Bead sliding on a uniformly rotating wire

[dyn]

Hi
I am working through some notes and came across this example. The wire rotates at angular frequency ω so the polar angle is given by θ = ωt. The generalised coordinate is r. Using the Euler-Lagrange equation leads to
d²r/dt² = rω²
The notes then state that this leads to the solution r = Ae^ωt meaning the bead moves exponentially outward. I have 3 questions regarding this ;
1 - The general solution should be Ae^ωt + Be^-ωt so what has happened to this 2nd term which means the bead moves exponentially inwards ?
2 - If the bead moves outwards , what force is causing this ? In an inertial frame there are no fictitious forces so there is no centrifugal force and centripetal force is directed inwards
3 - In which direction is the constraint force acting ?
Thanks

 

[Dale]

If the bead moves outwards , what force is causing this ? In an inertial frame there are no fictitious forces so there is no centrifugal force and centripetal force is directed inwards

It doesn’t require force to move in a direction, force is only required to accelerate in a direction. In an inertial frame the bead never accelerates outwards. The acceleration is inwards at all times.

 

[dyn]

It doesn’t require force to move in a direction, force is only required to accelerate in a direction. In an inertial frame the bead never accelerates outwards. The acceleration is inwards at all times.

If the acceleration is inwards at all times how does the bead move outwards ?

 

[A.T.]

If the acceleration is inwards at all times how does the bead move outwards ?

Because acceleration and movement are different things. In the inertial frame you don't need any acceleration to move outwards. A straight tangential path would also increase the distance to the center.

 

[Ibix]

If the acceleration is inwards at all times how does the bead move outwards ?

Presumably you are comfortable with the notion that if the bead were fixed to the wire so that it moved in a circle then its acceleration is towards the rotation axis? Qualitatively, what happens to the bead if you decrease the magnitude of that acceleration a tiny bit while maintaining the rotation rate?

I think that the key observation is that acceleration changes the velocity vector, and adding two non-parallel vectors doesn't mean that their magnitudes add. The velocity is initially tangential and the acceleration is perpendicular to that.

 

[etotheipi]

With $\mathcal{L}(r, \dot{r}) = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} m \dot{r}^2$ you will indeed determine that $\ddot{r} - \omega^2 r = 0$ and so $r = ae^{\omega t} + be^{-\omega t}$ is the general solution. To fix the constants you need initial conditions. Notice that the acceleration is purely tangential i.e. $\mathbf{a} = (\ddot{r} - r \dot{\theta}^2) \mathbf{e}_r + (2\dot{r} \dot{\theta} + r\ddot{\theta}) \mathbf{e}_{\theta} = 2\omega \dot{r} \mathbf{e}_{\theta}$. This you expect because in the absence of friction the contact force of the rod on the bead will always be orthogonal to the rod

In the uniformly rotating reference system of the rod with $\boldsymbol{\omega} = \omega \mathbf{e}_z$ the Newtonian equation of motion is $m\mathbf{a}' + 2m\boldsymbol{\omega} \times \mathbf{v}' + m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}') = \mathbf{F}$. This can be simplified further by noticing that in the rotating system $\mathbf{r}' = r \mathbf{e}_x$, then $\mathbf{v}' = \dot{r} \mathbf{e}_x$ and also $\mathbf{a}' = \ddot{r} \mathbf{e}_x$ so$$m \ddot{r} \mathbf{e}_x + 2m \omega \dot{r} \mathbf{e}_y - m\omega^2 r \mathbf{e}_x = F \mathbf{e}_y$$It's clear then that in the rotating frame the contact force $F \mathbf{e}_y$ balances the Coriolis term $2m \omega \dot{r} \mathbf{e}_y$

 

[vanhees71]

...and the centrifugal force accelerates the bead outwards (in the rotating frame of reference), as expected.

 

[Dale]

If the acceleration is inwards at all times how does the bead move outwards ?

