$A$ body of mass $m$ is moving with speed $V$ along a circular path of radius $r$. Now,the speed is reduced to $\frac{V}{2}$ and the radius is increased to $3r$. For this change,the initial centripetal force needs to be:

$A$ 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.

$A$ cyclist taking a turn bends inwards while a car passenger taking the same turn is thrown outwards. The reason is:

$A$ conical pendulum of length $1\,m$ makes an angle $\theta = 45^\circ$ with respect to the $Z$-axis and moves in a circle in the $XY$ plane. The radius of the circle is $0.4\,m$ and its centre is vertically below $O$. The speed of the pendulum,in its circular path,will be ..... $m/s$ (Take $g = 10\,ms^{-2}$)

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

$A$ vehicle is moving on a track with constant speed as shown in the figure. The apparent weight of the vehicle is: