# What happens when the centrifugal force and Coriolis force combine?

Imagine someone is walking on a merry-go-round as it spins. Let's say the merry-go-round has a painted circle on it that is concentric with the outer edge of the merry-go-round and the person walking is walking along the painted circle (grabbing onto whatever available handles there are to resist the forces trying to pull him off the circle). Their walking speed is constant and equal to the tangential velocity due to the spinning, so the person completes two circular paths in the time it takes the merry-go-round spins once. What would the total force on the person be? According to this equation:

$$a_B = a_A + \Omega \times r_{B/A} + \Omega \times \left(\Omega \times r_{A/B}\right) + 2\Omega \times \left(v_{B/A} \right)_{xy} + \left(a_{B/A} \right)_{xy}$$

it seems that the force should be $$3\Omega v$$ (or $$\frac{3v^2}{r}$$) by adding up the third and fourth terms. But when I try to calculate the force using vectors I get $$4\Omega v$$ ($$\frac{4v^2}{r}$$) since the total tangential velocity is double and the velocity in the equation is squared. Can anybody tell me where I'm going wrong?

• You'll have to explain the notation used in the equation, for instance the subscripts $xy$, and $B/A$. Commented Jan 16, 2021 at 18:09

## 2 Answers

For force, you need to include mass in your acceleration relationship. The total force in the inertial (ground) frame is different from the effective total force (including fictitious forces) in the non-inertial rotating (merry-go-round) frame. Here I calculate the total force in both frames.

Let the positive direction be radially inward. The rotating frame rotates with constant angular velocity $$\omega$$. In the non-inertial rotating frame the person with mass m at distance $$r$$ from the center of the merry-go-round is moving with constant speed $$v%$$ in the tangential direction, and experiences the sum of the following forces: $$F$$ the net force in the inertial ground frame, the centrifugal force $$-m \omega^2 r$$, and the coriolis force $$-2m \omega v$$ (both radially outward). (The centrifugal and coriolis forces reduce to these simple results from applying the vector cross produce algebra for your problem.) $$\omega = v/r$$ for this situation. So the centrifugal force is $$-mv^2/r$$ and the coriolis force is $$-2mv^2/r$$. In the rotating frame, the force balance is $$ma^{*2} = F -mv^2/r -2mv^2/r$$ where $$a^*$$ is the acceleration in the rotating frame. We know that $$a^{*2} = v^2/r$$, so $$mv^2/r = F -mv^2/r -2mv^2/r$$ and $$F = 4mv^2/r$$ is the net force in the inertial frame. We can verify this is correct since in the inertial frame the speed is $$2v$$ so the required centripetal force is $$m(2v)^2/R = 4mv^2/R$$ and this equals F.

The net force in the inertial frame is $$4mv^2/r$$ radially inward. The net force in the rotating frame is $$4mv^2/R -mv^2/r -2mv^2/r = mv^2/r$$ radially inward. The sum of the centrifugal and coriolis forces in the rotating frame is $$-3mv^2/r$$ radially outward.

Let me restate the setup as follows:
It is as if there is a second merry-go-round on top of the merry-go-round, so that the overall angular velocity of the second merry-go-round is twice that of the first.

So: for an object/person stationary with respect to the second merry-go-round the required centripetal force is four times the required centripetal force for the case of being stationary with respect to the first merry-go-round.

Of course you are interested in expressing this in terms of the angular velocity of the first merry-go-round.

Here are my thoughts:

First the case in terms of the angular velocity of the second merry-go-round:
If you are initially stationary with respect to the second merry-go-round, and you release, then your initial acceleration with respect to the second merry-go-round will be a function of how fast the point of release is accelerating away from you. (You have just released, you are moving in a straight line now; the point of release is accelerating away from you.)

Now the case in terms of the angular velocity of the first merry-go-round:
The first merry-go-round has a slower angular velocity. Your straight line velocity with respect to the first merry-go-round is faster than with respect to the second merry-go-round. Also, since the angular velocity of the first merry-go-round is slower the acceleration of the point of release away from you is slower.

I'm not going to track down the details, but it must be the case that the above differences make it so that the factor 4 versus factor 3 is explained.

More generally: about the centrifugal term, $$\Omega^2 r \$$, and the Coriolis term $$2 \Omega v \$$. Those terms express acceleration of the rotating coordinate system away from an object moving in a straight line.

It's not necessarily the case that you can equate that acceleration with a force. One case where you can is of course when you are stationary with respect to the rotating coordinate system. Then the vector for the required centripetal force (to remain stationary with respect to the rotating coordinate system) is equal in magnitude and opposite in direction to the centrifugal term.

But when your motion with respect to the rotating coordinate system is accelerated to begin with the case is more complicated.