$A$ 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 $R$)

The bob of a simple pendulum is of mass $10 \, g$. It is suspended with a thread of $1 \, m$. If we hold the bob so as to stretch the string horizontally and release it,what will be the tension at the lowest position? (Take $g = 10 \, m/s^2$)

$A$ point mass '$m$' attached at one end of a massless,inextensible string of length '$\ell$' performs a vertical circular motion and the string rotates in a vertical plane,as shown in the diagram. The increase in the centripetal acceleration of the point mass when it moves from point $A$ to point $C$ is $(g = \text{acceleration due to gravity})$:

$A$ small body of mass $m$ slides down from the top of a hemisphere of radius $r$. The surfaces of the block and the hemisphere are frictionless. The height at which the body loses contact with the surface of the sphere is

$A$ particle of mass $200 \,g$ is moving in a circle of radius $2 \,m$. The particle is just 'looping the loop'. What are the speed of the particle and the tension in the string at the highest point of the circular path? (Take $g = 10 \,m/s^2$)

$A$ 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.