# Problem involving a rotating frame. Help

1. Sep 21, 2012

### 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?

2. Sep 21, 2012

### Staff: Mentor

Right.

Right.

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?

3. Sep 21, 2012

### 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.

4. Sep 21, 2012

### 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.

5. Sep 21, 2012

### Staff: Mentor

The Coriolis force will be radially inward.

6. Sep 21, 2012

### ashishsinghal

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

7. Sep 22, 2012

### 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

8. Sep 22, 2012

### Staff: Mentor

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.

9. Sep 22, 2012

### Staff: Mentor

There must be a net centripetal force, acting radially inward.
The centrifugal force acts radially outward. So that's not enough to explain things. See my last post about the Coriolis force.

10. Sep 22, 2012

### 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
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} \cos (\omega t) & -\sin (\omega t) & 0\\ \sin (\omega t) & \cos (\omega t) & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\y_3 \end{pmatrix} \qquad (*).$$

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 :-).

11. Sep 22, 2012

### ashishsinghal

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.

12. Sep 22, 2012

### ashishsinghal

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.

13. Sep 22, 2012

### ashishsinghal

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