$A$ stone is tied to a string of length $L$ 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 its velocity as it reaches a position where the string is horizontal is

$A$ body crosses the topmost point of a vertical circle with critical speed. Its centripetal acceleration,when the string is horizontal,will be

$A$ body of mass $m$ connected to a massless and unstretchable string moves in a vertical circle of radius $R$ under gravity $g$. The other end of the string is fixed at the center of the circle. If the velocity at the top of the circular path is $n\sqrt{gR}$,where $n \geq 1$,then the ratio of the kinetic energy of the body at the bottom to that at the top of the circle is:

As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

For a particle moving in a vertical circle,the total energy at different positions along the path (the motion is under gravity) is:

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a horizontal velocity $v_{o}$ at the lowest point $A$ such that it completes a circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point,$C$. This is shown in the figure. Obtain an expression for: $(i) v_{o}$; $(ii)$ the speeds at points $B$ and $C$; $(iii)$ the ratio of the kinetic energies $(K_{B} / K_{C})$ at $B$ and $C$. Comment on the nature of the trajectory of the bob after it reaches the point $C$.