$A$ particle of mass $100 \ g$ is performing vertical circular motion as shown in the figure. Find the tangential acceleration at the given position in $m/s^2$. (Take $g = 10 \ m/s^2$)

$A$ $1\, kg$ stone at the end of a $1\, m$ long string is whirled in a vertical circle at a constant speed of $4\, m/s$. The tension in the string is $6\, N$ when the stone is at $(g = 10\, m/s^2)$.

$A$ rod of length $L$ is hung from one of its ends,and a mass $m$ is attached to its free end. What tangential velocity must be imparted to $m$ so that it reaches the top of the vertical circle? ($g$ = acceleration due to gravity)

$A$ stone of mass $m$ $kg$ is tied to a string of length $L$ $m$ and moved in a vertical circle of radius $49$ $cm$ in a vertical plane. If it completes $30$ revolutions per minute,the tension in the string when it is at the lowermost point is nearly. [Take $\pi^2=10$ and acceleration due to gravity,$g=10$ $m/s^2$]

$A$ 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

$A$ stone of mass $2 \,kg$ tied to a light inextensible string of length $\frac{5}{3} \,m$ is whirling in a circular path in a vertical plane. If the ratio of the maximum tension to the minimum tension in the string is $4$, then the speed of the stone at the highest point of the circle is $\left(g=10 \,ms^{-2}\right)$.