- #1
Andreas C
- 197
- 20
Just refer to my profile picture to see what the issue is!
Here's the problem: a ball of mass m is connected to a vertical pole with an inextensible, massless string of length r. The angle between the string and the pole is θ. The pole rotates around the z axis with a constant angular velocity $$\dotφ$$
We can easily determine the following equations:
$$x=r*sinθ*cosφ$$ $$y=r*sinθ*sinφ$$ $$z=r(1-cosθ)$$
Now I determined that the pesky Lagrangian is equal to the following:
$$L=\frac{mr^2}{2}(\dot\theta^2+\dot\phi^2sin^2\theta)+mgr(1-cos\theta)$$
So, where's the issue? Well, solving the Euler-Lagrange equations we find this:
$$mr^2\ddot\phi sin\theta+2mr^2\dot\phi\dot\theta cos\theta sin\theta=0$$
Which means that
$$\ddot\phi=-\frac{2\dot\phi\dot\theta}{tan\theta}$$
That's a really odd result. φ' is supposed to be constant, so φ''=0. But here we can see that φ'' can only be 0 if either φ' or θ' are 0, or if θ=π/4. Weird.
Am I missing something (I obviously am, it's a rhetorical question)?
Here's the problem: a ball of mass m is connected to a vertical pole with an inextensible, massless string of length r. The angle between the string and the pole is θ. The pole rotates around the z axis with a constant angular velocity $$\dotφ$$
We can easily determine the following equations:
$$x=r*sinθ*cosφ$$ $$y=r*sinθ*sinφ$$ $$z=r(1-cosθ)$$
Now I determined that the pesky Lagrangian is equal to the following:
$$L=\frac{mr^2}{2}(\dot\theta^2+\dot\phi^2sin^2\theta)+mgr(1-cos\theta)$$
So, where's the issue? Well, solving the Euler-Lagrange equations we find this:
$$mr^2\ddot\phi sin\theta+2mr^2\dot\phi\dot\theta cos\theta sin\theta=0$$
Which means that
$$\ddot\phi=-\frac{2\dot\phi\dot\theta}{tan\theta}$$
That's a really odd result. φ' is supposed to be constant, so φ''=0. But here we can see that φ'' can only be 0 if either φ' or θ' are 0, or if θ=π/4. Weird.
Am I missing something (I obviously am, it's a rhetorical question)?