Problem involving a rotating frame. Help

[ashishsinghal]

This is a conceptual problem which I am facing for many days.

If we convert a scenario in an inertial frame into a rotating frame, we apply a pseudo force i.e. centrifugal force radially outwards on the particle. Right?

Also if a particle is having a circular motion in any frame of reference it must have a centripetal force directed radially inwards in that frame. Right?

Now, if I have a particle A which is at rest with respect to the ground, it has no force on it with respect to the ground. Now if I see this particle in a frame of reference rotating about origin O, I will apply centrifugal force directed along OA. But since in this frame A seems to undergo circular motion it must have an acceleration directed along AO in the rotating frame. But this is not happening. Why?

Please help ASAP. If I am not able to show the situation clearly please tell me.

 

[Doc Al]

If we convert a scenario in an inertial frame into a rotating frame, we apply a pseudo force i.e. centrifugal force radially outwards on the particle. Right?

Right.

Also if a particle is having a circular motion in any frame of reference it must have a centripetal force directed radially inwards in that frame. Right?

Right.

Now, if I have a particle A which is at rest with respect to the ground, it has no force on it with respect to the ground. Now if I see this particle in a frame of reference rotating about origin O, I will apply centrifugal force directed along OA. But since in this frame A seems to undergo circular motion it must have an acceleration directed along AO in the rotating frame. But this is not happening. Why?

If it's going in a circle (in that frame) it certainly has a centripetal acceleration.

Realize that the outward centrifugal force is not the only pseudo-force in this scenario. There is also a Coriolis force. Combined, those two give a net force (pseudo-force) in the centripetal direction, as required.

Does that get at your concerns?

 

[ashishsinghal]

Even I thought the same way. But the coriolis force is not directed radially. It is a tangential force. Also coriolis force depends upon the rate of change of radial distance which here is zero.

 

[ashishsinghal]

Coriolis force is the tangential force experienced on the particle which is coming closer or going away from the center of circular motion. Just like the winds change directions.

 

[Doc Al]

Even I thought the same way. But the coriolis force is not directed radially. It is a tangential force. Also coriolis force depends upon the rate of change of radial distance which here is zero.

The Coriolis force will be radially inward.

 

[ashishsinghal]

Can you please explain why? As according to what I have read it is tangential.

 

[Aniket1]

There is no centripetal force in this case.
Since in the rotating frame, the object appears to be moving circularly, sum of all forces should be equal to mv^2/r
The forces acting on the object here is the pseudo force due to the rotation of my frame.
thus, Centrifugal force=mv^2/r
Does this solve your problem?

 

[Doc Al]

Can you please explain why? As according to what I have read it is tangential.

I have no idea why you think that the Coriolis force is tangential. It is always perpendicular to the velocity. It's given by -2m\vec{\Omega}\times\vec{v}
where Ω is the angular velocity vector of the rotating frame. (See Coriolis force or any classical mechanics textbook.)

Thus the Coriolis force is radially inward and equal to 2mω²r and the centrifugal force is radially outward and equal to mω²r. Thus the net force is inward and equal to mω²r, just as you would expect for circular motion.

 

[Doc Al]

There is no centripetal force in this case.
Since in the rotating frame, the object appears to be moving circularly, sum of all forces should be equal to mv^2/r

There must be a net centripetal force, acting radially inward.

The forces acting on the object here is the pseudo force due to the rotation of my frame.
thus, Centrifugal force=mv^2/r

The centrifugal force acts radially outward. So that's not enough to explain things. See my last post about the Coriolis force.

 

[vanhees71]

There is nothing very complicated involved. For a free particle in the inertial frame with cartesian coordinates x_j you have the Lagrangian
L=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} \dot{x}_j \dot{x}_j.
Here and in the following Einstein summation is implied.

In terms of the coordinates y_j fixed in the rotating frame (let's assume the rotation is around the three-axis) you have
\begin{pmatrix}<br /> x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}<br /> \cos (\omega t) &amp; -\sin (\omega t) &amp; 0\\<br /> \sin (\omega t) &amp; \cos (\omega t) &amp; 0 \\<br /> 0 &amp; 0 &amp; 1<br /> \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\y_3 \end{pmatrix} \qquad (*).

Now take the time derivative of this and pluck it into the above Lagrangian. You'll get after some algebra
L=\frac{m}{2} \left [\dot{y}_j \dot{y}_j+2 \omega (y_1 \dot{y}_2-y_2 \dot{y}_1)+\omega^2 (y_1^2+y_2^2) \right ].

With \vec{\omega}=\omega (0,0,1)^t this reads
L=\frac{m}{2} \left [\dot{\vec{y}}^2+2 \vec{\omega} \cdot (\vec{y} \times \dot{\vec{y}}) + (\vec{\omega} \times \vec{y})^2 \right].

With the Euler-Lagrange equations you'll get the correct equations of motion, which can be reinterpreted as the motion of a particle under the influence of the Coriolis and the centrifugal force:
m \ddot{\vec{y}}=-m \left [2 \vec{\omega} \times \dot{\vec{y}}-\vec{\omega} \times (\vec{\omega} \times \vec{y}) \right ].

Of course you could also simply derive Eq. (*) two times wrt. to time, but that's more cumbersome :-).

 

[ashishsinghal]

There is nothing very complicated involved.

Sir, thanks for all the effort but after the work you just showed me I would seriously ask you to consider your definition of complicated. But thanks for your effort.

 

[ashishsinghal]

I have no idea why you think that the Coriolis force is tangential. It is always perpendicular to the velocity. It's given by -2m\vec{\Omega}\times\vec{v}
where Ω is the angular velocity vector of the rotating frame. (See Coriolis force or any classical mechanics textbook.)

Thus the Coriolis force is radially inward and equal to 2mω²r and the centrifugal force is radially outward and equal to mω²r. Thus the net force is inward and equal to mω²r, just as you would expect for circular motion.

I went to the wikipedia link you posted. Yup,I admit I was mistaken. Coriolis force can be in any direction in the plane of circular motion. Thanks a lot.

 

[ashishsinghal]

Thanks everyone. Special thanks to Doc Al. This thread is closed.