$A$ 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).

$A$ 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 $C$.

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

$A$ weightless thread can bear tension up to $3.7 \ kg \ wt$. $A$ stone of mass $500 \ g$ is tied to it and revolved in a circular path of radius $4 \ m$ in a vertical plane. The maximum angular velocity of the stone will be (acceleration due to gravity,$g = 10 \ m/s^2$) (in $rad/s$)

$A$ heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when:

$A$ 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 $R$)