$A$ particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $\ell$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is

$A$ person with his hand in his pocket is skating on ice at the rate of $10 \ m/s$ and describes a circle of radius $50 \ m$. What is his inclination to the vertical? (Take $g = 10 \ m/s^2$)

$A$ particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance $r$. The speed of the particle is

"Momentum and changes in momentum are not always in the same direction." Explain with a suitable example.

$A$ 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.

$A$ 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.