Bead sliding on a uniformly rotating wire

[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