Suppose your initial velocity is tangential. Then to stay at a constant radius requires an inward acceleration of $v^2/r$. You will move outward for any inward acceleration in the range $(0,v^2/r)$

 

[kuruman]

To answer question 1 in the original post, the notes are misleading when they state that "this leads to the solution r = Ae^ωt meaning the bead moves exponentially outward." Both solutions are needed to match the initial conditions. For example, if you specify $r(0)=r_0,~\dot r(0)=v_0,~$ you get $$r(t)=\frac{1}{2}\left(r_0+\frac{v_0}{\omega}\right) e^{\omega t } + \frac{1}{2}\left(r_0-\frac{v_0}{\omega}\right)e^{-\omega t }=r_0 \cosh(\omega t)+\frac{v_0}{\omega}\sinh(\omega t).$$

 

[vanhees71]

Suppose your initial velocity is tangential. Then to stay at a constant radius requires an inward acceleration of $v^2/r$. You will move outward for any inward acceleration in the range $(0,v^2/r)$

But there is no force radially inward (assuming no friction in gliding of the bead along the rotating wire, or better rod such that you can neglect deformation).

 

[Dale]

But there is no force radially inward (assuming no friction in gliding of the bead along the rotating wire, or better rod such that you can neglect deformation).

I was making a general statement, not one specific to this scenario.

 

[dyn]

...and the centrifugal force accelerates the bead outwards (in the rotating frame of reference), as expected.

In an inertial frame looking down on the rotating wire and bead there is no centrifugal force yet the bead still accelerates outwards ?

 

[etotheipi]

In an inertial frame looking down on the rotating wire and bead there is no centrifugal force yet the bead still accelerates outwards ?

You just need to be careful what you mean by the word accelerate. On the one hand you do have $\ddot{r} > 0$, but on the other hand the acceleration $\mathbf{a} \parallel \mathbf{e}_{\theta}$ has no radial component.

 

[dyn]

You just need to be careful what you mean by the word accelerate. On the one hand you do have $\ddot{r} > 0$, but on the other hand the acceleration $\mathbf{a} \parallel \mathbf{e}_{\theta}$ has no radial component.

Yes , i mean $\ddot{r} > 0$ which is an outward acceleration. It cannot be caused by centrifugal force in an inertial frame so what is the force causing this outward acceleration ?

 

[etotheipi]

Yes , i mean $\ddot{r} > 0$ which is an outward acceleration. It cannot be caused by centrifugal force in an inertial frame so what is the force causing this outward acceleration ?

There isn't actually any radial acceleration, because $\mathbf{F} \cdot \mathbf{e}_r = m\mathbf{a} \cdot \mathbf{e}_r = m(\ddot{r} - r\dot{\theta}^2) = 0$

 

[Dale]

Yes , i mean $\ddot{r} > 0$ which is an outward acceleration.

$\ddot{r}>0$ is not outward acceleration in an inertial frame. In an inertial frame $\ddot{r}>0$ is any acceleration that is less inward than $v^2/r$. I already explained this above.

 

[hutchphd]

Yes , i mean r¨>0 which is an outward acceleration. It cannot be caused by centrifugal force in an inertial frame so what is the force causing this outward acceleration

I believe the problem you are having is that by your definition, an object moving at constant speed in a straight line not passing through the origin will have a "outward acceleration".

 

[dyn]

I'm totally confused now ! The basic issue was to solve the problem to obtain the position of the bead given by r. The general solution is given by r = Ae^wt + Be^-wt. Focussing on the 1st term gives $\ddot{r} > 0$ which to me looks exactly like an acceleration outwards in the radial direction, It says that the 2nd derivative of the radial coordinate with respect to time is a positive quantity , ie the bead is moving outwards at a faster and faster rate ; ie accelerating

 

[etotheipi]

@dyn a good exercise would be to consider @hutchphd's example, e.g. take for instance a free particle moving at constant velocity along a trajectory $\mathbf{x}(t) = (K,t)$ in Cartesian coordinates for $t \in (-\lambda, \lambda)$ where $K, \lambda$ are constants, and compute the quantity $\ddot{r}$

 

[hutchphd]

ie the bead is moving outwards at a faster and faster rate

The same can be said for my hypothetical uniform velocity object! At closest approach the radial speed is zero an then it increases asymptotically $\ddot r >0$ . Zero force

This is a 3D problem not a 1D problem
.

