$A$ weightless thread can support a tension of up to $30 \, N$. $A$ stone of mass $0.5 \, kg$ is tied to it and is revolved in a circular path of radius $2 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

$A$ 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 $C$.

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).

$A$ stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:

Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

$A$ stone of mass $1 \ kg$ is tied to a string $2 \ m$ long and is rotated at a constant speed of $40 \ ms^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 \ ms^{-2}$].

In a circus,a stuntman rides a motorbike on a vertical circular track of radius $r$. Find the minimum speed he must maintain at the highest point of the track.