In order to just complete the vertical circular motion,the ratio of kinetic energy of a particle at the highest point to that at the lowest point is

The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

$A$ particle tied to one end of a string is being rotated in a vertical circle with constant frequency. The tension in the string at points $A, B, C$ and $D$ are $T_1, T_2, T_3$ and $T_4$ respectively. Then:

Consider a particle moving with a minimum speed $v$ at the highest point of a vertical circle of radius $R$. If the radius of the circle is doubled, the corresponding minimum speed will be:

$A$ block of mass $m$ is given a velocity $20\ m/s$ on a horizontal rough surface of length $x\ m$. After the rough surface,there is a vertical circular path of radius $R = 2\ m$ which is frictionless. The block enters the circular path with the minimum velocity required to complete the path. Find the length of the rough surface $x$ if the coefficient of friction is $\mu = 0.5$.

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. The minimum value of $H$ required so that the particle makes a complete vertical circle is given by (in $R$)