How Is Centrifugal Force Calculated for a Particle in a Rotating Frame?

[decentfellow]

Homework Statement

A particle of mass $m$ rotates about $Z-$axis in a circle of radius $a$ with a uniform angular speed $\omega$. It is viewed from a frame rotating about the $Z-$axis with a uniform angular speed $\omega_0$. Find the centrifugal force on the particle.

Homework Equations

$$\vec{F}=m\vec{a}$$

The Attempt at a Solution

I am a little confused by the language of the question. On reading the first sentence it looks like the particle is being observed from the ground and the second sentence is just given to confuse one who is solving the question. So, if that is the case then the answer should be $m\omega^2a$.

But if what I assumed was wrong then that means that the particle is being observed from the reference frame which is rotating at a constant angular speed $\omega_0$, if that is so then the angular speed of the particle in the reference frame of the ground is $(\omega+\omega_0)$, so we get $m(\omega+\omega_0)^2a$ as the centrifugal force.

But the answer is neither, it is $m\omega_0^2a$

 

[Orodruin]

The centrifugal force only depends on the angular velocity of the rotating frame, not on how anything moves in it.

 

[haruspex]

An observer in a rotating reference frame cannot make sense of F=ma except by inventing a fictitious force to include in the sum of forces. We call this centrifugal force.
The observer sees the mass circling at some rate (which you can easily calculate) and can compute a centripetal force to account for that motion. But if the observer also knows the actual forces on the mass there will be a discrepancy. It is the discrepancy that the observer considers as centrifugal force.

 

[decentfellow]

@haruspex @Orodruin Oh no! So I was supposed to calculate the pseudo force that had to be included in the FBD of the particle and what I was calculating in the second case, that I had considered in my 1st post, was the centripetal acceleration of the particle am I right?

 

[Orodruin]

So I was supposed to calculate the pseudo force that had to be included in the FBD

No, just the centrifugal part. Since the object is also moving in the rotating frame, it will also be subjected to a Coriolis force.

 

[decentfellow]

No, just the centrifugal part. Since the object is also moving in the rotating frame, it will also be subjected to a Coriolis force.

Can you tell me what is Coriolis force?

 

[Orodruin]

Can you tell me what is Coriolis force?

A fictitious force in a rotating coordinate system that is proportional to the object's velocity.

 

[haruspex]

A fictitious force in a rotating coordinate system that is proportional to the object's velocity.

I thought it was proportional to the radial component of velocity. Both the mass and the observer are rotating about the Z axis, so the mass has no radial velocity in respect of the rotational axis of the observer's frame.

 

[Orodruin]

I thought it was proportional to the radial component of velocity. Both the mass and the observer are rotating about the Z axis, so the mass has no radial velocity in respect of the rotational axis of the observer's frame.

The Coriolis force is equal to $-2m\vec\omega\times\vec v$. It will be non-zero whenever velocity is not parallel to the angular velocity. So to be more specific, it isproportional to the speed and cross product of direction and angular velocity so "linearly dependent on the velocity" would have been a better choice of words.