The minimum velocity $v$ required for a simple pendulum to complete one full rotation in a vertical plane is given. If the length of the string is reduced to one-fourth,what will be the new required velocity?

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and is set to perform vertical circular motion as shown in the figure. Initially,by applying a horizontal velocity $v_0$ at point $A$,the string becomes slack when the bob reaches point $D$. The ratio of the kinetic energy of the bob at points $B$ and $C$ is:

$A$ body of mass $m$ attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ($g =$ gravitational acceleration).

$A$ stone of mass $m$,tied to a string,is being whirled in a vertical circle with a uniform speed. The tension in the string is

$A$ body is placed as shown in the figure. To complete one full vertical circular loop,the height $h$ must be: