$A$ particle of mass $m$ is rotating in a plane in a circular path of radius $r$. Its angular momentum is $L$. The centripetal force acting on the particle is:

$A$ $500 \,kg$ car takes a round turn of radius $50 \,m$ with a velocity of $36 \,km/h$. The centripetal force is (in $\,N$)

$A$ thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{g / R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega = \sqrt{2g / R}$? Neglect friction.

$A$ smooth wire of length $2\pi r$ is bent into a circle and kept in a vertical plane. $A$ bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter $AB$,as shown in the figure,the bead is at rest with respect to the circular ring at position $P$ as shown. Then the value of $\omega^2$ is equal to

$A$ man carrying a monkey on his shoulder is cycling smoothly on a circular track of radius $9 \,m$ and completes $120$ revolutions in $3$ minutes. The magnitude of the centripetal acceleration of the monkey is (in $m/s^2$):

$A$ particle rotates in a horizontal circle of radius $R$ in a conical funnel with constant speed $V$. The inner surface of the funnel is smooth. The height of the plane of the circle from the vertex of the funnel is (where $g$ is the acceleration due to gravity):