$A$ body of mass $1\, kg$ is moving in a vertical circular path of radius $1\, m$. The difference between the kinetic energies at its highest and lowest position is ......... $J$.

$A$ 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 $g = 10 \ ms^{-2}$].

$A$ bob of mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in the figure. If the string becomes slack at some point $P$ making an angle $\theta$ with the horizontal,the ratio of the speed $v_p$ of the bob at point $P$ to its initial speed $v_0$ is:

The minimum horizontal speed with which a body must be projected, so that it goes around a smooth vertical circular track of radius $4 \,m$ is $(g=9.8 \,ms^{-2})$. (in $\,ms^{-1}$)

$A$ body slides down a frictionless track which ends in a circular loop of diameter $D$. The minimum height $h$ of the body in terms of $D$ so that it may just complete the loop is:

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then: