hangs at one end of a string of length 'a', the other end of which is fixed. It is given a horizontal velocity 'u' at it's lowest position so that the string would just become slack, when it makes an angle of 60 degrees with the upward drawn vertical line. Find the tension in the string at point of projection ?

Answer 1

   At 60° angle the string becomes slack. That means that The tension in the string becomes zero and negative subsequently.  So we need to get the formula for tension in the string at an angle Ф. Let the speed of the body be v at angle Ф of the string with the vertical.

Initial KE = 1/2 m u²
Gain in PE at angle Ф : m g (1 - cos Ф)
Energy conservation:    1/2 m u² = 1/2 m v² + m g a (1 - cosФ)
=>                    v² = u² - 2 a g (1 - cosФ)      ---(1)

The forces acting on the body are the tension T along the string and the weight mg. As the body moves along a circular path, the forces are balanced along the radius.

          T = mv²/a + mg cosФ
             = m u²/a - 2 m g (1 - cosФ) + mg cosФ
         T  = m u² /a  - m g (2 - 3 cosФ)             -------  (2)

If the string becomes slack,  then T= 0 and later negative.  So at Ф = 60°,
T = 0
=>   u²/a = g (2 - 3 cos60°) = g /2 
=> u = √(ag/2)

Tension at initial point of projection Ф = 0°,  (using equation (2)
      T = m g/2 - mg (2 -3) = 3 mg/2

Answer 2

Answer:

Explanation:In this question we will apply the concept of circular motion in vertical plane.