# How to understand what causes the centripetal force in this situation and what would be the direction of those forces and the centripetal acceleration

A table with smooth horizontal surface is turning at an angular speed $$ω$$ about its axis. A groove is made on the surface along the radius and a particle is gently placed at a distance $$a$$ from the center. Find the speed of the particle as its distance from the center becomes $$L$$.

This exact same question was asked here.

The solution there used a non-inertial frame of reference. My question is: what are the forces acting on the particle which cause the centripetal force and what is its direction?

In the non-inertial frame, the centrifugal force acts in a outward direction to the particle which causes the particle to move outward.

But what forces cause the particle to move in an outward direction if this problem is solved in an inertial frame? What is the direction of the centripetal force?

Here is a diagram for understanding the question better.

Note: in this image, $$x$$ is used instead of $$a$$.

There is no centripetal force here at all, actually, as there is no force pointing towards the center of the disk. In an inertial frame of reference there isn't a force pointing radially outwards either.

So why does the particle move outwards? It's actually due to the tangential force from the groove, as this is the only force acting in the plane of the disk.

For more mathematical detail, the general equations of motion in a plane in polar coordinates is

$$\mathbf F=m(\ddot r-r\dot\theta^2)\,\hat r+m(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta$$

In our case in the plane our force is just the one due to the groove: $$\mathbf F=F\,\hat\theta$$ (this force changes magnitude over time; $$F$$ is not constant). So we have then

$$\ddot r-r\dot\theta^2=0$$ $$r\ddot\theta+2\dot r\dot\theta=F/m$$

With $$\dot\theta=\omega$$ and $$\ddot\theta=0$$ we end up with

$$\ddot r-r\omega^2=0$$ $$2\omega\dot r=F/m$$

and so you can see $$r$$ and $$\dot r$$ changes even if $$\mathbf F\cdot\hat r=0$$

Of course, we can't go much farther without knowing what $$F$$ is here. And this is why we analyze this system using either non-inertial reference frames or Lagrangian mechanics. From there you can easily determine $$r(t)$$ and then plug into the above equations to determine what $$F$$ should be to cause this motion to occur.

In the NON INERTIAL FRAME The centrifugal force acts in a outward direction to the particle which causes the particle to move outward.

Better forget about the centrifugal force and always stick to inertial reference frames. Non-existent forces such as centrifugal force are introduced to make first and second Newton's laws of motion work in non-inertial reference frame. Since these forces actually do not exist, they do not have a reaction pair from the third Newton's law of motion.

but what forces cause the Particle to move in an outward direction if this problem is solved in an inertial frame ? What is the direction of the centripetal force ?

When an object sits on a stationary disk, there are two forces acting on the object:

• gravitational force exerted by the Earth; points towards center of the Earth, and
• normal force exerted by the disk surface; perpendicular to the contact surface and points away from the surface.

Since object does not accelerate (move) in the vertical direction, these two forces are equal in magnitude and opposite in the direction.

Let's first imagine the objects sits on a disc surface with no groove. When the disk starts rotating about axis that goes perpendicular to the surface through its center, if there were no friction between the object and disk surface, the object would remain at rest. If there is some friction, then this friction force opposes relative motion between the disk surface and the object.

The net (resultant) force acting on the body is a vector sum of all forces that act on the body. The net force component that points towards the center of rotation is called radial (centripetal) force, and the component that points tangentially to rotation direction is called tangential force. These two are not like gravitational, normal, or friction force - they are just components of the net (resultant) force acting on the body.

In your particular case, given that the surface is smooth, there is no friction which means there is nothing to provide radial (centripetal) component of the net force. But there is tangential component exerted by the groove made on the disk surface. It works on a similar principle as a car seat in your car - what gives acceleration to your body is actually the car seat pushing you forward, and this you can feel when accelerating.

• The surface is smooth, there is no friction Commented Mar 14, 2022 at 11:19
• Firstly the surface is smooth. Commented Mar 14, 2022 at 11:20
• @BioPhysicist Thanks, I missed that from the original question. Commented Mar 14, 2022 at 11:26
• If the surface is smooth, as I have already discussed, the body will never start rotating. This is not true. The groove exerts a tangential force which is responsible for the particle moving outwards Commented Mar 14, 2022 at 11:37
• @BioPhysicist You are right, I completely neglected the groove when I was first writing my answer. Commented Mar 14, 2022 at 11:40

You need acceleration to move in a circle. More specifically you need an inward/radial acceleration of strength $$a=v^2/r$$. On a surface with friction the frictional force would provide the necessary force to move in a circle, but since that is lacking the particle will try to move in a straight line.

Imagine you have a ball attached to a string and you swing it around you head. Suddenly the string breaks will cause the ball to fly away in a straight path that is 90$$^\circ$$ to where the string was. Something similar happens in the case of this frictionless ball. The groove provides a force that makes the particle rotate but since there is no inward/radial force at all the particle will get flung out.

but what forces cause the Particle to move in an outward direction if this problem is solved in an inertial frame?

The issue isn't the non-inertial frame, the issue is the "outward direction". The outward direction is not one specific direction, it varies from point to point. That means that when you are speaking of the outward direction you are speaking of some non-Cartesian basis vectors, and you must account for the fact that your basis vector direction changes from point to point.

The standard approach for non-Cartesian basis vectors is to use Lagrangian mechanics with polar coordinates. So, here we have your radial coordinate, $$r$$, and your angular coordinate, $$\theta$$. Then the kinetic energy is $$T=\frac{1}{2}m (\dot r^2 + r^2 \dot \theta^2)$$, where due to the problem setup we have $$\dot \theta = \omega$$ which is a constant. And since there is no potential we have $$V=0$$, but for circular motion we generally have some central potential $$V=V(r)$$ which we will include here for generality and just set to 0 later. So the Lagrangian is $$L= T - V = \frac{1}{2}m (\dot r^2 + r^2 \omega^2) -V$$

Now, if we solve the Euler equations for this Lagrangian we get $$\ddot r = -\frac{1}{m} \frac{\partial V}{\partial r}+r \omega^2$$

Now, the outward acceleration is $$\ddot r$$ and the outward velocity is $$\dot r$$ so we see that there is a real centripetal force of $$-\partial V/ \partial r$$ which in this case is $$0$$. But there is also a second term which is $$r \omega^2$$. This term is similar to a centrifugal force term since it is responsible for accelerating the particle outward even in the absence of a real force from $$V$$.

So, even if you are using an inertial frame, because you are using non-Cartesian coordinates in your description, you still get something similar to a centrifugal force. Even in an inertial frame, there is an "outward" acceleration of $$r\omega^2$$ that is simply due to describing the situation in terms of the "outward" direction which changes from place to place.