One end of a light inextensible string of length $2.4\,\text{m}$ is fixed at point $A$. Its other end is fastened to a particle $P$ of mass $0.2\,\text{kg}$. $P$ travels at constant speed around a horizontal circle whose centre is vertically beneath $A$, while the string remains taut and inclined at an angle of $60^\circ$ to the vertical.
(i)[4]
Determine the speed of $P$.
(ii)[4]
Now remove the string of length $2.4\,\text{m}$, and let $P$ be attached to $A$ by a light inextensible string of length $1.2\,\text{m}$. The particle $P$ moves with angular speed $4\,\text{rad s}^{-1}$ in a horizontal circle whose centre is vertically below $A$. Calculate the angle between the string and the vertical.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Resolve vertically to obtain $T\cos60 = 0.2g$” …