- #36
Dale
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To explicitly find the constraint force you can also modify the excellent approach of @vanhees71 shown above as follows: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.
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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