# Centrifugal force and inertia

• I
• Chenkel

#### Chenkel

I believe I understand centripetal force, acceleration is necessary for something to spin in a circle because things normally want to continue moving in a straight line (Newton's first law), so a force is necessary to keep something rotating. If you have an object fastened to a rotating disk it will have a pushing force, the centripetal force, but I do not see a force causing something to want to move outward, I don't even see any outward movement! So centrifugal force is really confusing me... Something is either attached to the rotating disk or not; when something's attached, there's a pushing force; when the object is not attached to the spinning disk, the disk doesn't effect it. I can imagine a disk with a solid rim that prevents an object from escaping the rotating disk, the rim pushes against the object keeping it inside the circle, but the object pushes back against the rim (Newton's third law), the push against the rim is equal and opposite that of the centripetal force, is the force against the rim a centrifugal force?

I hope someone can help me understand this, thank you!

• vanhees71 and Delta2

centrifugal force is a fictitious force that appears when you want to describe the system in the frame of reference of the rotating thing. This frame of reference is non-inertial because it is rotating. Due to this an extra force appears, which is the centrifugal force. But in an inertial frame of reference there will be no centrifugal force.

• Meir Achuz, malawi_glenn, vanhees71 and 2 others
Yes a rigorous treatment of centrifugal force requires to do math in a rotating frame of reference. It is a fictitious force, not an interaction force, however regarding the math is full treated like any other interaction force. Don't know if I should replace the term "interaction force" with the term "real force", this will cause debate here as in some other threads here in PF. Interaction forces are for example gravitational forces, or electrostatic forces, or friction forces , air resistance , water resistance, buoyancy e.t.c

• vanhees71 and Chenkel
In short, fictitious forces are forces that appear in non-inertial reference frames due to the frame being non-inertial. The centrifugal force is one such force.

In an inertial frame, an object moving in a circle (regardless whether this is caused by being on a rope or being attached to a rotating disk or whatever other mechanism) requires a centripetal force to provide it with the necessary acceleration. The object moves in a circle due to the acceleration provided by the force.

In a frame rotating about the center of rotation with the same angular frequency, the object is stationary. In such a frame, the fictitious centrifugal force is a force pointing away from the rotational center. It exactly cancels the centripetal force and therefore the object remains at rest in the rotating frame.

• malawi_glenn, vanhees71, Chenkel and 1 other person
Just to say something, in some introductory physics books, high school level, they use the term centrifugal force as the force that the body undergoing the circular motion exerts to the body on the center of the circle as reaction to the centripetal force, that is due to Newton's 3rd Law. This is not a fictitious force.

The centrifugal force that we say it is fictitious is exerted on the body that undergoes circular motion together with the centripetal force (I think post #4 explains it very nicely).

• vanhees71, Arjan82, Lnewqban and 2 others
I'd prefer the term "inertial force" over the term "fictitious force", because "fictitious" somehow sounds as if this thing is somehow not "real".

The point is that the terms called "inertial force" in the equation of motion as formulated in non-inertial reference systems rather belong to the left-hand side, ##m \vec{a}'##, rather than on the right-hand side of the equation, but on the other hand it's more intuitive to bring everything to the right-hand side except the term ##m \ddot{\vec{x}}'##, where ##\vec{a}'## and ##\vec{x}'## are the components of the acceleration and position vector wrt. the Cartesian bases fixed in the non-inertial frame of reference.

Take, for example, a frame of reference rotating somehow wrt. an arbitrary inertial frame of reference. Then one can derive that the time derivative for the vector components in this non-inertial frame for any vector are
$$\mathrm{D}_t \vec{A}'=\dot{\vec{A}}' + \vec{\omega}' \times \vec{A}',$$
where ##\vec{\omega}'## are the components of the momentary angular velocity of the non-inertial frame wrt. the inertial frame.

Applying this to the equation of motion for a point particle you need
$$\vec{v}'=\mathrm{D}_t \vec{x}' = \dot{\vec{x}}' + \vec{\omega}' \times \vec{x}'$$
and
$$\vec{a}'=\mathrm{D}_t \vec{v}'=\ddot{\vec{x}}' + \vec{\omega}' \times \dot{\vec{x}}' + \frac{\mathrm{d}}{\mathrm{d} t} (\vec{\omega}' \times \vec{x}') + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}')$$
or
$$\vec{a}'=\ddot{\vec{x}}' + 2 \vec{\omega}' \times \dot{\vec{x}}' + \dot{\vec{\omega}}' \times \vec{x}' + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}').$$
The equation of motion wrt. the rotating reference frame thus reads
$$m \vec{a}' = \vec{F}',$$
where ##\vec{F}'## are the components of the "true forces" acting on the body (forces like gravitational or electromagnetic forces due to the presence of other masses or electric charges and currents).