 

[etotheipi]

$\ddot{r}>0$ is not outward acceleration in an inertial frame. In an inertial frame $\ddot{r}>0$ is any acceleration that is less inward than $v^2/r$. I already explained this above.

I might just add that in this explanation $v^2/r$ ought to be replaced with $r\dot{\theta}^2$ or ${v_{\bot}}^2/r$ because in general the velocity will not be purely tangential

Also there's nothing special about inertial reference systems compared to non-inertial reference systems in this context, i.e. the kinematic equations are identical and $\ddot{r} > 0$ does not necessarily correspond to $a_r > 0$ for general motion referred to an arbitrary non-inertial reference system

 

[kuruman]

There is no outward acceleration. Here is an bare bones repeat to what has already been tried in post #6. One can easily show that in polar coordinates $$\frac{d^2\vec r}{dt^2}=(\ddot r -r\dot \theta^2)~\hat r+(2\dot r \dot \theta+r\ddot \theta)~\hat \theta.$$It is given that $\dot \theta = \text{const.}=\omega$ and, according to OP, $\ddot r=r\omega^2.$ Then this reduces to $$\frac{d^2\vec r}{dt^2}=2\dot r \omega~\hat \theta.$$

 

[etotheipi]

although I have written that formula in post #6

 

[kuruman]

although I have written that formula in post #6

So you did. I edited my post to reflect this fact.

 

[etotheipi]

So you did. I edited my post to reflect this fact.

No need to do that! Yours is stated more clearly anyway. I just thought it was funny

 

[Dale]

I might just add that in this explanation v2/r ought to be replaced with rθ˙2 or v⊥2/r because in general the velocity will not be purely tangential

Yes, you are right. I specified tangential velocity in the first post, but got sloppy with the repeat.

 

[A.T.]

It says that the 2nd derivative of the radial coordinate with respect to time is a positive quantity , ie the bead is moving outwards at a faster and faster rate ; ie accelerating

The radial vector doesn't have a constant direction in the inertial frame. To get individual acceleration components in the inertial frame you need to use basis vectors that are fixed in that frame.

 

[Dale]

Focussing on the 1st term gives r¨>0 which to me looks exactly like an acceleration outwards in the radial direction, It says that the 2nd derivative of the radial coordinate with respect to time is a positive quantity , ie the bead is moving outwards at a faster and faster rate ; ie accelerating

Not every second derivative is an acceleration in an inertial frame. In particular, $r$ is not a coordinate in an inertial frame. So its second derivative does not give you an acceleration in an inertial frame.

Perhaps it will help to think of this a bit backwards. If you have $\ddot r =0$ that does not mean that you are inertial, and in fact it means that you have an inward acceleration of $v^2/r$. So clearly the second derivative of $r$ is not acceleration.

 

[dyn]

The general solution to the problem was r = Ae^wt + B^-wt. If i just consider the 1st term as my notes suggest ; does $r$ and $\dot{r}$ and $\ddot{r}$ tell me anything about the bead's position and it's motion ?

 

[Ibix]

The general solution to the problem was r = Ae^wt + B^-wt. If i just consider the 1st term as my notes suggest ; does $r$ and $\dot{r}$ and $\ddot{r}$ tell me anything about the bead's position and it's motion ?

I haven't thought about whether or not you can ignore one of the terms. It's either impossible or just a matter of picking initial conditions carefully. But of course $r$ and its derivatives tell you about position and motion. The functions $r(t)$ and $\phi(t)$ are literally a list of positions at different times. But the derivatives don't necessarily correspond to directly measurable quantities like acceleration.

You should analyse the inertial case - you don't even need to do any maths. Pick a straight line that passes near but not through the origin, and have an object travel inertially along it. When the object is initially far from the origin, which way is $r$ changing? When it's at the point of closest approach to the origin, which way is $r$ changing? When it's leaving again, which way is $r$ changing? What does that sequence of signs of $\dot{r}$ tell you about whether $\ddot{r}=0$?

 

[Dale]

does $r$ and $\dot{r}$ and $\ddot{r}$ tell me anything about the bead's position and it's motion ?

