A light inextensible string has one end fixed at point $A$. It runs through a smooth bead $B$ of mass $0.3\,\text{kg}$, and the other end is fixed at point $C$, directly beneath $A$. The bead $B$ moves at constant speed in a horizontal circle of radius $0.6\,\text{m}$ whose centre lies between $A$ and $C$. The string is at an angle of $30^{\circ}$ to the vertical at $A$ and at an angle of $45^{\circ}$ to the vertical at $C$ (see diagram).
(i)[5]
Calculate the speed at which $B$ moves.
(ii)[3]
The lower end of the string is removed from $C$, and $B$ is then attached to that end of the string. The other end of the string stays fixed at $A$. The bead is put into motion so that it travels with angular speed $3\,\text{rad s}^{-1}$ in a horizontal circle whose centre is vertically below $A$. Calculate the tension in the string.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Vertical resolution gives $T\cos30 - T\cos45 = 0.3g$” …