One end of a light inextensible string is fixed to the top point $A$ of a solid sphere that is held in place, with centre $O$ and radius $0.6\text{ m}$. The string’s other end is joined to a particle $P$ of mass $0.2\text{ kg}$, which is in contact with the sphere’s smooth surface. The angle $AOP$ is $60^\circ$ (see diagram). The sphere applies a contact force of magnitude $R\text{ N}$ on $P$, and the tension in the string is $T\text{ N}$.
(i)[2]
By resolving vertically, show that $R + (\sqrt{3})T = 4$.
(ii)[2]
$P$ is then put into motion and travels with angular speed $\omega\text{ rad s}^{-1}$ in a horizontal circle on the surface of the sphere. Find an equation in $R$, $T$ and $\omega$.
(iii(a))[2]
Hence calculate $R$ for $\omega = 2$.
(iii(b))[4]
Hence determine the greatest possible value of $T$ and the matching speed of $P$.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Vertical resolution gives $R\cos60+T\cos30=0.2g$” …