Arun
Last Activity: 6 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