Now it is more intuitive to write the equation of motion in the form that it looks as if we were working in an inertial frame of reference, i.e.,
$$m \ddot{\vec{x}}' =\vec{F}' - m [2 \vec{\omega}' \times \dot{\vec{x}}' + \dot{\vec{\omega}}' \times \vec{x}' + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}')].$$
Then it looks as if there are additional forces acting on the particle, the said inertial (or in many books called fictitious) forces: the Coriolis force, the force due to angular acceleration (I don't know, if there is a common name for this term), and the centrifugal force.

• malawi_glenn, Chenkel and Delta2
I think the force due to angular acceleration is called Euler force.

• • weirdoguy, malawi_glenn, Chenkel and 1 other person
• Chenkel and Delta2
@vanhees71 ##D_t## is the time derivative in the inertial frame and doting is the time derivative in the non inertial frame or the other way around?

• Chenkel
##\mathrm{d}_t## or a dot is just the usual time derivative. ##\mathrm{D}_t## is the covariant time derivative, taking into account the time-dependence of the non-inertial basis, describing the rotation of the non-inertial frame relative to the inertial frame. Primed vectors are the vector components wrt. the non-inertial frame.

• Chenkel and Delta2
##\mathrm{d}_t## or a dot is just the usual time derivative. ##\mathrm{D}_t## is the covariant time derivative, taking into account the time-dependence of the non-inertial basis, describing the rotation of the non-inertial frame relative to the inertial frame. Primed vectors are the vector components wrt. the non-inertial frame.
I thought wikipedia had it the other way around but now I see you are in agreement with wikipedia.

https://en.wikipedia.org/wiki/Rotating_reference_frame#Time_derivatives_in_the_two_frames

• Chenkel
The derivation is as follows: Let ##\vec{e}_j## be the time-independent right-handed Cartesian basis in the inertial frame and ##\vec{e}_k'## the right-handed Cartesian basis in the rotating frame. Then there's a rotation matrix ##D_{jk}(t)## such that (Einstein summation convention in use from now on)
$$\boldsymbol{e}_k'=D_{jk} \boldsymbol{e}_j.$$
The transformation for vector components follows from
$$\boldsymbol{V}=V_k' \boldsymbol{e}_k' = V_{k}' D_{jk} \boldsymbol{e}_j \; \Rightarrow \; V_j=D_{jk} V_k'$$
Since ##\hat{D}=(D_{jk})## is a rotation matrix we have ##\hat{D}^{-1} = \hat{D}^{\text{T}}## and thus
$$\vec{e}_j=D_{jk} \boldsymbol{e}_k'.$$
The time derivative of the vector ##\boldsymbol{V}## is
$$\dot{\boldsymbol{V}} = \dot{V}_k' \boldsymbol{e}_k' + V_k' \dot{\boldsymbol{e}}_k'.$$
Now
$$\dot{\boldsymbol{e}}_k'=\dot{D}_{jk} \boldsymbol{e}_j = \dot{D}_{jk} D_{jl} \boldsymbol{e}_l',$$
i.e.,
$$\dot{\boldsymbol{V}} = (\dot{V}_l' + V_k' \dot{D}_{jk} D_{jl}) \boldsymbol{e}_l'.$$
In matrix-vector notation the components of the time derivative of the vector, which I denote by ##\mathrm{D}_t \vec{V}'##, thus is (written in matrix-vector notation, where from now on unprimed symbols like ##\vec{V}## are column vectors of the vector components with respect to the ##\boldsymbol{e}_j## and ##\vec{V}'## those wrt. to ##\boldsymbol{e}_k'##.)
$$\mathrm{D}_t \vec{V}' = \dot{\vec{V}}' + \hat{D}^{\text{T}} \dot{D} V_{k}.$$
Now because of ##\hat{D}^{\text{T}} \dot{D}=\hat{1}=\text{const}## you have
$$\dot{\hat{D}}^{\text{T}} \hat{D}+\hat{D}^{\text{T}} \dot{\hat{D}}=0 \; \rightarrow \; \hat{D}^{\text{T}} \dot{\hat{D}}=-\dot{\hat{D}}^{\text{T}} \hat{D} = -(\hat{D}^{\text{T}} \dot{\hat{D}})^{\text{T}},$$
i.e., ##\hat{D}^{\text{T}} \dot{\hat{D}}## is an antisymmetric matrix. For its matrix elements you can thus write
$$(\hat{D}^{\text{T}} \dot{\hat{D}})_{lk} = D_{jl} \dot{D}_{jk} =-\epsilon_{lkm} \omega_m',$$
such that
$$\mathrm{D}_t V_l'=\dot{V}_l' -\epsilon_{lkm} \omega_m' V_k' = \dot{V}_l' + \epsilon_{lmk} \omega_m' V_k' = \dot{V}_l' +(\vec{\omega}' \times \vec{V}')_l,$$
or in vector notation
$$\mathrm{D}_t \vec{V}' = \dot{\vec{V}}' + \vec{\omega}' \times \vec{V}'.$$