Yes. But telling you “anything about the bead’s ... motion” is not the same as “acceleration”.

 

[vanhees71]

In an inertial frame looking down on the rotating wire and bead there is no centrifugal force yet the bead still accelerates outwards ?

Let's describe the bead on the wire rotating in the $xy$ plane by
$$\vec{r}=\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos (\omega t) \\ r \sin(\omega t) \end{pmatrix}.$$
Here $r=r(t)$ is the unknown function we want to solve for. To get the equations of motion we use Hamilton's principle of least action. The Lagrangian is
$$L=\frac{m}{2} \dot{\vec{x}}^2.$$
So we need
$$\dot{\vec{r}}=\begin{pmatrix} \dot{r} \cos(\omega t) -r \omega \sin(\omega t) \\ \dot{r} \sin(\omega t) + r \omega \cos(\omega t) \end{pmatrix}.$$
After some algebra you get
$$L=\frac{m}{2} (\dot{r}^2 + r^2 \omega^2).$$
The equation of motion is given by the Euler-Lagrange equation,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{r}}=m \ddot{r} = \frac{\partial L}{\partial r} = m \omega^2 r.$$
The variable $r$ is obviously with respect to the reference frame rotating wrt. the inertial frames, and thus we have a centrifugal force (the right-hand side of the equation) when we describe it in this rotating frame. The solution is
$$r(t)=r_0 \cosh(\omega t) + \frac{v_0}{\omega} \sinh(\omega t),$$
where $r_0$ and $v_0$ are the initial position and velocity along the wire as measured in the rotating reference frame.

The acceleration as observed in the inertial frame thus is
$$\ddot{\vec{r}} = \begin{pmatrix} (\ddot{r}-r \omega^2) \cos(\omega t) -2 \dot{r} \omega \sin(\omega t) \\ (\ddot{r}-r \omega^2) \sin(\omega t) + 2 \dot{r} \omega \cos(\omega t) \end{pmatrix}.$$
Using the equation of motion for $r(t)$ this simplifies to
$$\ddot{\vec{r}}=2 \omega \dot{r} \begin{pmatrix} -\sin(\omega t) \\ \cos(\omega t) \end{pmatrix}.$$
So we have only an acceleration perpendicular to the wire as it must be, because there's no force in direction of the wire, because we assumed no friction. The force on the bead is the coercive force of the wire which is perpendicular to the wire. The opposite force on the wire as seen from the rotating frame is the corresponding Coriolis force.

 

[dyn]

So we have only an acceleration perpendicular to the wire as it must be, because there's no force in direction of the wire, because we assumed no friction. The force on the bead is the coercive force of the wire which is perpendicular to the wire. The opposite force on the wire as seen from the rotating frame is the corresponding Coriolis force.

My notes also state that the constraint force is not perpendicular to the motion and thus it does work on the bead ( energy is not conserved ). This doesn't seem to agree with the statement above.

Also in simple terms can someone tell me what happens to the bead ; does it move outwards or inward ? Does it speed up , slow down or move at the same rate as it travels along the radial distance ?

 

[hutchphd]

Also in simple terms can someone tell me what happens to the bead ; does it move outwards or inward ? Does it speed up , slow down or move at the same rate as it travels along the radial distance ?

@vanhees71 provided a complete solution in #32. Can you not draw a graph or two? Good exercise for the brain

 

[kuruman]

Also in simple terms can someone tell me what happens to the bead ; does it move outwards or inward ? Does it speed up , slow down or move at the same rate as it travels along the radial distance ?

Here is a polar plot of $r(t)=r_0 \cosh(\omega t)+\dfrac{v_0}{\omega}\sinh(\omega t).$
The plot parameters are $r_0 = 1$, $\omega = 0.006 ~\pi$, and $\dfrac{v_0}{\omega}=-3 r_0$. If you are wondering why this particular choice, it's trial and error until I got something interesting. The negative initial velocity causes the bead to spiral in and then spiral back out. The starting point is at (1,0}. As you can see from this plot, the trajectory depends on the initial conditions.