$A$ vehicle of mass $200\,kg$ is moving along a levelled curved road of radius $70\,m$ with an angular velocity of $0.2\,rad/s$. The centripetal force acting on the vehicle is $.........\,N$.

$A$ string of length $L$ fixed at one end carries a body of mass $m$ at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The string makes an angle $\theta$ with the vertical. The angular frequency of the body is $\omega$. The tension in the string is

$A$ 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 doubled,keeping the radius the same,the new force will 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.

$A$ motorcycle driver doubles their velocity when taking a turn. The force exerted outwardly will be

$A$ car is moving with velocity $v$ at the top of a semi-circular hill of radius $40 \,m$ such that the normal force on it is zero. Find the velocity $(v)$ of the car. [Use $g=10 \,ms^{-2}$] (in $\,ms^{-1}$)