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I have read that it would slide off the table, because apparently in the frame of the table it experiences a centrifugal force outwards, but I can't seem to agree with that logic.

In the ground frame of reference there is no force in the direction of the groove (let me call it the instantaneous x axis). Centrifugal force is just a psuedo force right? So how is it that in the frame of the table it experiences an outward force?

According to me it is just like keeping a block on a frictionless table. Causing the table to move would have no effect on the block which would eventually fall off the table because it would eventually run out of table. But here the table is only rotating so no chance of that happening.

Perhaps it's some deep rooted misconception about circular motion .. please help!

EDIT: THE GROOVE IS DIAMETRICAL.

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  • $\begingroup$ Please describe the grove. Is it radial, from the center to outer circumference, or is it a spiral, from the center to the outer circumference? $\endgroup$ – Thomas Lee Abshier ND Nov 8 at 12:44
  • $\begingroup$ Okay , done that $\endgroup$ – Scilife Nov 8 at 12:56
  • $\begingroup$ I will mention that the currently highest voted answer there says there is no force responsible for the object moving outwards, and that the object moves outward at a constant speed. Both of these ideas are incorrect. $\endgroup$ – Aaron Stevens Nov 8 at 13:20
  • $\begingroup$ I read your answer to the other question, and I am satisfied by that explanation. Thanks a lot. $\endgroup$ – Scilife Nov 8 at 14:21
  • $\begingroup$ Walls of groove make block start moving, right? Once it's moving it slides off because there are no forces to make it do otherwise. What would keep it accelerating towards the center? The groove wall? No, it applies a tangential force; never towards the center. To stay at one radial position, something has to push it inwards. Normally that's friction. No friction means no accelerating towards the center, which means moving outwards $\endgroup$ – Jim Nov 8 at 14:24
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The groove applies a normal reaction on the block, which is the reason that the block rotates along with the table with the same angular velocity as the table. Now since this normal force is tangential to the table it causes the tangential speed of the block to increase. Tangential velocity is equal to angular velocity cross radius and angular velocity remains constant (same as of table), therefore the radius (distance of block from center of table) increases, which is manifest as the centrifugal force.

The given link may help (pg 127 from the cover, question 13). Hope this bridges the gap in understanding: https://drive.google.com/file/d/1cpuVbLtqdYiT_p5BAaUiWgKWFi2Gty1l/view

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If the groove is sufficiently deep/steep-sided to retain it, and assuming the body starts off somewhere in the groove that is not at the exact centre of rotation, then the body will slide to the end of the groove. That will happen because the wall of the groove will press against the body and accelerate it and there is no centripetal frictional force to make the body follow in a circle.

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The body would keep its state, as no force is acting on it (first law): if the body is at rest with respect to the ground frame, it would remain at rest; if it has some speed, it would keep that speed.

However, as seeing from the rotating reference frame, in both of these cases the body is accelerating. So, in order to describe the motion from that non-inertial frame, you need an inertial force $\vec F_i = m\vec A$, where $\vec A$ is the acceleration of the rotating frame as measured from the ground frame, so you can write Newton's second law as $\vec F + \vec F_{i} = m\vec a'$, where $\vec a'$ is the acceleration of the body as described from the rotating reference frame.

$\vec F$ depends on interactions of the body with the table, the air, the Earth etc. In your idealised scenario, $\vec F = \vec 0$ (ignoring even gravity, as the motion of interest is supposed to be in the horizontal). Nevertheless, $\vec a' \ne \vec 0$. Indeed, $\vec F_i = m\vec a'$, which is the inertial force you need to describe that motion.

For the case of a reference frame which rotates with constant angular velocity, $\vec F_i$ can be decomposed in two contributions (not two components): a centrifugal force and a Coriolis force. So, in your case both centrifugal and Coriolis accelerations are present, though no forces act on the body!

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    $\begingroup$ The force of the groove acts on the body. In the rotating frame the groove force is cancelled out by the Coriolis force, and the centrifugal force pulls body outwards. $\endgroup$ – Aaron Stevens Nov 8 at 13:27
  • $\begingroup$ So you are supposing the body is already rotating with the table, maybe in such a way that it is at rest with respect to the table (I did not get that)... In this case, $\vec a'= \vec 0$, which means $\vec F_i = -\vec F$. But $\vec F = -mR\omega^2\hat r$, which is the centripetal force (inward) needed to keep the motion, as describe from the ground reference frame. So $\vec F_i$ is outward. $\endgroup$ – mmagos Nov 8 at 14:48
  • $\begingroup$ Anyway, my bad. My answer was wrong: I tried to approach the case in which the body is placed at rest on the table, but forgot the groove when doing that :s But I hope the comment above helps. $\endgroup$ – mmagos Nov 8 at 15:02
  • $\begingroup$ There is no centripetal force here. The body is not at rest relative to the table. It moves outwards along the groove. $\endgroup$ – Aaron Stevens Nov 8 at 15:15
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In the ground frame of reference we have a single force acting on body. This is the normal force from the side wall of the groove. This force always acts tangentially so it accelerates the body tangentially so as to keep it in the groove as the table rotates. However, there is no radial centripetal force to keep the body moving in a circle. So in the ground frame of reference the body spirals outwards while maintaining a constant radial speed - the speed of rotation of the table.

In the (non-inertial) table frame of reference, the normal force from the side wall of the groove has the same magnitude as in the ground frame of reference, although it now acts in a constant direction because the groove is stationary. However, there are also two fictitious forces acting on the body - centrifugal force and coriolis force. The coriolis force is equal and opposite to the side wall force (to simplify this part, I am assuming the groove has vertical sides and a flat bottom). So the body does not move perpendicular to the groove. The centrifugal force acts radially outwards and is the only radial force acting on the body, so in the table frame of reference the body accelerates radially outwards in a straight line.

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