$A$ particle moves with constant angular velocity in a circular path of a certain radius and is acted upon by a certain centripetal force $F$. If the angular velocity is kept the same but the radius of the path is halved,the new force will be

Assume a proton is rotating along a circular path of radius $1 \,m$ under a centrifugal force of $4 \times 10^{-12} \,N$. If the mass of the proton is $1.6 \times 10^{-27} \,kg$, then its angular velocity of rotation is

$A$ particle is released on a vertical smooth semicircular track from point $X$ so that $OX$ makes an angle $\theta$ from the vertical (see figure). The normal reaction of the track on the particle vanishes at point $Y$ where $OY$ makes an angle $\phi$ with the horizontal. Then

In a conical pendulum,the bob of mass $m$ moves in a horizontal circle of radius $r$ with uniform speed $V$. The string of length $L$ describes a cone of semi-vertical angle $\theta$. The centripetal force acting on the bob is $(g = \text{acceleration due to gravity})$

$A$ wire of length $2.5 \ m$ is fixed at one end and a box of mass $4 \ kg$ is tied at the other end. If the wire rotates in a horizontal circle about the fixed end with $\frac{2}{\pi} \ rev/s$ rotations per second,then the tension in the wire is (in $N$)

$A$ small $100 \ g$ sleeve $B$ can slide on a smooth,circular and rigid wire frame $A$ of radius $5 \ m$ placed in a vertical plane. The wire frame is rotating about its vertical diameter at $2 \ rad/s$. When the sleeve is brought to a particular angular position $\theta$ (measured from the vertical downward axis) other than the bottom and the top of the ring,the sleeve will not slide on the wire frame. Calculate the normal force $N$ (in Newtons) of interaction between the sleeve and the wire frame at this position.