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A narrow groove is cut along a diameter in the surface of a horizontal disc with center $O$. Particles $P$ and $Q$, of masses $0.2\,\text{kg}$ and $0.3\,\text{kg}$ respectively, lie in the groove, and the coefficient of friction between each of the particles and the groove is $\mu$. The particles are attached to opposite ends of a light inextensible string of length $1\,\text{m}$. The disc rotates with angular velocity $\omega\,\text{rad/s}$ about a vertical axis passing through $O$ and the particles move in horizontal circles (see diagram).

Given that $\mu=0.36$ and that both $P$ and $Q$ move in the same horizontal circle of radius $0.5\,\text{m}$, calculate the greatest possible value of $\omega$.

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The official answer starts with: $$ \begin{align*} T + \mu N_P &= m_P r \omega^2\\ T - \mu N_Q &= m_Q r \omega^2 \end{align*} $$

and (correctly) calculates $\omega = 6\,\text{rad/s}$. I infer that at the limiting instant $P$ is pulling $Q$ out (hence the negative sign for frictional force pointing outwards).

Question:

I don't see the basis for such an assumption. Suppose there is no string and each particle sits on its own. From,

$$ F = N = \mu mg = mr\omega^2 $$

the limiting angular speed (without sliding) is same for both particles, and after that (attached with a string) they must both be pulling out. Where am I wrong?

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When the masses start to move, they will move in the same direction, because the string is inextensible. The friction forces will then be in the same direction. Q is heavier so Q will move outward and P (which is on the opposite side of the centre) will be dragged inward. On Q the friction force will be inwards, while on P it will be outwards.

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