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

$A$ 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:

$A$ mass attached to one end of a string crosses the topmost point on a vertical circle with critical speed. Its centripetal acceleration when the string becomes horizontal will be ($g =$ gravitational acceleration).

In the case of vertical circular motion of a particle attached to a thread of length $r$, if the tension in the thread is zero at an angle of $30^{\circ}$ with the horizontal as shown in the figure, the velocity at the bottom point $(A)$ of the circular path is ($g =$ gravitational acceleration).

$A$ bucket filled with water is revolved in a vertical circle of radius $4 \ m$ and the water just does not fall down. The time period of revolution will be ........ $s$.

$A$ body attached to one end of a string performs motion along a vertical circle. Its centripetal acceleration,when the string is horizontal,will be [ $g$ = acceleration due to gravity]