Oscillations of load with spring after rod is suddenly stopped

[alalalash_kachok]

Homework Statement

The rod is rotating with constant angle velocity \omega, load what has weight m may slide on it. The load is held at certain distance by spring with stiffness k and startring length r_0.

Relevant Equations

Find depending r on t if stop the rotating of rod.

I understand that after stopping of rotating I should consider second Newton's law:
m d^2r/dt^2 = k(r-r_0)
And using the law of energy conservation I can propose that energy of circular motion I (\omega)^2/2, where I = mr^2 - moment of intertia will be converted into spring's oscillation. But I not understand what I can do with Newton's law, if there is a constant that disturbs me to consider this motion as oscillations and use the fact that square of oscillation frequency is k/m. Sorry for my bad English, I hope that you can understand this task

 

[haruspex]

Homework Statement: The rod is rotating with constant angle velocity \omega, load what has weight m may slide on it. The load is held at certain distance by spring with stiffness k and startring length r_0.
Relevant Equations: Find depending r on t if stop the rotating of rod.

I understand that after stopping of rotating I should consider second Newton's law:
m d^2r/dt^2 = k(r-r_0)
And using the law of energy conservation I can propose that energy of circular motion I (\omega)^2/2, where I = mr^2 - moment of intertia will be converted into spring's oscillation. But I not understand what I can do with Newton's law, if there is a constant that disturbs me to consider this motion as oscillations and use the fact that square of oscillation frequency is k/m. Sorry for my bad English, I hope that you can understand this task

I don’t understand how the rotational energy is relevant. If the rotation of the rod stops then that energy is immediately lost.
Perhaps I have not understood your description. Can you post the original wording in whatever language?

 

[alalalash_kachok]

Sorry for my misleading, I agree with you. But I have remembered about that I can consider Newton's law for moment after stopping rotating: m $$\frac{d^2r}{dt^2} = k(r-r_0)$$. And try solve this equation, taking $$z = c*e^{i\omega t}$$. After that I get r as superposition of a general solution of the homogeneous equation and the partial non-uniform equation $$r = a \cdot \cos{\omega_0 t + \phi} + r_0$$, where \omega_0 = \sqrt{k/m}. Can I suppose that $\phi$ is 0 just after stopping rotating?

 

[haruspex]

Sorry for my misleading, I agree with you. But I have remembered about that I can consider Newton's law for moment after stopping rotating: m $$\frac{d^2r}{dt^2} = k(r-r_0)$$. And try solve this equation, taking $$z = c*e^{i\omega t}$$. After that I get r as superposition of a general solution of the homogeneous equation and the partial non-uniform equation $$r = a \cdot \cos{\omega_0 t + \phi} + r_0$$, where \omega_0 = \sqrt{k/m}. Can I suppose that $\phi$ is 0 just after stopping rotating?

You mean $$r = a \cdot \cos(\omega_0 t + \phi) + r_0$$. Curly braces, {}, have special meaning in LaTeX.
The value of $\phi$ depends on how you define t=0. If that is the instant the rod stops rotating then you need a value of $\phi$ for which $t=0$ gives $r=r_0$.

 

[alalalash_kachok]

You mean $$r = a \cdot \cos(\omega_0 t + \phi) + r_0$$. Curly braces, {}, have special meaning in LaTeX.
The value of $\phi$ depends on how you define t=0. If that is the instant the rod stops rotating then you need a value of $\phi$ for which $t=0$ gives $r=r_0$.

Okay, assume that t=0 is moment when the rotating rod is stopped. After that the load starts spring fluctuations, don't you think? Can I consider Newton's law: $$m (\omega)^2 r_1= k(r_1-r_0)$$ for the moment just before stopping and take from this starting value of r(t)? Then put it in equation r(t) as solution of the partial non-uniform equation

 

[haruspex]

Okay, assume that t=0 is moment when the rotating rod is stopped. After that the load starts spring fluctuations, don't you think? Can I consider Newton's law: $$m (\omega)^2 r_1= k(r_1-r_0)$$ for the moment just before stopping and take from this starting value of r(t)? Then put it in equation r(t) as solution of the partial non-uniform equation

Sounds right. Please post what you get.