Bead sliding on a uniformly rotating wire

[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.

 

[Dale]

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.

To explicitly find the constraint force you can also modify the excellent approach of @vanhees71 shown above as follows:

In polar coordinates the Lagrangian for a free particle is a slight modification of his: $$L=\frac{m}{2}\left( \dot r^2 + (r \dot \theta)^2\right)$$ and for this problem there is a holonomic constraint $$f=\theta-\omega t = 0$$

Then the Euler Lagrange equations are: $$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot r}\right)-\frac{\partial L}{\partial r} - \lambda \frac{\partial f}{\partial r} = m \ddot r - m r \dot \theta^2 = 0 $$ $$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot \theta}\right)-\frac{\partial L}{\partial \theta} - \lambda \frac{\partial f}{\partial \theta} = m r^2 \ddot \theta + 2 m r \dot r \dot \theta - \lambda = 0 $$ but with the constraint $f=0$ we immediately know $\ddot \theta = 0$ and $\dot \theta = \omega$ so the above simplify to $$\ddot r=\omega^2 r$$ $$\lambda = 2m \omega r \dot r$$ The first of those, unsurprisingly, has the same solution as found above $$r(t)= r_0 \cosh(\omega t) + \frac{v_0}{\omega}\sinh(\omega t)$$ but now we can additionally evaluate the constraint forces: $$F_r=\lambda \frac{\partial f}{\partial r}=0$$ $$F_\theta = \lambda \frac{\partial f}{\partial \theta} = 2 m \omega r \dot r$$

So the constraint force is neither inward nor outward, but is at all times directed entirely in the $\theta$ direction. Thus @vanhees71 is correct. Note that your notes are also correct as the motion is not purely radial so indeed even though the constraint force is always perpendicular to $r$ it is not always perpendicular to the motion.

 

[vanhees71]

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 ?

Energy is conserved, because the Lagrangian is not explicitly time dependent. The conserved energy is given by the Hamiltonian
$$H=p r-L=\frac{m}{2} (\dot{r}^2-\omega^2 r^2).$$
The constraint force is not perpendicular to the motion (I guess you mean velocity) but to the wire.

 

[dyn]

Thank you everyone