• Chenkel and Delta2
the push against the rim is equal and opposite that of the centripetal force, is the force against the rim a centrifugal force?
This is sometimes called a reactive centrifugal force. It is a completely different thing from what in physics we normally refer to as a centrifugal force (sometimes an inertial centrifugal force).

• jbriggs444 and Chenkel
Yes a rigorous treatment of centrifugal force requires to do math in a rotating frame of reference. It is a fictitious force, not an interaction force, however regarding the math is full treated like any other interaction force. Don't know if I should replace the term "interaction force" with the term "real force", this will cause debate here as in some other threads here in PF. Interaction forces are for example gravitational forces, or electrostatic forces, or friction forces , air resistance , water resistance, buoyancy e.t.c
I'm coming back to this thread as I feel I now have an intuitive understanding of centrifugal force, I believe when I spin and my arms go outward, there is no centripetal force counteracting the fictitious 'centrifugal' force, and that is what causes the arms to move outward, after they are fully extended there's a tension in the arms which counter the fictitious force so that relative to my frame of reference the arms are not moving. I thought I might never be able to understand centrifugal force, but here I am with some intuition :)

• Delta2 and jbriggs444
after they are fully extended there's a tension in the arms which counter the fictitious force so that relative to my frame of reference the arms are not moving
Close, but the thing you have called "tension in the arms" is real but the fictitious force is not, so it doesn't need to be 'countered' (fictitious forces do not have any equal and opposite reaction).

When you are rotating with your arms out straight there is NO (fictitious) centrifgugal force, but if you imagine you have a watch on with a loose strap and it slides off your wrist while you are spinning, THAT is centrifugal force.

• Chenkel
Close, but the thing you have called "tension in the arms" is real but the fictitious force is not, so it doesn't need to be 'countered' (fictitious forces do not have any equal and opposite reaction).

When you are rotating with your arms out straight there is NO (fictitious) centrifgugal force, but if you imagine you have a watch on with a loose strap and it slides off your wrist while you are spinning, THAT is centrifugal force.
So I was thinking about this, and in a the non inertial reference frame of the rotating body you need the net total of force on an elemental mass in the rotating reference frame to equal 0 when the elemental mass is not moving in the rotating reference frame, this is what you need to make Newton's laws work again; when you remove the centripetal force, you start to see the object move outward, you can always remove the centripetal force, imagine a tube rotating about it's center, and a ball inside the tube close to the center, the ball will have no centripetal force, so it will start to radially move outward because it wants to move in a straight line. Suppose there are caps on the tube, the moment it comes in contact with the cap, a centripetal force will be applied and it will counteract the centrifugal force, causing it to have no net acceleration in the rotating reference frame. Notice that I'm not saying the fictitious force is a reaction force to the centripetal force, I'm saying it is there to give a net total of 0 force to a mass not accelerating with respect to the non inertial reference frame. I feel my understanding isn't perfected yet, but hopefully that will come in time.

• Delta2
When you are rotating with your arms out straight there is NO (fictitious) centrifgugal force, but if you imagine you have a watch on with a loose strap and it slides off your wrist while you are spinning, THAT is centrifugal force.
Huh? Where did you get this from? In the rotating frame there certainly is a centrifugal force on the arms. This is no different from the centrifugal force acting on the clock you mentioned. The only difference is that the force balance on the arms add up to zero and they therefore do not extend further whereas the force balance on the clock is non-zero and it therefore accelerates outward. All of this in the rotating frame of course.

In a non-rotating frame, there is no centrifugal force. The tension in the arms keep the arms moving in a circle and the absence of a force on the clock means it slides off because it will move in a straight line.

• • • Chenkel, vanhees71, malawi_glenn and 1 other person
The only difference is that the force balance on the arms add up to zero and they therefore do not extend further
what if it spins REALLY fast!
physicists version of dismemberment by tying arms to horses who pull them off • Delta2
what if it spins REALLY fast!
physicists version of dismemberment by tying arms to horses who pull them off Don’t worry, it is just a ficticious force … Physicist method of execution: Spin head so fast that it actually explodes from internal pressure.
(Bonus problem: How fast would that be?)

• vanhees71