# A 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 its lowest position so that string would just become slack when it makes angle of 60 with upward drawn verticle line find the tension in the string at point of projection

Arun
25757 Points
4 years ago
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