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Suppose there is a disc which is rotating about its centre with constant angular velocity, the surface of disc is rough and there is a small block at some distance away from centre (at rest initially), now if one observes what happens in the foregoing motion to block in rotating frame it can be explained using centrifugal and coriolis force but in ground frame how to explain what force is causing the block to move in a straight path leaving the disc finally?

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TL;DR When both the box and the disc are at rest, there are two (significant) forces acting on the box: (i) weight from the gravitational force, and (ii) normal force from the contact with the disc surface. If disc starts rotating, there are two possible scenarios: (i) if there is no friction force, the box remains at the same position as seen by ground observer; (ii) if there is friction force, the box rotates together with the disc up until the disc (angular) velocity is large enough to overcome static friction force between the box and the disc surface, after which the box leaves the disc.

Below I discuss these two scenarios, both seen by a ground observer. I would suggest always to analyze free-body diagrams (forces that act on the body) from an inertial reference frame. Otherwise, you will just get confused by non-existent forces such as a centrifugal force.


No friction between box and disc surface

The only two forces acting on the box are the gravitational force (the weight) exerted by the Earth and normal force exerted by the disc surface. Since the box does not move in vertical direction, these two forces are equal in magnitude and opposite in direction. Even if the disc rotates, the box remains at rest, because there are still only two forces exerted on the box.


There is friction between box and disc surface

Imagine both box and disc rotate at some angular velocity $\omega$. There are three forces exerted on the object: (i-ii) weight and normal force as already discussed, and (iii) static friction force exerted by the disc surface in direction parallel to the surface. Since the box rotates about an axis that goes through disc center, the net force on the box is (always) directed towards the center of rotation (study uniform circular motion to understand why):

$$\vec{F}_\text{net} = m \frac{v^2}{R} \hat{r} = m \omega^2 R \hat{r}$$

where $v$ is linear velocity, $\omega$ is angular velocity, $R$ is distance of the box from the center of rotation, and $\hat{r}$ is a unit vector that points towards the center of rotation. This net force equals vector sum of all forces that are exerted on the object:

$$\vec{F}_\text{net} = \vec{w} + \vec{n} + \vec{f}_s$$

where $\vec{w}$ and $\vec{n}$ are weight and normal force which cancel as already discussed, and $\vec{f}_s$ is static friction force. From this it is obvious that the static friction provides a radial force component which makes the box rotate together with the disc. In other words, the static friction opposes relative motion between the box and the disc surface.

However, the static friction force has a maximum value it can take which depends on the coefficient of static friction $\mu_s$. If the velocity ($v$ or $\omega$) is too large such that the net force is larger than the maximum static friction force, the box starts slipping on the disc surface and eventually falls off.

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  • $\begingroup$ Sir consider the situation of friction being there then is it possible to explain the effect in ground frame ? $\endgroup$
    – Orion_Pax
    Feb 14, 2022 at 17:29
  • $\begingroup$ Understood Sir thanks $\endgroup$
    – Orion_Pax
    Feb 14, 2022 at 17:34
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the surface of disc is smooth and there is a small block at some distance away from centre (at rest initially) means that in the ground frame the box does not move.

In the rotating frame which travels with the disc the box will be seen to undergo a circular motion with acceleration $r\omega^2$ towards the centre of rotation of the disc.

In the accelerated (relative to ground) rotating frame one would include a fictitious centripetal force of magnitude $mr\omega^2$ towards the centre of rotation if one wanted the circular motion of the box to be consistent with Newton's laws of motion.

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what force is causing the block to move in a straight path

It's not a force, it's the absence of force. By Newton's first law, an object in motion will naturally continue to move in a straight line. The only reason the block moves in a circle in the first place is because friction provides sufficient centripetal force to keep the block moving in a circle. If the disc spins fast enough, friction cannot provide sufficient force to keep the block moving in a circle, at which point it slips. It is the sudden absence of the frictional force that results in the block moving in a straight line.

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