Question

A stone tied 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 lowest position and has a speed u . Find the magnitude of the cgange in its velocity as it reaches a position, where the string is horizontal.

Answer 1

The magnitude of the change in its velocity as it reaches a position, where the string is horizontal is:

• Using Work-Energy Theorem, $\frac{1}{2} mu^2 - \frac{1}{2} mv^2 = mgL$

        Here u = Speed at the lowest point

                  v = Speed at horizontal point

                 m = mass of the particle

                  g = acceleration due to gravity

                  L= length of the string

• Solving the equation, we get: $v^{2} =u^2 -2gL$

• Initial velocity vector = u i

        Final velocity vector = v j =$\sqrt{u^2 -2gL}$ j

        Δv=$\sqrt{u^2 -2gL}$ j - u i

• Magnitude of the change in velocity = $\sqrt{(u^2 -2gL) + u^2} = \sqrt{2u^2 -2gL}$