What will be the minimum speed of the roller-coaster so that the passenger at the top, when becomes upside down, does not fall out (in $m \ s^{-1}$)? Consider the acceleration due to gravity as $10 \ m \ s^{-2}$, and the radius of curvature of the roller coaster is $10 \ m$.

$A$ bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and is set to perform vertical circular motion as shown in the figure. Initially,by applying a horizontal velocity $v_0$ at point $A$,the string becomes slack when the bob reaches point $D$. The ratio of the kinetic energy of the bob at points $B$ and $C$ is:

In order to just complete the vertical circular motion,the ratio of kinetic energy of a particle at the highest point to that at the lowest point is

$A$ simple pendulum has a length of $75 \, cm$. It describes a vertical semi-circle. What will be the velocity of the bob when it passes through the mean position?

$A$ particle of mass $200 \,g$ is whirled in a vertical circle of radius $80 \,cm$ using a massless string. The speed of the particle when the string makes an angle of $60^{\circ}$ with the vertical line is $1.5 \,m/s$. The tension in the string at this position is ........... $N$.

$A$ particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance $h$ below the highest point such that