Bead sliding on a uniformly rotating wire

In summary, the wire rotates at angular frequency ω and the polar angle is given by θ = ωt. The generalised coordinate is r and using the Euler-Lagrange equation leads to d2r/dt2 = rω2. The notes state that this leads to the solution r = Aeωt meaning the bead moves exponentially outward. The bead moves outwards without any inward force, as acceleration and movement are different things. In the inertial frame, the bead never accelerates outwards, but its acceleration is inwards at all times. In the uniformly rotating reference system of the rod, the force of gravity balances the Coriolis term and the centrifugal force accelerates the bead outwards. In the rotating
  • #36
dyn said:
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.
 
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  • #37
dyn said:
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.
 
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  • #38
Thank you everyone
 
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