Magnitude of Change in Velocity in Vertical Circle – Rankers Physics

Vertical Circular Motion: Practice Problem & Solution

A stone is tied to a string of length '$\ell$' and 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 the change in velocity as it reaches a position where the string is horizontal ($g$ being acceleration due to gravity) is: (2004)
$\sqrt{u^2 - g\ell}$
$u - \sqrt{u^2 - 2g\ell}$
$\sqrt{2g\ell}$
$\sqrt{2(u^2 - g\ell)}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Vertical Circular Motion. Understanding the underlying formula is key to arriving at the correct answer below:

Velocity at horizontal position is $v_2 = \sqrt{u^2 - 2g\ell}$. Since velocities are perpendicular, magnitude of change in velocity is $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.

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