A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u . The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x u 2 - g L . The value of x is

Step-by-step solution

Applying conservation of mechanical energy at the point A and B , we get 1 2 m u 2 + 0 = 1 2 m v 2 + m g L ⇒ v = u 2 - 2 g L Now in vector form. v → i = u   i ^ and v → f = v   j ^ Therefore, change in velocity will be, ∆ V → = v j ^ - u i ^ and Δ V → = u 2 + v 2 ⇒ Δ V = u 2 + u 2 - 2 g L = 2 u 2 - g L Hence, x = 2