# Particle inside a smooth groove performing circular motion

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1. Nov 17, 2014

1. The problem statement, all variables and given/known data

A circular table of radius rotates about its center with an angular velocity 'w'. The surface of the table is smooth. A groove is dug along the surface of the table at a distance 'd' from the centre of the table till the circumference. A particle is kept at the starting point of groove and then released. Find the velocity of the particle when it reaches the end of the groove.
2. Relevant equations

3. The attempt at a solution

Taking the table as the frame of reference, a pseudo force mw2x (x being the distance from the center of rotation) acts on the particle and solving the differential equation vdv/dx = mw2x will get us the answer.

But my question is, how can we solve it seeing the motion as an observer on ground? With respect to ground there is no visible force (except the normal forces by the two surfaces of the groove it is exposed to; but they cancel each other out, don't they?) that acts on it but the ball still moves. My friend says it is coriolis force. If it is, can you please provide an intuitive explanation regarding coriolis force?

2. Nov 17, 2014

### BruceW

doesn't the mass cancel from that equation? I don't think there should be an 'm' on the right hand side of the equation vdv/dx = mw2x It might be good to check that equation. Also, I think it is a centrifugal force, which is a fictitious force similar to the Coriolis force.
Yes, you're right, with respect to the ground, some very weird fictitious forces appear. And this is because the ground is in acceleration. It is a non-inertial frame of reference, and so the laws of physics are a bit different.
In truth, I think this is only truly explained by general relativity, where these weird fictitious forces start to make sense. So until you are learning general relativity, it might be best to just stay with the intuition that inertial frames of reference are the correct ones, and the fictitious forces are simply artifacts that appear due to writing the coordinates in an accelerating frame of reference. At least, this is the intuition that I have been most comfortable with.

3. Nov 18, 2014

Thank you BruceW, for your assistance. Yes, that 'm' should not have been there- put that by mistake.

4. Nov 18, 2014

### BvU

You have the right hunch here: they don't cancel

5. Nov 18, 2014

### Orodruin

Staff Emeritus
GR is certainly not required to make sense of accelerating reference frames. It is perfectly possible to do so already in Newtonian mechanics and inertial forces will arise due to non-inertial frames being used. Inertial forces appearing will always be due to writing things in non-inertial frames, regardless of whether the theory is Newtonian mechanics, special relativity, or general relativity.

6. Nov 18, 2014

### BruceW

hmm. you mean the nonspinning inertial frames in general relativity are analogous to the inertial frames of Newtonian mechanics? I don't think it's such a strong analogy, since if I use a nonspinning inertial frame in general relativity, then generally I still get weird things like distant stars moving around in circles. So in a sense, all frames are weird in general relativity.

7. Nov 18, 2014

### Orodruin

Staff Emeritus
All I am saying is that there is no reason to involve GR in order to solve the problem at hand. It can be perfectly well understood using Newtonian mechanics in a rotating coordinate frame (or a non-rotating one as well for that matter - it is only a matter of properly taking the constraints into account). I believe the reference to GR will simply act to confuse the OP. General coordinate frames are also perfectly viable in Newtonian mechanics and it is easy to derive their relation to any inertial frame and the resulting inertial forces.

8. Nov 19, 2014

### BruceW

true. the OP'er was looking for an intuitive explanation of accelerated reference frames. I'm saying that in my opinion, the most intuitive explanation is from general relativity. And until I learned a bit about general relativity, I found the most intuitive explanation was to just think of the inertial reference frames as the 'good frames', and the accelerated frames as the 'bad frames', so that if you want to know what kind of fictitious forces will appear in an accelerated frame, you will need to look at how the accelerated frame is related to some inertial frame.

9. Nov 19, 2014

### haruspex

If the distance from the centre is x, the acceleration in the radial direction is $\ddot x - {\dot x}^2/x$. What is the force in the radial direction?

10. Nov 19, 2014

The force will be mass times the acceleration you have mentioned and that will be equal to zero from where we shall get the answer. Same explanation was provided by my professor but my point why are we including x˙2/x\ddot x when no visible forces are acting on the body to provide the centripetal acceleration.

Why don't they cancel? Both the normal reactions are equal and vectorially opposite to each other.

11. Nov 19, 2014

### DocZaius

Out of curiosity, has anyone actually found the solution to the original problem? I am getting that:

x(t) = d e^(w t)
that thus tfinal = ln(R/d) / w where R is the radius of the table
and that v(tfinal) = wR which seems wrong, since it doesn't depend on d.

12. Nov 19, 2014

### Orodruin

Staff Emeritus
Your solution does not fulfil the initial condition (particle kept at rest at d before release). Your general solution is missing one part, which is necessary to fulfil both initial conditions (you have a second order ODE and should have two independent solutions which can be fixed through the initial conditions).

13. Nov 19, 2014

### DocZaius

Ah thanks. Since the OP's question was about interpreting this problem in an inertial reference frame, rather than seeking the solution to the problem, I imagine it is safe to put this here in case anyone is curious. I now get:
$x(t)=d\ \text{cosh}(\omega t)\\ x'(t)=d\ \omega\ \text{sinh}(\omega t)\\ t_{\text{final}}=\frac{\text{arcosh}(\frac{R}{d})}{\omega}\\ x'(t_{\text{final}})=d\ \omega\ \sqrt{{(\frac{R}{d})}^{2}-1}$

Last edited: Nov 19, 2014
14. Nov 19, 2014

### BvU

In a rotating frame of reference there also is a Coriolis force that depends on the radial speed. The particle wants to "stay behind" and needs to be accelerated in the tangential direction. The pushing wall of the groove provides that force. Since the Coriolis force is perpendicular to vr, it doesn't enter in your differential equation.

Your assumption that the normal reaction forces are equal and opposite is not correct: they aren't opposite (the sum offsets mg) and they aren't equal.

See e.g. here

15. Nov 19, 2014

### rcgldr

I'm assume the groove is a straight line along a radial. Angular momentum is not conserved because an external torque will be required to keep the table revolving at constant angular velocity as the ball moves outwards. All points on the groove experience centripetal acceleration, but there's no force in the direction of the groove since it's assumed to be frictionless. Even though the ball only experiences tangental acceleration from a unbalanced force applied by the trailing side of the groove at any moment in time, at a later moment in time, the velocity related to that prior acceleration will have a radial component (moving away from the center). The situation is similar to a ball in a frictionless tube that is kept spinning at constant velocity.

16. Nov 19, 2014

### BruceW

yeah. In an accelerating reference frame, there will be fictitious forces that cannot be attributed to any real force. probably the best way to interpret this is that the inertial (non-accelerating) reference frames are 'good frames' and by choosing an accelerating frame, we introduce these fictitious forces. So in Newtonian mechanics, we can only talk about accelerating frames as long as we refer them to some inertial frame. In the inertial frame, there will be only real forces, no fictitious ones.

17. Nov 19, 2014

### haruspex

18. Nov 19, 2014

### BruceW

agreed! :)
and you can use
$$x = r \cos{\theta}$$
and
$$y = r \sin{\theta}$$
and
$$\hat{r} = \frac{x \hat{\imath} + y \hat{\jmath}}{\sqrt{x^2+y^2}}$$
and
$$\hat{\theta} = \frac{x \hat{\jmath} - y \hat{\imath}}{\sqrt{x^2+y^2}}$$
all these equations to check the equation haruspex gave. But keep in mind that $x$ here is different to the $\vec{x}$ that haruspex has mentioned. The one haruspex mentioned is equal to $x \hat{\imath} + y \hat{\jmath}$ (i.e. the position vector of the object).