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

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- Thread starter Chenkel
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I hope someone can help me understand this, thank you!

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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.

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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).

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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.

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I think the force due to angular acceleration is called Euler force.

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- #11

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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

- #12

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$$\boldsymbol{e}_k'=D_{jk} \boldsymbol{e}_j.$$

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

$$\mathrm{D}_t \vec{V}' = \dot{\vec{V}}' + \vec{\omega}' \times \vec{V}'.$$

- #13

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This is sometimes called athe push against the rim is equal and opposite that of the centripetal force, is the force against the rim a centrifugal force?

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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 :)

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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).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

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.

- #16

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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.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.

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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.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.

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.

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what if it spins REALLY fast!The only difference is that the force balance on the arms add up to zero and they therefore do not extend further

physicists version of dismemberment by tying arms to horses who pull them off

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Don’t worry, it is just a ficticious force …what if it spins REALLY fast!

physicists version of dismemberment by tying arms to horses who pull them off

Physicist method of execution: Spin head so fast that it actually explodes from internal pressure.

(Bonus problem: How fast would that be?)

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