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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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2 Answers 2

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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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First consider just P in the groove on the circular disc. When the disc begins to rotate, there is the tendency that the particle P will move away from the center of the disc (or center of rotation). The frictional force is the force that provides the centripetal force to keep the body in the circular path. When the friction peaks and the angular frequency is increased, the object P now slides off the circular path. So both particles P and Q have the tendency to move away from the center of the circle when the disc rotates. Due to the higher mass of Q(but same linear and angular velocity and same distance from the center) Q has a higher momentum and at the point of sliding away from the disc, P would also be at the point of sliding in the direction of Q (because Q is more massive). This implies that for Q, the tension (T) and the frictional force are acting towards the center of the circle, as it is at the point of sliding away from the center of the circle. For P however, because of the influence of Q, it will be at the point of sliding towards the center of the disc so the tension(T) is acting towards the center of the circle while the frictional force acts to oppose the tendency to slide towards the center of the disc.

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